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7/23/2019 Ergonomi - Bab 4 Pengolahan Data - Aldi Dwi Apriliana - Universitas Komputer Indonesia
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Bab 4
Pengumpulan dan pengolahan data
4.1. Pengumpulan data4.1.1. Data anthropometri
4.1.1.1. Data anthropometri posisi berdiri
Untuk membuat suatu produk yang ergonomis kita memerlukan dimensi tubuh yang sesuai dengandimensi produk yang akan dibuat. Data anthropometri posisi berdiri sangat berguna untuk
membuat produk yang sesuai saat kita berdiri.a) Data tinggi badan berdiri
Tabel 3.4.1. Data anthropometri TBB
1 2 3 4 5 6 7 8 9 10
1 154,0 1!,4 151, 15," 1"#, 1"1," 14!,1 14,! 1",0 15#,1
21"$,# 1#,1 1"!,$ 1"5, 15#,$ 1"!,4 1"!,! 10,0 155,1 154,3
3 1",# 1$,3 1!0," 1!1,1 1"3,$ 1"0," 1$,0 1!0,3 1"3,0 154,!
4 1$, 1#,5 1!1,3 1!3,$ 1"4,4 1!3,3 1$,4 1!0,1 1"0,! 1"0,3
5 10,5 1"5,0 153,0 13,3 151,4 14!,# 14$,$ 1"$,5 14,0 1"5,!
6 14",1 15,0 15$,# 155,1 1"#,$ 1!3,5 154, 154," 15!,! 1#,!
7 1"5,5 151," 10,$ 1"4, 15!,1 1#," 153, 15#,5 155, 15,1
8 1!#,! 1!,4 1"3,5 1!0," 1"3,3 1"4,0 15$,4 1"$,4 11,# 14",$
9 1"4,4 1#,4 150, 1"3,0 15,# 10," 1!3,5 1!3,# 1!3,4 1!0,3
10 10,# 1$,$ 1!0,$ 1!1,5 151,0 151,0 1"0,! 1"1,$ 1"!,0 15,"
b) Data tinggi mata berdiriTabel 3.4.#. Data anthropometri T%B
1 2 3 4 5 6 7 8 9 10
1 11,4 1#,4 13!,5 1#4,! 150,! 11,# 143,$ 13!,$ 1$,! 14$,
2 134,# 1##,0 1$,4 14$," 13,0 1#",$ 1#4,! 14$,! 1#!,4 1#,"
3 1#5,1 14!," 14", 15!,5 15," 133,$ 1#3,0 1#",4 15,# 1"1,3
4 1#"," 10," 15#,3 1#,4 153,# 10, 13,0 10,4 15!," 13,0
5 15, 1#!,$ 1#",1 15,# 150,! 131,5 143, 14,$ 13",# 153,3
6 131,1 14!,$ 1#$,$ 1!," 141,3 134,3 130," 10," 13,5 154,1
7 134,! 134,$ 1#,4 14#, 130,4 131,# 14,$ 133,3 150,! 14#,4
8 134,4 1#$,$ 133,5 14#,$ 13$,1 1#3,3 1"1,3 15,1 143,1 15$,$
9 10,# 150, 15#,! 131,0 130,5 153,0 1#,0 1#4,5 1, 13!,$
10 11,4 14!,1 14,0 143," 1#",$ 13,0 13!,4 150,3 13#, 134,!
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7/23/2019 Ergonomi - Bab 4 Pengolahan Data - Aldi Dwi Apriliana - Universitas Komputer Indonesia
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&) Data tinggi bahu berdiri
Tabel 3.4.3. Data anthropometri T'B
1 2 3 4 5 6 7 8 9 10
1 15#." 14#.3 14!.# 140.5 13!." 14$ 14#.3 1#$. 13.5 13".#
2 143.3 133." 135.! 14.3 1#5 14$." 150.5 13!.3 13.4 151.4
3 153.5 150.5 153.# 13#.! 130 130.4 151.1 1#5.# 143." 145.
4 143.4 13.4 14#. 145.3 150.1 1#4 1#!.1 135 13.$ 134.#
5 141.$ 1#5.$ 1##.! 1#!.4 144." 150." 135 1#5.5 143.# 14.#
6 1#5. 135 154.# 130.4 143.4 1#4.$ 145.5 150." 1#.1 1#4."
7 140." 13.4 143.$ 14! 1##. 15#.5 154.$ 155 133.# 13!
8133.! 14. 13$.# 133 153." 1##.1 154.3 131." 13$.3 1##.#
9 130.1 13!.5 14".1 1#4 145 14.# 151.$ 1#$.$ 133.4 143.
10 130.4 154.$ 145 13$.3 1#!.3 153.4 14.! 13".5 14!." 13.4
d) Data tinggi (angkauan tangan
Tabel 3.4.4. Data anthropometri TT
1 2 3 4 5 6 7 8 9 10
1 #04.3 #0!." ##$.! #1.5 #13. ##5.$ #04. #3!.1 1$!.4 ##4.!
2 1$. ##5.$ ###.4 #31.! #04.# 1$1.1 ###." #10. 1$4 1$."
3 1$4.5 #10.5 ##3.4 #0$. #13 ##.1 ##0.# 1$1.$ #1$.$ 1$#.1
4 #01." ##5." 1!!.3 ##".5 #01.1 #30.1 #04.# ##4.! #40.# 1!$.$
5 #01." #35. 1$!.5 1$3. #3$.4 ##!.# ##. #35.1 #03." #0.!
6 ##1.5 #3!." #41.5 ##! #11." ##1.! 1$. #11. #30.# 1!.!
7 #3." #33.4 #0". ###.# 1!".1 1$3. 1$0.! ##0.$ #05.# 1$4.
8 #1!." 1$#. #40. #1!.4 #34 #0$.4 #04." #11." ##4.5 ##$.#
9 1$!.! 1$!.# #11.# #0 1$".5 ##!.$ #03.5 #3.$ 1!.5 #1
10 1!".! #1!.3 ##".3 ##5.5 #3!.# 1!".4 ##0.3 1$!.1 #04.1 1!.$
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e) Data (angkauan tangan
Tabel 3.4.5. Data anthropometri T
1 2 3 4 5 6 7 8 9 10
1 "" ,$ $,3 "4,0 4,3 5," ",4 !,$ $,3 "4,$ !0,!
2 5, 5,0 3,0 "#,! 3," 5,4 "5,3 !0,# ", !0,0
3 "0,0 3," "1,0 "4,4 "!,3 "!,4 "!," ""," "$,0 $,$
4 "4,4 "1,$ "5, "3," ",# ,0 3,4 3,# "!,# "5,!
5 "3," 1, "0," !0,1 ",5 $,0 "5,0 !1,0 ,3 3,3
6 "!,1 1," "5,$ "",! 1,$ #,3 "1,! "1,3 !,! ",5
7 $,1 $," "0," ",5 "1,0 $,3 #,3 "",# "4,0 !,!
8 !, "4,! "4,4 "1, #,3 "#,! ",5 ",5 "3,0 "$,!
9 "",5 ",$ ,# $," "3, "0,1 "!," 1,4 "0,$ 5,$
10 #,$ ",1 "#,5 !0," $,0 ,1 "1,5 !0,0 5,$ #,#
*) Data rentang tanganTabel 3.4.. Data anthropometri +T
1 2 3 4 5 6 7 8 9 10
1 1!5," 153, 1!," 11," 14$,# 1$#,4 1", 15!,1 150,$ 14,"
2 1!",1 15$," 1"0,4 10,4 10,0 153,# 154," 1!1,$ 11,0 1$0,$
3 1"!,0 1$3,! 1!",! 1!4,$ 1!5,0 1"#,4 1!4,$ 1"#,5 15",3 1$,"
4 1!#, 154,1 1$#,$ 1$4,0 1"3,3 1",1 1, 1"4,0 1"$,# 15","
5 1"4,3 10,# 1$0,4 10,$ 1$#,5 15$," 1"4, 1!,! 15," 1"1,$
6 1"3," 153,$ 14",$ 1"3,$ 1!3,0 15,4 1"", 11,1 1",$ 1$1,
7 14",# 14!,$ 1!3,4 1"5,# 155,1 14!, 1!, 1!",! 154,0 1",5
8 1"$,0 15,0 1," 15$,$ 1!$,5 13,5 1!1,0 1"3,0 1$#,$ 1!3,4
9 13,4 1$3,5 1"," 1!$,3 1!0,4 1",0 1!0,# 1!3,0 1!,3 15$,0
10 1"!,1 1!4,3 1!0," 10,3 1!1,1 1",! 1$3,4 15$,! 1$0,1 1$#,0
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g) Data tinggi siku berdiri
Tabel 3.4.". Data anthropometri TB
1 2 3 4 5 6 7 8 9 10
1 10$,# 110,5 11#,# 10, 111 105,4 $,4 $" 10",4 11#
2 100,4 101,3 103,$ 110,4 11#,4 10#, 10#," 10",! 10,4 $!,4
3 10$,3 10!,5 10," 11#,4 $,4 $",# 101,! 103,! 110,4 105
4 10$,# 10,! $$," 103," $,$ 111 10#,# $$,1 10#,$ 10",3
5 103,! 10!,1 10!,! 10#," 10!, 10! 105,$ 110 10!,3 $",4
6 10,4 110,$ 101,4 101,5 110,# 10!,4 10," 101," 105,$ 101
7 $",4 10",$ 104,! 100,! $" 10!," 10$," 110,! 111,1 105,5
8 101,$ 104, 10$,3 11#,5 $",# 10," $$," 10#, 10! 10#,#
9 $, 104 10,# 11#,# 110," 105,5 $",! 101,# $!,1 10$,"
10 104,4 10"," 104," 11#,4 $",! 104,5 10",3 10",1 $$,$ $,4
h) Data tebal badanTebal 3.4.!. Data anthropometri TB
1 2 3 4 5 6 7 8 9 10
1 44,3 43," 51,1 54,1 5#, 40,1 50,! 54,0 4$,0 41,5
24#,3 50,! 43, 40,1 41,$ 4,3 4,# 54,1 4!,4 4$,5
3 3$,4 4",3 4",! 53,1 3$,5 5#,3 41,0 3$, 51,$ 53,!
4 4!,3 40,3 5#,1 43,5 50,! 45,$ 53,4 4#,4 4",3 43,1
5 50," 51,3 40, 3$,$ 53,0 41, 4", 4!,1 40,$ 41,4
6 4!,1 4#,3 40, 53,3 3$, 44," 50," 41,0 44,4 40,#
7 44," 43,$ 43,# 4,4 44,$ 41,3 4!," 40,0 4",0 51,3
8 44,0 4!," 5#,0 44,1 45,3 54,0 4!,! 51,1 44,1 4,4
9 4,# 51,5 4", 4#,3 43,3 4#,! 53," 51,! 4#,3 40,5
10 4$,1 4",0 43,5 50,4 54,4 53,1 51,# 45,1 4",$ 51,1
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i) Data pan(ang lengan ba-ah
Tebal 3.4.$. Data anthropometri /B
1 2 3 4 5 6 7 8 9 10
1 #5,3 #"," #5,5 #4 #",1 #!,$ #", #!," #,1 #4,
2 #,$ #$,5 #4,1 #4,3 #$, #,! #4,1 #",! #",1 #",1
3 #$,1 #4,5 #,1 #4,4 #$,3 #4 #",! #5,# #!,! #5
4 #,! #",5 #!,# #4,4 #$ #5,1 #",! #5, #," #,$
5 #4,$ #$,# #5," #,# #!,$ #5,# #!,5 #!,$ #$,4 #,4
6 #5 #! #,4 #5 #5,! #5,# #5,# #4,4 #$,1 #5,#
7 #,$ #5,! #,$ #!,$ #",1 #$," #$,1 #, #!,5 #4,4
8 #",1 #!,$ #4,5 #!,! # #4,5 #",$ #4 #",4 #5,5
9 #!,! #, #",# #" #4,5 #4, #$ #!,# #!,# #!,3
10 #", #4,4 #!,3 #!,! #,3 #", #! #4,3 #4,3 #4,!
4.1.1.2. Data anthropometri posisi dudu
Data anthropometri posisi duduk berhubungan dengan perhitungan dimensi tubuh manusia dalam posisi duduk. Dengan data ini kita dapat membuat produk yang sesuai pada saat tubuh manusia
sedang duduk, dan bisa membuat produk lebih nyaman.a) Data tinggi duduk tegak
Tabel 3.4.10. Data anthropometri TDT
1 2 3 4 5 6 7 8 9 10
1 !0,0 "!,3 $0,1 !3,1 !3,# !, !1,# "!,# !#,0 !",#
2 !#,$ !4,1 "$,1 $0,# $1," !"," "!,4 $1,$ "!,# !#,!
3 !0,4 "$,0 !4,$ !5,! !1,3 !3,1 !!,# $0,4 "!,5 !#,3
4 "!,# !!,$ "$,! !!,4 !, "$," $1,# !3,1 $#,0 !0,#
5 "!,! !,3 !,4 "$,$ !",# $0,3 !$,5 !1, !1," $0,3
6 !$,# !0, !3, !",1 "!,0 "$,! !0,3 !3,# !1,# !!,0
7 $1,1 $1,$ !5,5 !",0 "!,0 !3,# !3,5 $0, !#,# $1,
8 !0,1 !",0 !!,# "$,4 $1,# !5,$ $1, !4,1 !1, !,0
9 !", !!,$ $1,3 !3,$ !4,5 !4,3 !#,! !5,0 "!,$ $1,#
10 $0, !#,# !1,# !0,1 !,3 !,5 "$,$ !3,3 !,5 !#,!
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b) Data tinggi mata duduk
Tabel 3.4.11. Data anthropometri T%D
1 2 3 4 5 6 7 8 9 10
1 "#,5 !#,3 $,0 !#,0 "4,5 !,0 "$, ",! "",1 !#,3
2 "!,1 !, !#,! !,4 $,$ "5,5 !1,! "3,$ !3," ",1
3 !0," !, "1,$ !,4 !1,0 "3,4 "5,0 ," ,$ !3,0
4 "5,3 !3,0 "4, ",3 "3,5 !1,# "!,1 "4,! "5,0 !#,4
5 "5,$ ",3 !#,# !#,0 "4,1 ,3 "4,4 $,4 "1, $,4
6 !, "#, "5,0 "3,4 !#,5 $,3 !0,# ",$ "0,0 "$,0
7 $, ",# "4,! ",5 !0, !#,4 "#,3 $,# !3,1 !#,1
8"$,1 "5," !1,! !0,! "#,5 "3,5 "0,# !#,4 $,1 "1,3
9 "," "1,$ !1,# "!,3 !3,1 "5,! ",1 !0,3 "1,1 "#,"
10 !#," "4,3 !0,! "0,1 "0,3 !0,1 $,5 "$,! "#,# "!,#
&) Data tinggi bahu duduk
Tabel 3.4.1#. Data anthropometri TBD
1 2 3 4 5 6 7 8 9 10
1 0,$ 55,3 1,$ #,1 55,1 , ",4 "," 5$,! 5$,5
2
4, " 1,! 4 1 ,! 3,# #,# #,# 4,#
3 "," 1,1 1,4 53,3 5",5 3," 3,4 5,! 53," 3
4 #, ! ,# #,# 53 5",! 5,5 5$,! 4,! 0,"
5 , ,# 3,# 4,$ #," 4,3 53, ",3 53,5 1
6 5,$ 5!," 5$,5 54," 5,4 4,5 ,5 1,# 4 !,3
7 5!," !,3 5,5 !,1 5", 53,3 0,3 53 55,! 0,
8 !,5 !, 5$,$ !,5 5",5 !,! 1,1 5",1 # 53,$
9 3,5 5,1 4,5 !,4 55,$ 0,# 3 54,4 !,!
10 ",1 0," 5,# 53, 5!,5 ,! ,1 0, 0,5 55
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d) Data lebar bahu
Tabel 3.4.13. Data anthropometri /B
1 2 3 4 5 6 7 8 9 10
1 40, 41,4 3$, 40, 4,# 3!,3 35,4 44," 3",! 3,5
2 45,1 3",# 3,3 3,# 40,5 3,0 3!,! 44,1 41,! 45,0
3 44,4 3$,# 41," 3!,1 4#,1 3!, 35,! 34, 41,! 34,$
4 3!,3 41,1 3,1 3!,5 3$,1 3,4 3",# 34,0 41,3 3!,"
5 34,$ 3,1 41,4 4#, 45,3 3$,$ 3,1 40,! 35,3 45,0
6 3,4 3!,! 3", 35, 40,# 44,3 44,5 3!," 3",! 3$,3
7 45,0 44," 3$,4 3,! 40,0 40,# 44,! 34,! 40,! 35,0
8 3!,! 34,1 43,1 3$,5 40," 3,# 34," 4#,$ 3!,0 35,3
9 4,# 34,! 4,0 3,$ 40," 3,# 4#," 43,5 41," 3",5
10 35,# 35,1 3",1 4#,5 4,0 4#,# 3$,4 45,! 44,5 3",5
e) Data lebar sandaranTabel 3.4.14. Data anthropometri /
1 2 3 4 5 6 7 8 9 10
1 #5,$ ##,# #4,4 #4," #1,0 1!," #0,4 #,0 #0," 1$,#
2 1!,5 #0,! #1,$ #5,$ ##,3 ##," #5," 1$, #1,! #4,3
3 ##,5 #3,5 #4, #0,0 ##,5 #1,5 #3," #3, ##,4 #4,#
4 #5,5 #3, ##,3 #0,4 #3,5 #3,0 #5, #5," #5, #1,0
5 1!,5 ##,! #4,4 #5,5 #5,5 ##,4 ##,0 #0,5 ##,0 #3,0
6 #5,5 #4,! ##,3 #3,5 #3," #4,0 #3,3 #0,0 #4," #4,1
7 1!,4 ##,0 ##, #0,5 ##,0 ##,0 #1,# 1$,$ #1,3 #3,
8 1$,5 #0,$ #1,! #0,# #3,# 1!,! #3,5 #3, 1!," #5,3
9 1$,1 1!,! 1$,! 1$,4 #4,$ #1,4 #0,5 1!, #1,$ #3,$
10 #0," #1, #3,1 ##,! #3,# #3,# #4," #5,! #0,5 #4,1
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*) Data siku ke siku
Tabel 3.4.15. Data anthropometri
1 2 3 4 5 6 7 8 9 10
1 33,$ 3,! 45,3 43,3 34,3 4",1 44,1 3$,1 3#," 3#,!
2 3,$ 35,3 41," 3#,1 41, 43,! 4#,$ 41,5 3#,$ 34,#
3 3$,5 33," 45,3 43,4 3$,$ 34,$ 45, 34,$ 44," 4#,4
4 34,$ 4#," 3$,4 40,0 4", 45,5 45,# 4,# 45," 4,0
5 3#,3 3!,5 3$," 44,1 4,5 34,! 4,0 3!, 34,4 33,!
6 45, 44,# 3$,4 40, 45,3 35," 3#,$ 33,$ 40,0 3",!
7 35,5 43,3 34, 44,3 43,4 40,0 44,0 4,1 3#,# 4",!
8 31,5 45,3 3!,0 34,1 3,0 3!,$ 4#,1 44,1 40, 31,5
9 41,$ 45,3 4",1 31, 45,4 4#,3 4#,4 33, 45,4 33,"
10 41,5 3$,0 45,0 4,# 3#, 35,! 34,$ 4#, 3#,! 31,#
g) Data lebar pinggulTabel 3.4.1. Data anthropometri /
1 2 3 4 5 6 7 8 9 10
1 40, 41,$ 3#, #, 34," 43,0 #!, 34,5 40," #!,
2 3, 3#,$ #,4 3!,5 #$,3 43,1 #",$ 3!,4 #!,0 30,"
3 4#,1 #5,! 3", 33, 35, 30,4 #!,$ 30," 3#,3 35,1
4 3#," 33,# #",0 40,4 30,! 31,5 41,3 #",# 30, #5,$
5 #5,3 41,5 3$,1 30,# #,5 34,3 30,5 30,4 34,4 #!,4
6 34,$ 4#,# 31,4 #!,1 #,0 #!, 35,# 43,5 33,! 34,5
7 3$,1 3,1 #$," 3"," 3#," 3!," 34,1 #!,1 #$,$ 41,#
8 40,5 4#,! 43,! 3!,5 35,5 3!,4 34," 3!,0 3#,0 3!,1
9 3",! 41,1 31, #,4 35,4 #$,$ 33,# 3",5 3$, 35,1
10 41,$ 33,4 43," 41,! #,! #",1 4#,5 #4," 3$,0 3",0
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h) Data tinggi sandaran
Tabel 3.4.1". Data anthropometri T
1 2 3 4 5 6 7 8 9 10
1 44,# 44,5 4,5 50,0 4,0 4",1 43,3 4!,0 3$,0 3!,0
2 4,4 3!,0 5#,5 4$,5 43,0 44,3 40,5 41,0 4,5 4",0
3 4,0 4",1 43,3 4!,0 3$,0 3!,0 50,5 4!,5 4#,! 41,0
4 43,0 44,3 40,5 41,0 4,5 4",0 4#,0 40,5 4!,0 4,5
5 3$,0 3!,0 50,5 4!,5 4#,! 41,0 41,0 4,0 4#, 53,0
6 4,5 4",0 4#,0 40,5 4!,0 4,5 41,5 41,0 44,0 4,0
7 4#,! 41,0 41,0 4,0 4#, 53,0 4,0 44,# 44,5 4,5
8 4!,0 4,5 41,5 41,0 44,0 4,0 44,5 4,4 3!,0 5#,5
9 4#, 53,0 4,0 44,# 44,5 4,5 50,0 4,0 4",1 43,3
10 44,0 4,0 44,5 4,4 3!,0 5#,5 4$,5 43,0 44,3 40,5
i) Data tinggi pinggang
Tabel 3.4.1!. Data anthropometri T
1 2 3 4 5 6 7 8 9 10
1 1!,4 #0,# #0,0 #0,4 1!,4 #0,# 1",0 ##,! 1!,4 ##,
2 1$,# 1!,$ #1,! 1$,! #0,! #3,3 #5,0 #4,# #4,5 1!,1
3 1$,1 1!, 1!,0 #0,3 #1,0 #1, 1$," 1",5 #0,# 1!,3
4 #1,1 #0,# ##,0 #0,$ #1,$ #1, #4,0 ##,$ #0,! #1,4
5 1",1 #3,1 #4,4 1",5 #0,3 #0,# #3,5 1"," #1,! #1,3
6 1$,$ #0,4 1"," 1!,3 ##,4 #3," #0,$ #1,4 #0,# 1$,!
7 ##,$ 1$,3 1$, #0,$ 1",5 1!,4 1",$ #0,3 1",4 #4
8 #3," #1," 1$,3 1!,4 #3,$ #0,3 1$,# 1!,$ ##,$ #4,
9 1$,0 ##,$ #4,$ #4,1 1$,4 #0, #3,5 #4,5 #3,$ 1!,"
10 ##,1 ##,# #4,$ #0,5 1!,5 #1,4 #0,4 #0,3 #0, ##,!
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() Data pan(ang sandaran
Tabel 3.4.1$. Data anthropometri
1 2 3 4 5 6 7 8 9 10
1 #"," #4,0 #4,# #5,$ ##,0 #," #1," #$,! #1," #1,#
2 #, ##,$ ##,0 3#,# #4,3 #!,3 31, 31," 30,! #",#
3 3#,5 30,0 #",# ##,4 #$, #!,1 #,! 3#,5 #,! 31,0
4 #3,4 #",4 31,$ #1,! #5,0 #,5 #3,# #!, 3#,4 #",
5 #",4 #!,0 #,3 #4,5 #,5 31,# 3#,4 #$,0 #$,3 #3,#
6 #4,4 #$,$ 30,$ #3," #,3 #!,5 31," #4,1 #5, #5,"
7 #$,5 #,0 #4, #!,# #4,5 #!, #$,3 #1,! #3," #4,3
8 #3,5 #5,1 31,0 #1,5 31,4 31,0 #,4 #",# #",3 #,#
9 #4,0 #,4 #3,! #5,0 #3,1 #!,$ 3#,4 #1,0 #4,3 #$,1
10 30,3 31,0 3#,1 #", #,4 #3,# ##," ##,3 30, #5,3
k) Data tinggi siku dudukTabel 3.4.#0. Data anthropometri TD
1 2 3 4 5 6 7 8 9 10
1 1!,! 1",# #," #3,! #4, 1$,# ##,$ #0,3 ##,3 #1,"
2 #,1 ##,5 #5," ##," 1",4 ##,0 ##,$ ##,3 1$, ##,1
3 #",0 #,0 #3,1 1$,4 #,1 #1," #1,! #3,$ 1",$ #0,5
4 1!,# #4,3 #4,0 1,5 #,0 #5, #1,$ #1,# #1,$ #3,4
5 #5,4 #5,0 ##,0 1",# 1$,! #3,! #,# #,3 1!,# #4,$
6 #0,3 #5,$ 1$," ##,0 1$,# 1, ##,4 #0,3 #4,# #1,5
7 #4,# #1,3 #5,5 #4,! ##,5 1",1 #1," #5," ##," #3,
8 #4, #1, #0,4 #4,# #4," 1$,1 #, #0,$ 1!,# #4,4
9 #,# 1"," 1!,$ #0,$ 1",# 1!,1 1$,4 ##," #",0 #0,$
10 1," 1!, #0,3 1$, #0,# #0,$ #4,$ #3,5 #0,1 ##,
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l) Data tebal paha
Tabel 3.4.#1. Data anthropometri T'
1 2 3 4 5 6 7 8 9 10
1 #1,! #$,5 #5,$ #",! #1,1 #3,# #$,1 #$,$ #",5 #,!
2 #,# #,1 #!,3 ##,$ #3,5 #!,0 #$," #1,$ 1$,! #3,0
3 1$, #1,4 #3,$ ##,1 #0,0 #1,5 #$,1 #,3 1$,1 #",1
4 #4,4 #!,4 #3,0 #3,3 #5,1 ##,5 #3,# #0,5 #0,0 #",3
5 #0,$ #4, #5,5 #0,1 #3,# ##,0 #0,0 #1,$ #,5 #,!
6 #," #0, #5," #!,0 #3,$ #3,3 #!,! #$,$ #$,! #4,4
7 #5,$ #, #",1 #1," #3,1 #1,5 #1,4 ##," #1," #1,3
8 #!, 1$,4 #$, #0,# #4,4 #$,3 #!,$ #3, #",! #!,
9 #!, #3," #!,# #0,4 #0,3 #4, #$," #,5 ##,0 #1,#
10 #",3 #4,4 #3,5 #!,# #5," #3,4 #",5 #,! #",0 #0,5
m) Data tinggi poplitealTebal 3.4.##. Data anthropometri T/
1 2 3 4 5 6 7 8 9 10
1 44,$ 4#,! 4!,3 41,1 45, 4"," 41,! 3$,4 43,# 4",!
2 3$, 4#,5 44,5 44, 41,! 3$,4 3$,! 4,# 4,! 3$,4
3 3!,3 3!,# 43,! 40,1 3$,# 4!,5 43,0 4!,# 4",1 4",#
4 43, 4!,# 4#,3 43,$ 40,1 4#,0 3$,$ 43,1 41,0 41,0
5 4",3 40," 41,! 4,4 44,5 44,# 41,! 4,0 4!,# 40,$
6 41,! 4#,! 3!,# 4"," 4,4 41,5 4", 45,0 45,$ 4!,0
7 4",5 45,0 3$,4 4#,4 4",1 40,1 40,3 3!, 3!, 4",3
8 4!,$ 45,0 4",5 41, 3$,4 43,5 4#,$ 41,5 3!," 40,1
9 41,1 3$," 45," 45," 3!,3 3$,3 40, 4",0 43,! 3!,!
10 43,1 44,# 4#,$ 43, 45,# 4", 43, 43,5 44,0 4!,"
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n) Data pantat popliteal
Tebal 3.4.#3. Data anthropometri /
1 2 3 4 5 6 7 8 9 10
1 4#, 53,1 43, 41, 43,0 50, 53, 5,5 5, 43,1
2 4!,$ 4",1 41,3 50,5 3$,5 44,1 5,# 45,0 4#,1 44,0
3 5,3 5,$ 4!,4 4",! 4$,3 45," 4!,1 5, 53,3 3$,!
4 41,1 55,1 50,# 50,$ 4$,5 4!,# 40, 51,! 44,1 4,5
5 53,0 5#,5 55,4 53,0 51, 43,5 5#,0 40,! 4$,4 51,$
6 5, 4!,4 5#,$ 51,! 53," 41,3 41,$ 4#,1 4#, 45,!
7 44," 50,! 54,! 51,5 4",# 5#, 51," 4,# 4"," 45,#
8 4!,# 4,1 45,4 45, 51,! 41,1 50,1 55,! 50,# 54,
9 44,1 41," 43,4 4$,1 43,! 44,0 41, 4#,$ 54,! 5,0
10 40,5 53,3 54,3 40," 5, 41, 55,3 40,! 51,$ 4!,"
o) Data tebal lututTebal 3.4.#4. T/
1 2 3 4 5 6 7 8 9 10
1 1",0 1,3 1," 1,4 1",5 ##,1 1",$ 1!," #0, 1",#
2 ##,# #1,3 #1," #1,$ #0,3 1,0 1,# ##,$ ##,0 #0,4
3 1!,! #0,3 1,5 1$,0 ##,0 1!,1 #1,5 1, 1!,3 1!,#
4 ##,$ #0,5 1$,5 1", #0,4 #1,! ##,# 1", #0,0 ##,"
5 1!,0 1!,# 1$, #0,3 1$,5 #0,0 1$,$ ##, 1$,3 #0,!
6 1",3 #0,$ 1",0 #0, #1,1 #0,3 1," #1,! 1"," #1,4
7 1$,5 1$," #0,$ 1",0 ##,$ #0,0 #1,# 1$,$ #1,# #0,1
8 #1,0 1$,1 1",4 1,4 #1,1 #1,$ #0,! ##,# 1!,3 1!,#
9 ##,0 1",! 1,0 #0," #1,3 #1,! #0," 1!,$ 1,5 #0,0
10 1$," 1$,1 1",# 1!,5 1",3 #1,3 ##,$ 1,3 ##,0 1,0
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p) Data pantat ke lutut
Tebal 3.4.#5. Data anthropometri /
1 2 3 4 5 6 7 8 9 10
1 4,5 "3,5 "#,0 "1,# 4,5 "3,5 "0,0 4,0 "1,0 3,0
2 ,5 ",5 4,5 "5,0 ,5 ",5 3,0 ,0 "3,5 5,5
3 "1,0 3,0 "0,0 4,0 "1,0 3,0 ,4 ",5 4,5 4,5
4 "3,5 5,5 3,0 ,0 "3,5 5,5 #,0 !,5 "0,5 "#,0
5 4,5 4,5 ,4 ",5 4,5 4,5 "4,0 !,0 0,5 5,5
6 "0,5 "#,0 #,0 !,5 "0,5 "#,0 "$,0 ,5 ,0 1,5
7 0,5 5,5 "4,0 !,0 0,5 5,5 55,! "3,5 "#,0 ",5
8 ,0 1,5 "$,0 ,5 ,0 1,5 "3,5 ",5 5,0 5!,0
9 "#,0 ",5 55,! "3,5 "#,0 ",5 "1,# 4,5 "3,5 "0,0
10 5,0 5!,0 "3,5 ",5 5,0 5!,0 "5,0 ,5 ",5 3,0
4.1.1.3. Data anthropometri !ari tangan
Data anthropometri (ari tangan digunakan untuk membuat produk yang berhubungan dengan (aritangan sehingga produk bisa lebih nyaman dan tepat saat digunakan.
a) Data pan(ang (ari 1Tabel 3.4.#. Data anthropometri 1
1 2 3 4 5 6 7 8 9 10
1 ,4 4,$ ",# ,3 ,3 5, 4,$ ,# ",5 5,"
2 ,0 5,3 5," , ,1 ,1 ,0 5,0 5,$ ,3
3 5,1 ,0 5,5 ",1 5,1 ,$ ,5 4,! ",5 5,!
4 ",# 5,! ," ,5 ",# ,0 ", ,! ",# 5,$
5 ,! 5,1 ,3 ,0 ",3 4,! ",# 5,1 5,$ ,5
6 5,! 5,! ",4 ," 5,0 5,1 ,0 5,0 ",1 ",#
7 5,! ,3 ,1 ", ",0 ",3 ",5 5," 5,4 ",0
8 5,! 5,3 ,3 ",3 ,4 ",5 5,0 ,# ,# 5,
9 5,! , 5, ",1 5,3 ,5 5," 5,3 5,# 5,1
10 5,! ",3 ", ,! ," ,3 5,4 ,1 ",4 5,0
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b) Data pan(ang (ari #
Tabel 3.4.#". Data anthropometri #
1 2 3 4 5 6 7 8 9 10
1 ,! ,4 ,$ ,! ,4 ",3 ," ",1 ",4 ,4
2 ," ,4 ",0 ",$ ,# ",$ ",5 ",3 ",# ",0
3 ",$ ",$ ",# !,0 ,4 ",4 ",4 !,0 ",4 ",5
4 ",4 ",5 ",$ ",1 ,! ,! ,4 ",! ",0 ,4
5 , ,# ",4 ,5 "," ",5 ,# ",$ !,0 ","
6 ,$ , ,3 ,$ ,# ",4 ",! ",1 "," !,0
7 "," ",4 ",3 ",$ ",0 ",5 ",5 ",# ", ,4
8 "," ",4 ",$ ,$ ",# ",1 !,0 ", ",3 ,$
9 ,3 ,3 ,$ ,4 , ",# ",3 ,3 ",# !,0
10 ,! ",5 ",5 , ,5 "," ",0 ", ",# ,5
&) Data pan(ang (ari 3
Tabel 3.4.#!. Data anthropometri 3
1 2 3 4 5 6 7 8 9 10
1 !,# !,# !,$ ",1 !, !,1 ",3 $,0 !,3 ",
2 ",1 ",! !,$ ",1 ",3 ",4 !,0 $,0 ",$ !,!
3 ",# ",1 !,3 !,5 !,4 ",5 !,4 ",1 ",# ",#
4 !,3 !,! !,3 !,! !, $,0 ",3 ",1 ",0 ",#
5 ",$ "," !,# ",$ $,0 ",! ", !,4 ",1 ",#
6 !,0 !, ",5 ",! ",! !, !,3 !,! !,1 !,4
7 ",3 ",0 !," ",! !, "," ",$ ",# ",1 ",3
8 ",4 ",! ", !, !,5 ",# !,3 ",$ !,5 ",4
9 !,3 ", ", ",5 !,4 !,# ",# ",! ",1 !,$
10 !,1 !,$ ",$ "," !,1 !,0 ",0 "," !,1 ",3
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d) Data pan(ang (ari 4
Tabel 3.4.#$. Data anthropometri 4
1 2 3 4 5 6 7 8 9 10
1 !,4 !,4 ",4 ! ,$ !,4 ,3 ,5 !,# !,1
2 ,5 ,3 ",$ ,# "," ,# ",0 ",4 !,4 !,0
3 ,! ,1 ",$ !,# ,0 ,4 ",0 ,3 ", ,5
4 !,1 ",1 ",# !,3 ,$ !,5 ,$ ",0 ,$ ,5
5 ,0 ",1 ,3 ,$ ,# ",1 !,4 ",3 ,5 !,3
6 "," ",4 ",4 , ",0 , ",4 ,4 ,1 !,0
7 ",1 ",4 ",$ !,# !,4 !,3 ",5 ,# ,# !,0
8 ",4 ",$ , ",0 ",$ ",# !,5 !,4 ",4 ",5
9 !,3 ",3 ", ",$ ,! !,4 ," ",0 !,# !,3
10 ",5 ",5 ,$ !,# ",! ,1 !,# ,1 ",! ,#
e) Data pan(ang (ari 5
Tabel 3.4.30. Data anthropometri 5
1 2 3 4 5 6 7 8 9 10
1 , ,3 5," , 5,$ ,3 ,3 5, 4,$ ,5
2 , , ,4 ,# 5,0 ,! 5,3 5,5 ,4 ,5
3 , ,5 5," ,5 5,$ 5,1 ,0 ," ,# 5,$
4 5," 5,# 5,1 4,$ ,0 ," ,$ 4,$ , ,"
5 ,5 5,1 5, 5,5 5,$ ,4 5,3 5,0 ,4 ,
6 5,3 5,# 5,$ ,0 ," 5,1 ,3 , 5,5 5,4
7 5,1 4,$ 5,0 ,5 4,$ ,$ 5,! ,! 5,1 ,!
8 5," 5,! ,1 ,# 5," ," 5,1 , 5,0 5,1
9 ,3 ,5 , 5,! 5,! 4,$ 5, ,5 , ,
10 ,$ ,# 5, , 5,$ ,3 5,1 5, ,0 5,0
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*) Data pangkal ke tangan
Tabel 3.4.31. Data anthropometri pangkal ke tangan
1 2 3 4 5 6 7 8 9 10
1 10,4 10,$ $,0 $,4 $,0 $,5 10," $, 11,4 $,$
2 11,$ $,! 11," 10,1 10,$ 10,0 10,4 10,3 11,3 $,3
3 10," $,! $,$ 11,3 10,5 $,4 10,4 11,! 11,5 11,"
4 $, 10,0 10,# 10,# 11,# 11,# 11,# 11,! 11,! $,3
5 $,$ 11,0 11,3 $,0 11,! $,1 10," 11,0 $,# 10,4
6 10,4 $,4 $,$ $,! 10,# 11,5 $, 10,0 10," 11,!
7 $,3 $," 10,4 11,3 $,0 10,$ 1#,0 $,! $,4 $,5
8 $,3 $,! $,4 $,3 11," 11,1 10,# 11,! $,! 11,0
9 10,5 10,5 10,4 $,4 $,4 10," 10, $,# 11, 10,4
10 10,! 11,3 $,! 1#,0 10,0 11,$ $,3 $," 10," $,1
g) Data lebar #,3,4,5Tabel 3.4.3#. Data anthropometri lebar #,3,4,5
1 2 3 4 5 6 7 8 9 10
1 ,4 , ,! ",1 ",! ",$ ,4 ,3 ,3 ,3
2 ",5 ,5 ,# ",3 ",1 ,0 ", ",$ ", ,$
3 ",5 , ",4 ",1 ",# ,# ",4 "," ",3 ,5
4 ", ,0 ",3 ",! ",0 ",3 , "," ,# ",1
5 "," ",# ",0 ," , ",$ ,$ ,1 ,! ",
6 ",3 "," ",1 ",# ",$ ",3 ",0 ",4 ,$ ,0
7 ," ,# ",# ,4 ,0 ",! ,$ ",# ,! ,"
8 ",$ ,! ,5 ,0 ," , ",1 ," , ",!
9 ", ",! ,5 ,4 ,5 ",3 ,5 ",! ,3 ,3
10 ",! , ," ,# ",4 ,1 , ," "," ,"
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h) Data lebar tangan
Tabel 3.4.33. Data anthropometri lebar tangan
1 2 3 4 5 6 7 8 9 10
1 $,3 10,3 11,0 $,5 $,4 10,4 $,! $,# $,4 10,4
2 $,$ $,0 10,4 11,0 10, $,$ $,$ 10,5 $,# 11,0
3 $,$ 10, 10,! $,1 10,3 10," 10,$ $,! 10, $,#
4 10,0 10,0 $,# 10,! 10,0 $,0 $, $,4 $,# 10,#
5 10,$ 10,3 10,# $,! $,5 $, $,! $,! 10,0 10,"
6 $,! $,5 $,3 $,0 10,# 10,0 $," $," 10,$ $,!
7 $,3 10,1 10,1 $,0 $,0 $," $,0 10,0 10," 10,4
8 $,! 10,4 10," 10,4 $,1 $,4 10,5 $, $,4 10,4
9 $,# $,4 10,4 10,1 10,! $,5 10,$ $,3 10,4 10,$
10 $,0 $,3 10,! 10," 10,4 $,5 $,5 $,3 $," 10,
i) Data pan(ang telapak tangan
Tabel 3.4.34. Data anthropometri pan(ang telapak tangan
1 2 3 4 5 6 7 8 9 10
1 1$,# 1!,# 1!, 1!, 1",# 1$,$ 1",1 1",3 1,# 1,5
2 1$,4 1$,# 1!,5 #0,0 1$,3 1",4 1!," 1,3 1$,3 1,!
3 1," 1",3 1," 1$,4 1",# 1," 1,$ 1"," 1$,0 1",3
4 1,4 1,4 1$," 1$,$ 1!,0 1$,# 1$,# 1",3 1", 1,4
5 1",$ 1!,5 1",4 1!,$ 1$,4 #0,0 1!,3 1",5 1$,! 1$,$
6 1",# 1"," 1!,# 1,3 1",# 1!,1 1!," 1",0 1$,1 1,
7 1!,5 1$,! 1",$ 1$,5 1,# 1$,3 1!,5 1$,# 1$, 1,!
8 1!," 1!,1 1",$ 1!,# 1$, 1,4 1!,! 1,! 1",3 1",
9 1, 1$,3 1$,5 1!,1 1",5 1!,$ 1,5 1!, 1$ 1!,5
10 1$, 1$,1 1!,$ 1,4 1", #0,0 1!,5 1!,3 1$,# 1",4
4.1.2. Data display "arna
Tes display -arna ini dilakukan dengan &ara praktikan menuliskan (elas apa tidak (elasnya suatu-arna yang dikombinasikan dengan -arna lain. Tu(uan tes display -arna ini untuk mengetahuidisplay -arna yang (elas menurut praktikan yang selan(utnya akan diproses untuk membuat peran&angan display poster).
Tabel 3.4.35. Data display -arna
2o. /atar arna elas Tidak elas
1 bubu 6oklat
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# %erah 3 uning 4 7range 5 'i(au
Biru "
6oklat
buabu ! %erah $ uning 10 7range 11 'i(au 1# Biru
13
uning
buabu 14 6oklat 15 7range 1 %erah 1" 'i(au
1! Biru 1$
7range
6oklat #0 buabu #1 uning ## %erah #3 'i(au #4 Biru #5
%erah
buabu # 6oklat #" 7range #! 'i(au #$ uning 30 Biru 31
'i(au
buabu 3# 6oklat 33 uning 34 7range 35 %erah 3 Biru 3"
Biru
buabu 3! 6oklat 3$ 'i(au 40 %erah
41 7range 4# uning
4.1.3. Data tes buta "arna
Tes buta -arna ini dilakukan dengan &ara pratikan menuliskan angka atau gambar yang terdapat pada buku tes buta -arna yang dipegang oleh asisten dan diganti setiap 5 detik. Tes buta -arna ini ber*ungsi mengetahui apakah seseorang mengalami kelainan buta -ana atau tidak dan kemudianakan dibuat peran&angan display poster.
Tabel 3.4.3. Data hasil tes buta -arna
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ldi D-i 8ang Benar
1 1# 1#
# ! !
3 5 5
4 #$ #$
5 "4
" "
" 45 45
! # #
$ #
10 1! 1
11
1# 35 35
4.1.4. Data atual ruangan
a) Tingkat kebisingan dan pen&ahayaantingkat kebisingan dan intensitas &ahaya sangat berpengaruh dalam akti9itas disuatu ruangan.
Untuk mengetahui tingkat kebisingan dan tingkat intensitas ruangan diperlukan pengamatan
menggunakan lux meter dan sound level , diba-ah ini merupakan hasil pengamatan tes tingkat
kebisingan dan pen&ahayaan di ruang sekretariat (urusan sek(ur).
Tabel 3.4.3". Tes kebisingan dan pen&ahayaan ruang sekretariat (urusan
atuan kun(ungan : 5 menit
'ari : ertama enin, " Desember #015)
Tempat penelitian : +uang sekretariat (urusan 2o Bilangan random am Lux meter Sound level
1 1 0!:05 0.#$ ".00
# 3 0!:15 0.#! 3.3
3 " 0!:35 0.30 0.#
4 10 0!:50 0.## 1.
5 11 0!:55 0.30 1.$
14 0$:00 0.#3 "."
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" 15 0$:15 0.#$ 4.1
! 1" 0$:#5 0.#3 3.
$ 1! 0$:30 0.30 3.5
10 #0 0$:40 0.30 0.
11 #3 0$: 55 0.30 4.5
1# #4 10:00 0.# 3.!
13 # 10:10 0.30 .5
14 #! 10:#0 0.30 5.!
15 30 10: 30 0.30 0.3
1 3# 10:40 0.#! 1.!
1"34 10:50
0.30 3.1
1! 3 11:00 0.30 0.4
1$ 3" 11:05 0.#3 "."
#0 40 11:#0 0.#$ 1.$
#1 41 11:#5 0.30 0.#
## 4# 11:30 0.30 3.5
#3 44 11:40 0.30 .5
#4 4 11:50 0.## 5.!
#5 4$ 1# :00 0.# 4.5
# 50 1# :05 0.#! 3.1
#" 5# 1# :15 0.30 3.4
#! 53 1# :#0 0.30 "."
#$ 54 1# :#5 0.## 3.
30 5 1#:35 0.3# 1.!
31 5" 1#:45 0.30 0.4
3# 5$ 1#:55 0.#" 5.!
33 3 13:00 0.#3 .5
34 3 13:00 0.30 0.#
35 " 13:#0 0.31 1.!
3 ! 13:30 0.3# 3.
3" $ 13:35 0.## 4.5
3! "1 13:45 0.30 5.!
3$ "4 14:00 0.30 .5
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40 "4 14:00 0.#4 1.!
41 "! 14:#0 0.# 5.!
4# !0 14:30 0.#! 0.3
43 !1 14:35 0.30 3.5
44 !# 14:40 0.#$ 1.!
45 !3 14:45 0.30 .5
4 !4 14:50 0.30 0.3
4" !5 14:55 0.3# 5.!
4! ! 15:00 0.#5 4.5
4$ !$ 15:15 0.30 5.!
50$0 15:#0
0.## 4."
Tabel 3.4.3!. Tes kebisingan dan pen&ahayaan ruang sekretariat (urusan
Tes kebisingan dan pen&ahayaan
atuan kun(ungan : 5 menit
'ari : edua elasa, ! Desember #015):
Tempat penelitian : +uang sekretariat (urusan
2o Bilangan random am Lux meter Sound level
1 1 0!:05 0.34 "."
# 3 0!:15 0.34 $.#
3 " 0!:35 0.34 5.1
4 10 0!:50 0.3# "3.#
5 11 0!:55 0.35 !.3
14 0$:00 0.34 4.
" 15 0$:15 0.35 #.5
! 1" 0$:#5 0.35 4.5
$ 1! 0$:30 0.35 5.4
10 #0 0$:40 0.35 4.!
11 #3 0$: 55 0.35 .1
1# #4 10:00 0.35 .!
13 # 10:10 0.34 4.#
14 #! 10:#0 0.35 #.3
15 30 10: 30 0.35 3.0
1 3# 10:40 0.35 ".3
1" 34 10:50 0.3# 4.0
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1! 3 11:00 0.35 4.4
1$ 3" 11:05 0.35 .5
#0 40 11:#0 0.35 !.$
#1 41 11:#5 0.35 !.$
## 4# 11:30 0.3 3.3
#3 44 11:40 0.35 ".5
#4 4 11:50 0.34 5.3
#5 4$ 1# :00 0.34 .5
# 50 1# :05 0.35 4.5
#" 5# 1# :15 0.35 !."
#!53 1# :#0
0.35 ."
#$ 54 1# :#5 0.3# 5.4
30 5 1# :35 0.3 3.5
31 5" 1# :45 0.3 .!
3# 5$ 1# :55 0.35 !."
33 3 13:00 0.35 "#.1
34 3 13:00 0.35 4.
35 " 13:#0 0.3" $.3
3 ! 13:30 0.3 4.5
3" $ 13:35 0.3 #.3
3! "1 13:45 0.3" 4.1
3$ "4 14:00 0.35 1.!
40 "4 14:00 0.3# ".!
41 "! 14:#0 0.35 3."
4# !0 14:30 0.35 !.4
43 !1 14:35 0.35 5.5
44 !# 14:40 0.35 4.3
45 !3 14:45 0.31 .#
4 !4 14:50 0.35 .$
4" !5 14:55 0.3" $.!
4! ! 15:00 0.34 !.3
4$ !$ 15:15 0.35 5.1
50 $0 15:#0 0.3$ #.$0
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b) /uas ruang sekretariat (urusan/ebar ; 3,$ meter an(ang ; 3," meter Tinggi ; 3,# meter
<ambar 3.4.1. +uang sekretariat (urusan
&) apasitas ruanganapasitas ruangan sekretariat (urusan kurang lebih dapat menampung 3 sampai 5 orang.
d) =ungsi ruangan
=ungsi /aboratorium dalam proses pendidikan, yaitu:1) ebagai tempat untuk menyelesaikan administrasi perkuliahan, pengambilan kartu u(ian
dan pemin(aman buku.#) ekretariat (urusan sebagai -adah logistik ringan yang menun(ang proses perkuliahan.3) ekretariat (urusan sebagai tempat bagi mahasis-a yang butuh in*ormasi mengenai
perkuliahan ataupun beasis-a.
e) kti9itas dalam ruangan
kti9itas yang dilakukan di lab sekretariat (urusan adalah sebagai berikut:1) %elakukan pemin(aman buku.#) %enyelesaikan administrasi perkuliahan.3) Bertanya in*ormasi mengenai e9ent yang ada di kampus ataupun diluar kampus.4) %entranskip nilai mahasis-a.
4.2. Pengolahan data4.2.1. Peran#angan ruang seretariat !urusan
4.2.1.1. Peran#angan produ
4.2.1.1.1. $e!a omputer
ada praktikum kali ini produk yang akan diran&ang yaitu produk me(a komputer. gar dapatmenghasilkan produk me(a komputer yang ergonomis diperlukan pengukuran antrhopometri
dimensi tubuh berdiri dan dimensi tubuh duduk. engukuran yang dilakukan antara lain sebagai
berikut:a) angkauan tangan (t)
Tabel 3.4.3$ angkauan tangan
%
n1 2 3 4 5 6 7 8 9 10 ´ x
1 "",$ $,3 "4 4,3 5," ",4 !,$ $,3 "4,$ !0,! "1,#5
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2 5, 5 5 "#,! 3," 5,4 "5,3 !0,# ", !0 "0,$
3 "0 3," 3," "4,4 "!,3 "!,4 "!," ""," "$ $,$ "3,3!
4 "4,4 "1,$ "1,$ "3," ",# 3,4 3,# "!,# "5,! "0,5"
5 "3," 1, 1, !0,1 ",5 $ "5 !1 ,3 5,3 "1,01
6 "!,1 1," 1," "",! 1,$ #,3 "1,! "1,3 !,! ",5 $,1$
7 $,1 $," $," ",5 "1 $,3 #,3 "",# "4 !,! $,!
8 !, "4,! "4,! "1, #,3 "#,! ",5 ",5 "3 "$,! "1,#"
9 "",5 ",$ ",$ $," "3, "0,1 "!," 1,4 "0,$ 5,$ "#,1
10 #,$ ",1 ",1 !0," $ ,1 "1,5 !0 5,$ #,# $,#5
∑ ´ x "0!,$
1) %enghitung rararata subgroup
x = ∑ xk
x = ∑ 708,9
10
x =70,89
#) %enghitung standar de9iasi
σ =√∑ ( x1−´ x )2
N
σ = √ (77,9-70,89 )2 + .... + (62,2 – 70,89 )2
100
σ = 5,79
3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup
σx =σ
√ k
σx =5,79
√ 10σx = 1,83
4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!
Z = 1 – [1 - β
2 ]Z = 1 – [1 – 0, 98
2 ]
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Z = 1 – 0, 01
Z = 0,99
Berdasarkan tabel distribusi normal, nilai yang paling mendekati untuk harga > ; 0,$$
adalah #,0 maka >t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
B= x+(Z t x σx )
B =70,89+(2,06 x 1,!3)
B ; "4,ii) Batas kontrol ba-ah BB)
BB= x-(Zt x σ x )
BB = x-(2,06 x 1,83)
BB ; ",1#
1 # 3 4 5 " ! $ 10".1#
!.1#
$.1#
"0.1#
"1.1#
"#.1#
"3.1#
"4.1#
"1.#5"0.$
"3.3!
"0.5""1.01
$.1$
$.!
"1.#"
"#.1
$.#5
U(i keseragaman data
A bar B BB # bar
<ambar 3.4.#. U(i keseragaman data
) U(i ke&ukupan data
N'=[
Zt
α √N∑ x i
2
- (∑ x i )2
∑ xi ]2
N'=[
2,06
0,04 √ 505115,83 - 502539,21
7089 ]2
2C ; 13,0eterangan: 2C 2, data &ukup
") Distribusi *rekuensi
i) angkauan +)
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+ ; DmaA Dmin
+ ; !1,00 E 1,40+ ; 1$,0
ii) Banyak kelas
; 1 F 3,3 log n ; 1 F 3,3 log 100
; ",Dibulatkan ke atas, sehingga ; $
iii) Gnter9al G)
I=R
I=19,60
9
I=2,18
Tabel 3.4.40. Distribusi *rekuensi
!) %ean
x =!"i.#$
!"i
x =7079,080
100
x =70,79
$) osisi median
osisi median ;1
2 x N
osisi median ;1
2 x100
osisi median ; 50
10) %edian
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x =%&$+ [1
2N- (*$-1 )
"$] x I
~x =67,935+[12
x 100- (35 )
51 ] x 2,18
x =67,935+[50-35
51 ]x 2,18
x = 70,05
11) %odus
x =%&$) +
[ ( "$) - "$)-1 )( "$)- "$)+1 )+ (" $) - "$)-1) ] x I
x = 67,935+[17-9
(17-10) + (17-9) ] x 2,18
x =67,935+[87+8 ] x 2,18
x =69,30
1#) ersentili) /etak 5
%tk 5 =I
100 N
%(tk 5 =5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada interval ke1
ii) /etak 50
%tk 50 =I
100 N
%(tk 50 =50
100 100
%tk 50 = 50
ersentil 50 terletak pada data ke50 pada interval ke4
iii) /etak $5
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%tk 95 =I
100 N
%tk 95 =95
100
100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada interval ke$
i9) 2ilai 5
5 = %&5 + I [I
100N - *k5-1
"5]
5 = 61,395, + 2,18
[5
100100 - 0
13 ]5 = 61,89
9) 2ilai 50
50 = %&50 + I [I
100N - *k50-1
" 50]
50 = 67,935 + 2,18 [50100
100 - 35
16 ]50 = 70,05
9i) 2ilai $5
95 = %&95 + I [I
100N - *k95-1
" 95]
95 = 78,835 + 2,18 [95
100100 - 91
9 ]50 = 80,01
b) +entang tangan rt)
Tabel 3.4.41. +entang tangan
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%
n1 2 3 4 5 6 7 8 9 10 ´ x
1 1!5," 153, 1!, 11," 14$, 1$#, 1", 15!, 150, 14," 1,1
2 1!",1 15$, 10, 10,4 10 153, 154," 1!1, 11 1$0,$ 1,$3
3 1"! 1$3, 1!", 1!4,$ 1!5 1"#, 1!4,$ 1"#, 15", 1$," 1"!,3
4 1!#, 154, 1$#, 1$4 1"3, 1", 1, 1"4 1"$, 15"," 1"5,05
5 1"4,3 10, 1$0, 10,$ 1$#, 15$, 1"4, 1!, 15, 1"1,$ 1"#,!
6 1"3," 153, 14", 1"3,$ 1!3 15, 1"", 11, 1", 1$1, 1"0,5
7 14",# 14!, 1!3, 1"5,# 155, 14!, 1!, 1!", 154 1",5 1,33
8 1"$ 15 1, 15$,$ 1!$, 13, 1!1 1"3 1$#, 1!3,4 1"4,4$
9 153,4 1$3, 1", 1!$,3 1!0, 1" 1!0,# 1!3 1!, 15$ 1"4,1!
10 1"!,1 1!4, 1!0, 10,3 1!1, 1", 1$3,4 15$, 1$0, 1$# 1"$,
∑ ´ x 1."#4,"3
1) %enghitung ratarata subgroup
=∑ i
k
=1726,73
10
= 172,67
#) %enghitung standar de9iasi
σ =√∑(i- )
2
N
= √(185,7 – 172,67 )
2 + (153,6 – 172,67 )
2 + / + (192,0 – 172,67 )
2
100
; 13,4"
3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup
σ =σ
√ k
σ =13,47
√ 10
σ = 4,26
4) 'itung nilai > tabel >t), dengan ? ; 0,$
Zt = Z[1-[1- β
2 ]]
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Zt = 1 [1- [1-0,96
2 ]]Zt = 0,98
>t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
= + (Zt σ )
= 172,67+ (2,06 4,26 )
= 1!1,45
ii) Batas kontrol ba-ah BB)
= - (Zt σ) = 172,67- (2,06 4,26 )
= 163,90
1 # 3 4 5 " ! $ 10155.00
10.00
15.00
1"0.00
1"5.00
1!0.00
1!5.00
1.11".$3
1"!.31"5.05
1"5.051"#.!0
1"0.50
1.33
1"4.4$1"5.1!
U(i keseragaman data
B BB bar
+ata sbgrup
<ambar 3.4.3. eta kontrol u(i keseragaman data +T
) U(i ke&ukupan data dengan H ; 0,04
N'=[
Zt
α √N∑ i
2- (∑ i )
2
∑ i]
2
N'=
[
2,06
0,04 √ 100(2999751,95) – (1726,73 )2
1726,73
]
2
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N'=[
2,06
0,04 √ 299975195 – 298159649,29
1726,73 ]2
N'=[
51,5 1347,401726,73 ]
2
N'= 40,20
eterangan: 2C 2, data &ukup
") Distribusi *rekuensii) angkauan +)
R = $- $I
R = 194 - 147,20
R = 46,80
ii) Banyak kelas )
= 1 + 3,3
= 1 + 3,3 90
= 7,4 7
iii) Interval G)
I =
R
I =46,80
7
I = 6,69
Tabel 3.4.4#. Tabel distribusi *rekuensi rentang tangan +T)
Interval &i ' ()( *)( ()B *)B )$ &i.)$
14",#0 153,!! ! ! 14",#0 153,!! 14",1$5 153,!!5 150,54 1#04,3#
153,!$ 10,5" 1$ #" 153,!$ 10,5" 153,!!5 10,5"5 15",#3 #$!",3"
10,5! 1",# 11 3! 10,5! 1",# 10,5"5 1",#5 13,$# 1!03,1#
1",#" 1"3,$5 1# 50 1",#" 1"3,$5 1",#5 1"3,$55 1"0,1 #04",3#
1"3,$ 1!0,4 1 1"3,$ 1!0,4 1"3,$55 1!0,45 1"",30 #!3,!0
1!0,5 1!",33 1" !3 1!0,5 1!",33 1!0,45 1!",335 1!3,$$ 31#",!3
1!",34 1$4,0# 1" 100 1!",34 1$4,0# 1!",335 1$4,0#5 1$0,! 3#41,5
+&i 100 +&i.)$ 1"#4!,3#
!) Mean
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Mean =∑ " i &
∑ " i
Mean =
17248,32
100
Mean = 172,48
$) %ediani) /etak median
/etak median =1
2 N
/etak median =1
2 100
/etak median
= 50
%edian terletak pada nilai ke50 pada Interval ke4
ii) 2ilai median
%edian = %&$+ [1
2N - *k$-1
"$]×I
%edian = 167,26 + [50 - 38
12 ]× 6,69
%edian = 167,26 +
[1212 ]
×6,69
%edian = 167,26 + 6,69
%edian = 1"3,$5
10) %odus
%odus = %&$ + [
"$ - "$-1
( "$ - "$-1 )+( "$ - "$+1 ) ]× I
%odus = 167,26 + [
19 - 8(19 - 8 ) + (19 - 11) ] × ,$
%odus = 167,26 + [11
11 + 8 ] × ,$
%odus = 167,26 + [11
19 ]× ,$
%odus = 167,26 + 3,87
%odus = 171,14
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11) ersentili) /etak 5
%tk 5 =I
100 N
%(tk 5 =5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada Interval ke1
ii) /etak 50
%tk 50 =I
100 N
%(tk 50 =50100
100
%tk 50 = 50
ersentil 50 terletak pada data ke50 pada Interval ke4
iii) /etak $5
%tk 95 =I
100 N
%tk 95 =95
100 100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada Interval ke"
i9) 2ilai 5
5 = %&5 + [I
100N - *k5-1
"5]× I
5 = 147,195 +
[
5 - 0
8
] × ,$
5 = 147,195 + [58 ]× ,$
5 = 147,195 + 4,18
5 = 151,38
9) 2ilai 50
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50 = 167,26 + [I
100N - *k50-1
"50]×I
50 = 167,26 + [50 - 3812 ] × ,$
50 = 167,26 + [12
12 ]×6,69
50 = 167,26 + 6,69
50 = 173,96
9i) 2ilai $5
95 = %&95 +[I100
N - *k95-1
" 95 ]×I
95 = 187,33 + [95 - 100
17 ]×6,69
95 = 187,33 + [-517 ]×6,69
95 = 187,33 + (-1,97)
95 = 185,37
&) Tinggi siku duduk tsd)Tabel 3.4.43. Tabel data tinggi siku duduk tsd)
n
1 2 3 4 5 6 7 8 9 10 ´ x
11!,!
0
1",#
0
#,"
0
#3,!
0
#4,
0
1$,#
0
##,$
0
#0,3
0
##,3
0
#1,"
0#1,"5
2#,1
0
##,5
0
#5,"
0
##,"
0
1",4
0
##,0
0
##,$
0
##,3
0
1$,
0
##,1
0##,33
3
#",0
0
#,0
0
#3,1
0
1$,4
0
#,1
0
#1,"
0
#1,!
0
#3,$
0
1",$
0
#0,5
0 ##,"4
41!,#
0
#4,3
0
#4,0
0
1,5
0
#,0
0
#5,
0
#1,$
0
#1,#
0
#1,$
0
#3,4
0##,30
5#5,4
0
#5,0
0
##,0
0
1",#
0
1$,!
0
#3,!
0
#,#
0
#,3
0
1!,#
0
#4,$
0##,!!
6#0,3
0
#5,$
0
1$,"
0
##,0
0
1$,#
0
1,
0
##,4
0
#0,3
0
#4,#
0
#1,5
0#1,#1
7#4,#
0
#1,3
0
#5,5
0
#4,!
0
##,5
0
1",1
0
#1,"
0
#5,"
0
##,"
0
#3,
0##,$1
8#4,
0
#1,
0
#0,4
0
#4,#
0
#4,"
0
1$,1
0
#,
0
#0,$
0
1!,#
0
#4,4
0##,4"
9#,#
0
1","
0
1!,$
0
#0,$
0
1",#
0
1!,1
0
1$,4
0
##,"
0
#",0
0
#0,$
0#0,$0
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101,"
0
1!,
0
#0,3
0
1$,
0
#0,#
0
#0,$
0
#4,$
0
#3,5
0
#0,1
0
##,
0#0,"4
∑ ##0,#
3
1) %enghitung ratarata subgroup
=∑ i
k
=220,23
10
= 22,02
#) %enghitung standar de9iasi
σ =
√∑
(i- )2
N
= √(18,8 – 22,02 )2 + (17,2 – 22,02 )2 + / + ( 48,7 – 22,02)2
100
; #,!5
3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup
σ =σ
√ k
σ = 2,85√ 10
σ = 0,90
4) 'itung nilai > tabel >t), dengan ? ; 0,$
Zt = Z[1-[1- β
2 ]]Zt = 1 [1- [1-0,96
2 ]]Zt = 0,98
>t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
= + (Zt σ )
= 22,02+ (2,06 0,90 )
=#3,!"
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ii) Batas kontrol ba-ah BB)
= - (Zt σ)
= 22,02 - (2,06 0,90 )
= 20,16
) U(i ke&ukupan data dengan H ; 0,04
N'=[
Zt
α √N∑ i
2- (∑ i )
2
∑ i]
2
N
'
=
[
2,06
0,04 √ 100(49315,83) – (2202,30 )2
2202,30
]
2
N'=[
2,06
0,04 √ 493158300 – 4850125,29
2202,30 ]2
N'=[51,5 285,40
2202,30 ]2
N'= 44,5
eterangan: 2C 2, data &ukup
<ambar 3.4.4. eta kontrol u(i keseragaman data
") Distribusi *rekuensi
i) angkauan +)
R = $- $ I
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R = 26,60 - 16,50
R = 10,10
ii) Banyak kelas )
= 1 + 3,3
= 1 + 3,3 100
= 7,6
Dibulatkan men(adi ;!
iii) Interval G)
I =R
I =10,10
8
I = 1,3 0
Tabel 3.4.44. Tabel distribusi *rekuensi tinggi siku duduk TD)
Interval &i & ()( *)( ()B *)B )$ &i.)$
1,50 1",!# $ $ 1,50 1",!# 1,4$5 1",!#4 1",1 154,44
1",!3 1$,15 $ 1! 1",!3 1$,15 1",!#4 1$,153 1!,4$ 1,40
1$,1 #0,4! 15 33 1$,1 #0,4! 1$,153 #0,4!# 1$,!# #$",#
#0,4$ #1,!1 13 4 #0,4$ #1,!1 #0,4!# #1,!11 #1,15 #"4,$0
#1,!# #3,13 1! 4 #1,!# #3,13 #1,!11 #3,140 ##,4! 404,55
#3,14 #4,4 1# " #3,14 #4,4 #3,140 #4,4$ #3,!0 #!5,5
#4,4" #5,"$ 1# !! #4,4" #5,"$ #4,4$ #5,"$! #5,13 301,0
#5,!0 #",1# 1# 100 #5,!0 #",1# #5,"$! #",1#" #,4 31",55
+&i 100 +&i.)$ ##0#,34
!) Mean
Mean =∑ " i &
∑ " i
Mean =2202,34
100
Mean = 22,02
$) %edian
i) /etak median
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/etak median =1
2 N
/etak median =1
2
100
/etak median = 50
%edian terletak pada nilai ke50 pada Interval ke5
ii) 2ilai median
%edian = %&$()+ [1
2N - * k$()-1
"$()] I
%edian
= 21,81 +
[50 - 46
18
]× 1,30
%edian = 21,81 + [418 ]×1,3 0
%edian = 21,81 + 0,29
%edian = ##,11
10) %odus
%odus = %&$) + [
"$) - "$)-1
( "$)- "$)-1)+ ( "$) - "$)+1 ) ]× I
%odus = 21,81 + [18 - 13
(18 - 13 ) + (18 - 12 ) ] × 1,30
%odus = 21,81 + [5 5 + 6 ] × 1,30
%odus = 21,81 +
[
5
11
]× 1,30
%odus = 21,81 + 0,59
%odus = 22,41
11) ersentili) /etak 5
%tk 5 =I
100 N
%tk 5 =5
100
100
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%tk 5 = 5
ersentil 5 terletak pada data ke5 pada Interval ke1
ii) /etak 50
%tk 50 =I100
N
%tk 50 =50
100 100
%tk 50 = 50
ersentil 50 terletak pada data ke50 pada Interval ke5
iii) /etak $5
%tk 95 =I
100
N
%tk 95 =95
100 100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada Interval ke!
i9) 2ilai 5
5 = %&5 +
[
I
100N - *k5-1
"5
]× I
5 = 16,49 + [5 - 0
9 ] × 1,3 0
5 = 16,49 + [59 ]× 1,30
5 = 16,49 + 0,72
5 = 17,23
9) 2ilai 50
50 = %& 50 + [I
100N - * k50-1
"50] I
50 = 21,81 + [50 - 46
18 ] × 1,30
50 = 21,81 + [418 ]×1,3 0
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50 = 21,81+ 0,29
50 = 22,11
9i) 2ilai $5
95 = %&95 +[I
100N - * k95-1
" 95] I
95 = 25,79 + [95 - 88
12 ]×1,30
95 = 25,79+[712 ]×1,3 0
95 = 25,79 + 0,76
95 = 26,55
4.2.1.1.2. (emari abinet
%eran&ang lemari kabinet yang sesuai data anthropometri praktikan mata kuliah IJ. DataanthropometrI yang digunakan untuk meran&ang lemari kabinet adalah tinggi bahu berdiri, lebar bahu, (angkauan tangan, an(ang ari 3 dan lebar (ari #, 3, 4 dan 5. erhitungannya sebagai berikut:a) Tinggi bahu berdiri
Tabel 3.4.45. Data pengolahan anthropometri T'B
n
1 2 3 4 5 6 7 8 9 10 ´ x
115#,"
0
14#,3
0
14!,#
0
140,5
0
13!,"
0
14$,0
0
14#,3
0
1#$,
0
13,5
0
13",#
0 141,"0
2143,3
0
133,"
0
135,!
0
14,3
0
1#5,0
0
14$,"
0
150,5
0
13!,3
0
13,4
0
151,4
0 141,04
3153,5
0
150,5
0
153,#
0
13#,!
0
130,0
0
130,4
0
151,1
0
1#5,#
0
143,"
0
145,
0 141,0
4143,4
0
13,4
0
14#,
0
145,3
0
150,1
0
1#4,0
0
1#!,1
0
135,0
0
13,$
0
134,#
0 13",0
5141,$
0
1#5,$
0
1##,!
0
1#!,4
0
144,"
0
150,"
0
135,0
0
1#5,5
0
143,#
0
14,#
0 13,43
61#5,
0
135,0
0
154,#
0
130,4
0
143,4
0
1#4,$
0
145,5
0
150,"
0
1#,1
0
1#4,"
0 13,05
7140,"
0
13,4
0
143,$
0
14!,0
0
1##,
0
15#,5
0
154,$
0
155,0
0
133,#
0
13!,0
0 14#,5#
8133,!
0
14,
0
13$,#
0
133,0
0
153,"
0
1##,1
0
154,3
0
131,"
0
13$,3
0
1##,#
0 13",5$
9130,1
0
13!,5
0
14",1
0
1#4,0
0
145,0
0
14,#
0
151,$
0
1#$,$
0
133,4
0
143,
0 13!,$"
1
0130,4
0
154,$
0
145,0
0
13$,3
0
1#!,3
0
153,4
0
14,!
0
13",5
0
14!,"
0
13,4
0 14#,0"
∑ 13$5,5
"
1) %enghitung ratarata subgroup
=∑ i
k
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=1395,57
10
= 139,56
#) %enghitung standar de9iasi
σ =√∑(i- )
2
N
= √(152,70 – 139,56 )2 + (142,30 – 139,56 )2 + / + (136,40 – 139,56 )2
100
; $,5#
3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup
σ =σ√ k
σ =9,52
√ 10
σ = 3,01
4) 'itung nilai > tabel >t), dengan ? ; 0,$
Zt = Z
[1-
[
1- β
2
]]Zt = 1 [1- [1-0,96
2 ]]Zt = 0,98
>t ; #,05) U(i keseragaman data
i) Batas kontrol atas B)
= + (Zt σ)
= 139,56 + (2,06 3,01 ) = 145,"
ii) Batas kontrol ba-ah BB)
= - (Zt σ)
= 139,56 + (2,06 3,01 )
= 133,35
) U(i ke&ukupan data dengan H ; 0,04
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N'=[
Zt
α √N∑ i
2- (∑ i )
2
∑ i]
2
N'=[
2,06
0,04 √ 100(1956693) – (13955,7 )2
13955,7 ]2
N'=[
2,06
0,04 √ 195669261 – 194761562,5
13955,7 ]2
N'=[51,5 952,73
13955,7 ]2
N'= 12,36
eterangan: 2C 2, data &ukup
1#5
130
135
140
145
150
141." 141.04 141.
13".13.43 13.05
14#.5#
13".5$13!.$"
14#.0"
U(i keseragaman data
B BB bar
+ata subgroup
<ambar 3.4.5. eta kontrol u(i keseragaman data T'B
") Distribusi *rekuensii) angkauan +)
R = $- $I
R = 155,0 - 122,1
R = 32,9
ii) Banyak kelas )
= 1 + 3,3
= 1 + 3,3 100
= 7,6 K !
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iii) Interval G)
I =R
I = 32,98
I = 4,12
Tabel 3.4.4. Tabel distribusi *rekuensi T'B)
Interval &i ' ()( *)( ()B *)B )$ &i.)$
1##,10 1#,#1 14 14 1##,10 1#,#1 1##,0$5 1#,#15 1#4,1 1"3!,1"
1#,## 130,33 " #1 1#,## 130,33 1#,#15 130,335 1#!,#! !$",$3
130,34 134,45 11 3# 130,34 134,45 130,335 134,455 13#,40 145,35
134,4 13!,5" 15 4" 134,4 13!,5" 134,455 13!,5"5 13,5# #04","3
13!,5! 14#,$ 10 5" 13!,5! 14#,$ 13!,5"5 14#,$5 140,4 140,35
14#,"0 14,!1 1! "5 14#,"0 14,!1 14#,$5 14,!15 144," #05,5$
14,!# 150,$3 11 ! 14,!# 150,$3 14,!15 150,$35 14!,!! 13",3
150,$4 155,05 14 100 150,$4 155,05 150,$35 155,055 153,00 #141,$3
+&i 100 +&i.)$ 13$31,
!) Mean
Mean =∑ " i &
∑ " i
Mean = 13931,66100
Mean = 13$,3#
$) %ediani) /etak median
/etak median =1
2 N
/etak median =1
2 100
/etak median = 50
%edian terletak pada nilai ke50 pada Interval ke5
ii) 2ilai median
%edian = %&$+
1
2N - *k$-1
"$
× G
%edian = 138,53+ [50 - 47
10 ]× 4,12
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%edian = 138,53 + [310 ]×4,12
%edian = 138,53 + 1,24
%edian = 13$,""
10) %odus
%odus = %&$ + [
"$ - "$-1
( "$ - "$-1 )+( "$ - "$+1 ) ]× G
%odus = 142,65 + [18 - 10
(18 - 10 ) + (18 - 11 ) ] × 4,1#
%odus = 142,65 +
[8 8 + 7 ] × 4,1#
%odus = 142,65 + [815 ]× 4,1#
%odus = 142,65 + 2,20
%odus = 144,85
11) ersentili) /etak 5
%tk 5 =I100
N
%tk 5 =5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada Interval ke1
ii) /etak 50
%tk 50 =I
100 N
%tk 50 =50
100 100
%tk 50 = 50
ersentil 50 terletak pada data ke50 pada Interval ke5
iii) /etak $5
%tk 95 =I
100 N
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%tk 95 =95
100 100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada Interval ke!
i9) 2ilai 5
5 = %&5 + [I
100N - *k5-1
"5]× G
5 = 122,055 + [5 - 0
14 ] × 4,1#
5 = 122,055 +
[5
14
]×
4,1#
5 = 122,055 + 1,47
5 = 123,57
9) 2ilai 50
50 = %& 50 + [I
100N - *k50-1
"50]×I
50 = 138,535 + [50 - 47
10 ] × 4,1#
50 = 138,535 + [310 ]×4,12
50 = 138,535 + 1,24
50 = 139,81
9i) 2ilai $5
95 = %&95 +[I
100N - *k95-1
" 95]×I
95 = 150,895 + [95 - 86
14 ]×4,12
95 = 150,895 +[914 ]×4,12
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95 = 150,895 + 2,65
95 = 153,58
b) /ebar bahuTabel 3.4.4. Data pengolahan lebar bahu
n
1 2 3 4 5 6 7 8 9 10 ´ x
1 40,0 41,40 3$,0 40,0 4,#0 3!,30 35,40 44,"0 3",!0 3,50 40,11
2 45,10 3",#0 3,30 3,#0 40,50 3,00 3!,!0 44,10 41,!0 45,00 40,10
3 44,40 3$,#0 41,"0 3!,10 4#,10 3!,0 35,!0 34,0 41,!0 34,$0 3$,1#
4 3!,30 41,10 3,10 3!,50 3$,10 3,40 3",#0 34,00 41,30 3!,"0 3!,0"
5 34,$0 3,10 41,40 4#,0 45,30 3$,$0 3,10 40,!0 35,30 45,00 3$,"4
6 3,40 3!,!0 3",0 35,0 40,#0 44,30 44,50 3!,"0 3",!0 3$,30 3$,3#
7 45,00 44,"0 3$,40 3,!0 40,00 40,#0 44,!0 34,!0 40,!0 35,00 40,15
8 3!,!0 34,10 43,10 3$,50 40,"0 3,#0 34,"0 4#,$0 3!,00 35,30 3!,33
9 4,#0 34,!0 4,00 3,$0 40,"0 3,#0 4#,"0 43,50 41,"0 3",50 40,#
10 35,#0 35,10 3",10 4#,50 4,00 4#,#0 3$,40 45,!0 44,50 3",50 40,53
L ´ x 3$,0$
1) %enghitung ratarata subgroup
=∑ i
k
=396,09
10
= 39,61
#) %enghitung standar de9iasi
σ =√∑(i- )
2
N
=
√(40,60 – 39,61
)
2 +
(41,40 – 39,61
)
2 + / +
(37,50 – 39,61
)
2
100
; 3,4"
3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup
σ =σ
√ k
σ =3,47
√ 10
σ = 1,10
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4) 'itung nilai > tabel >t), dengan ? ; 0,$
Zt = Z[1-[1- β
2 ]]Zt = 1 [1- [1-0,96
2 ]]Zt = 0,98
>t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
= + (Zt σ )
= 39,60 + (2,06 1,10 ) = 41,!"
ii) Batas kontrol ba-ah BB)
= - (Zt σ)
= 39,60 + (2,06 1,10 )
= 37,35
) U(i ke&ukupan data dengan H ; 0,04
N'=[
Zt
α √N∑ i
2- (∑ i )
2
∑ i]
2
N'=[
2,06
0,04 √ 100(158093,8) – (3960,9 )2
3960,9 ]2
N'
=
[2,06
0,04 √ 15809383 – 15688729
3960,9
]
2
N'=[51,5 347,35
3960,9 ]2
N'= 20,42
eterangan: 2C 2, data &ukup
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1 # 3 4 5 " ! $ 103.00
3".00
3!.00
3$.00
40.00
41.00
4#.00
40.11 40.10
3$.1#
3!.0" 3!.0"
3$."43$.3#
40.15
3!.33
40.#
U(i keseragaman data
B BB bar
+ata subgroup
<ambar 3.4.. eta kontrol u(i keseragaman data /B
") Distribusi *rekuensii) angkauan +)
R = $- $I
R = 46,2 - 34
R = 12,2
ii) Banyak kelas )
= 1 + 3,3
= 1 + 3,3 100
= 7,6 K !
iii) Interval G)
I =R
I =12,2
8
I = 1,53
Tabel 3.4.4". Tabel distribusi *rekuensi lebar bahu
Interval &i ' ()( *)( ()B *)B )$ &i.)$
34,00 35,5# 14 14 34,00 35,5# 33,$$5 35,5#5 34," 4!,4
35,53 3",05 15 #$ 35,53 3",05 35,5#5 3",055 3,#$ 544,35
3",0 3!,5! 13 4# 3",0 3!,5! 3",055 3!,5!5 3",!# 4$1,
3!,5$ 40,11 15 5" 3!,5$ 40,11 3!,5!5 40,115 3$,35 5$0,#5
40,1# 41,4 13 "0 40,1# 41,4 40,115 41,45 40,!! 531,44
41,5 43,1" 11 !1 41,5 43,1" 41,45 43,1"5 4#,41 4,51
43,1! 44,"0 ! !$ 43,1! 44,"0 43,1"5 44,"05 43,$4 351,5#
44,"1 4,#3 11 100 44,"1 4,#3 44,"05 4,#35 45,4" 500,1"
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+&i 100 +&i.)$ 3$#,54
!) Mean
Mean =∑ " i &
∑ " i
Mean =3962,54
100
Mean = 39,63
$) %ediani) /etak median
/etak median =1
2 N
/etak median =1
2 100
/etak median = 50
%edian terletak pada nilai ke50 pada Interval ke4
ii) 2ilai median
%edian = %&$()+ [1
2N - *k$()-1
"$()]×I
%edian = 38,545 + [50 - 42
15 ]× 1,53
%edian = 38,545 + [815 ]×1,53
%edian = 38,545 + 0,82
%edian = 3$,3
10) %odus
%odus = %&$) + [
"$) - "$)-1
( "$)- "$)-1)+ ( "$) - "$)+1 ) ]× I
%odus = 38,545 + [15 - 13
(15 - 13) + (15 - 13 ) ] × 1,53
%odus = 38,545 + [2 2 + 2 ] × 1,53
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%odus = 38,545 + [24 ]× 1,53
%odus = 38,545 + 0,77
%odus = 39,31
11) ersentili) /etak 5
%tk 5 =I
100 N
%tk 5 =5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada Interval ke1
ii) /etak 50
%tk 50 =I
100 N
%tk 50 =50
100 100
%tk 50 = 50
ersentil 50 terletak pada data ke50 pada Interval ke4
iii) /etak $5
%tk 95 =I
100 N
%tk 95 =95
100 100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada Interval ke!
i9) 2ilai 5
5 = %&5 + [I
100N - *k5-1
"5]× I
5 = 33,955 + [5 - 0
14 ] × 1,53
5 = 33,955 + [514 ]× 1,53
5 = 33,955 + 0,55
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5 = 34,50
9) 2ilai 50
50 = %& 50 + [I
100 N - *k50-1
"50 ]×I
50 = 38,545 + [50 - 42
15 ] × 1,53
50 = 38,545 + [815 ]×1,53
50 = 38,545 + 0,82
50 = 39,36
9i) 2ilai $5
95 = %&95 +[I
100N - *k95-1
" 95]×I
95 = 44,665 + [95 - 89
11 ]×1,53
95 = 44,665 + [611 ]×1,53
95 = 44,665 + 0,83
95 = 45,49
&) an(ang (ari 3Tabel 3.4.4". Data pengolahan anthropometrI 3
n
1 2 3 4 5 6 7 8 9 10 ´ x
1 !,#0 !,#0 !,$0 ",10 !,0 !,10 ",30 $,00 !,30 ",0 !,13
2 ",10 ",!0 !,$0 ",10 ",30 ",40 !,00 $,00 ",$0 !,!0 ",$3
3 ",#0 ",10 !,30 !,50 !,40 ",50 !,40 ",10 ",#0 ",#0 ",$
4 !,30 !,!0 !,30 !,!0 !,0 $,00 ",30 ",10 ",00 ",#0 !,04
5 ",$0 ","0 !,#0 ",$0 $,00 ",!0 ",0 !,40 ",10 ",#0 ",!!
6 !,00 !,0 ",50 ",!0 ",!0 !,0 !,30 !,!0 !,10 !,40 !,1$
7 ",30 ",00 !,"0 ",!0 !,0 ","0 ",$0 ",#0 ",10 ",30 ",
8",40 ",!0 ",0 !,0 !,50 ",#0 !,30 ",$0 !,50 ",40 ",$#
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9 !,30 ",0 ",0 ",50 !,40 !,#0 ",#0 ",!0 ",10 !,$0 ",!
10 !,10 !,$0 ",$0 ","0 !,10 !,00 ",00 ","0 !,10 ",30 ",!!
∑
"$,1!
1) %enghitung ratarata subgroup
=∑ i
k
=79,18
10
= 7,92
#) %enghitung standar de9iasi
σ =√∑ (i- )
2
N
= √(8,20 – 7,92 )2 + (8,20 – 7,92 )2 + / + (7,30 – 7,92 )2
100
; 0,5$
3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup
σ =σ
√ k
σ =0,59
√ 10
σ = 0,19
4) 'itung nilai > tabel >t), dengan ? ; 0,$
Zt = Z[1-[1- β
2 ]]
Zt = 1
[1-
[1-0,96
2 ]]Zt = 0,98
>t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
= + (Zt σ )
= 7,92 + (2,06 0,19 )
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= !,31
ii) Batas kontrol ba-ah BB)
= - (Zt σ)
= 7,92 - (2,06 0,19 )
= 7,53
) U(i ke&ukupan data dengan H ; 0,04
N'=[
Zt
α √N∑ i
2- (∑ i )
2
∑ i]
2
N'=[
2,060,04 √ 100(6305,32) – (791,80 )2
791,80 ]2
N'=[
2,06
0,04 √ 630532 – 626947,24
791,80 ]2
N'=[51,5 59,87
791,80 ]2
N'
= 15,16
eterangan: 2C 2, data &ukup
1 # 3 4 5 " ! $ 10".00".#0
".40
".0
".!0
!.00
!.#0
!.40
!.13
".$3
".$
!.04 !.04
".!!
!.1$
".
".$# ".!
U(i keseragaman data
B BB bar
+ata ,bgrup
<ambar 3.4.". eta kontrol u(i keseragaman data 3
") Distribusi *rekuensii) angkauan +)
R = $- $I
R = 9 - 7
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R = 2
ii) Banyak kelas )
= 1 + 3,3
= 1 + 3,3 100
= 7,6≈7
iii) Interval G)
I =R
I =2
7
I = 0,29
Tabel 3.4.4!. Tabel distribusi pan(ang (ari 3
Interval &i ' ()( *)( ()B *)B )$ &i.)$
",00 ",#! #0 #0 ",00 ",#! ,$$5 ",#!5 ",14 14#,!0
",#$ ",5" 1# 3# ",#$ ",5" ",#!5 ",5"5 ",43 !$,1
",5! ",! 1 4! ",5! ",! ",5"5 ",!5 ","# 1#3,5#
",!" !,15 14 # ",!" !,15 ",!5 !,155 !,01 11#,14
!,1 !,44 1 "! !,1 !,44 !,155 !,445 !,30 13#,!0
!,45 !,"3 10 !! !,45 !,"3 !,445 !,"35 !,5$ !5,$0
!,"4 $,0# 1# 100 !,"4 $,0# !,"35 $,0#5 !,!! 10,5
+&i 100 +&i.)$ "$#,!!
8) Mean
Mean =∑ " i &
∑ " i
Mean =792,88
100
Mean = 7,93
$) %ediani) /etak median
/etak median =1
2 N
/etak median =1
2 100
/etak median = 50
%edian terletak pada nilai ke50 pada inter9al ke4ii) 2ilai median
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%edian = %&$()+ [1
2N - *k$()-1
"$()]×I
%edian = 7,87 + [50 - 4814 ]× 0,29
%edian = 7,87 + [214 ]×0,29
%edian = 7,87 + 0,04
%edian = ",$1
10) %odus
%odus = %&$ +
["$ - "$-1
( "$- "$-1 )+ ( "$ - "$+1 ) ]× I
%odus = 6,99 + [20 - 0
(20 - 0 ) + (20 - 12 ) ] × 0,#$
%odus = 6,99 + [20
20 + 8 ] × 0,#$
%odus = 6,99 + [20
28 ]× 0,#$
%odus = 6,99 + 0,20
%odus = 7,20
11) ersentili) /etak 5
%tk 5 =I
100 N
%tk 5 =5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada inter9al ke1
ii) /etak 50
%tk 50 =I
100 N
%tk 50 =50
100 100
%tk 50 = 50
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ersentil 50 terletak pada data ke50 pada inter9al ke4
iii) /etak $5
%tk 95 =I
100 N
%tk 95 =95
100 100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada inter9al ke"
i9) 2ilai 5
5 = %&5 +
[
I
100N - *k5-1
"5
]× I
5 = 6,99 + [5 - 0
20 ] × 0,#$
5 = 6,99 + [520 ]× 0,#$
5 = 6,99 + 0,07
5 = 7,06
9) 2ilai 50
50 = %& 50 + [I
100N - *k50-1
"50]×I
50 = 7,87 + [50 - 48
14 ] × 0,#$
50 = 7,87+ [214 ]×0,29
50 = 7,87 + (0.04)
50 = 7,91
9i) 2ilai $5
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95 = %&95 +[I
100N - *k95-1
" 95]×I
95 = 8,74 + [95 - 8812 ]×0,29
95 = 8,74 +[712 ]×0,29
95 = 8,74 + 0,16
95 = 8,90
d) /ebar (ari #, 3, 4 dan 5
Tabel 3.4.4$. Data pengolahan anthropometri lebar (ari #,3,4,5 n
1 2 3 4 5 6 7 8 9 10 ´ x
1 ,4 , ,! ",1 ",! ",$ ,4 ,3 ,3 ,3 ,"$
2 ",5 ,5 ,# ",3 ",1 ,0 ", ",$ ", ,$ ",0
3 ",5 , ",4 ",1 ",# ,# ",4 "," ",3 ,5 ",0$
4 ", ,0 ",3 ",! ",0 ",3 , "," ,# ",1 ",0
5 "," ",# ",0 ," , ",$ ,$ ,1 ,! ", ",05
6 ",3 "," ",1 ",# ",$ ",3 ",0 ",4 ,$ ,0 ",1!
7 ," ,# ",# ,4 ,0 ",! ,$ ",# ,! ," ,"$
8 ",$ ,! ,5 ,0 ," , ",1 ," , ",! ,!"
9 ", ",! ,5 ,4 ,5 ",3 ,5 ",! ,3 ,3 ,$0
10 ",! , ," ,# ",4 ,1 , ," "," ," ,!5
∑ $.4
1) %enghitung ratarata subgroup
= ∑ ik
=69,64
10
= 6,97
#) %enghitung standar de9iasi
σ =√∑(i- )
2
N
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= √(6,40 – 6,97 )
2 + (6,60 – 6,97 )
2 + / + (6,70 – 6,97 )
2
100
; 0,5"
3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup
σ =σ
√ k
σ =0,57
√ 10
σ = 0,18
4) 'itung nilai > tabel >t), dengan ? ; 0,$
Zt = Z
[1-
[1- β2 ]]
Zt = 1 [1- [1-0,96
2 ]]Zt = 0,98
>t ; #,0
5) %enentukan B dan BB untuk u(i keseragaman data
i) Batas kontrol atas B)
= + (Zt σ) = 6,97 + (2,06 0,18 )
= ",33
ii) Batas kontrol ba-ah BB)
= - (Zt σ)
= 6,97 + (2,06 0,18 )
= 6,56
) U(i ke&ukupan data dengan H ; 0,04
N'=[
Zt
α √N∑ i
2- (∑ i )
2
∑ i]
2
N'=[
2,06
0,04 √ 100(4881,82) – (696,4 )2
696,4 ]2
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N'=[
2,06
0,04 √ 488182 – 484973
696,4 ]2
N'=[
51,5 56,65696,4 ]
2
N'= 17,55
1 # 3 4 5 " ! $ 10.#0
.40.0
.!0
".00
".#0
".40
."$
".0 ".0$ ".0 ".05".1!
."$.!" .$0 .!5
U(i keseragaman data
B BB bar
+ata subgroup
eterangan: 2C 2, data &ukup
<ambar 3.4.!. eta kontrol u(i keseragaman data lebar (ari #, 3, 4 dan 5
") Distribusi *rekuensii) angkauan +)
R = $- $I
R = 7,90 - 6,00
R = 1,$0
ii) Banyak kelas )
= 1 + 3,3
= 1 + 3,3 100
; ", 8
iii) Interval G)
I =R
I =1,9
8
I = 0,24
Tabel 3.4.50. Tabel distribusi *rekuensi lebar (ari #, 3, 4 dan 5
Interval &i ' ()( *)( ()B *)B )$ 'i.)$
,00 ,#3 1# 1# ,00 ,#3 5,$$5 ,#35 ,1# "3,3!
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,#4 ,4" $ #1 ,#4 ,4" ,#35 ,4"5 ,3 5",#0
,4! ,"1 ## 43 ,4! ,"1 ,4"5 ,"15 ,0 145,0$
,"# ,$5 ! 51 ,"# ,$5 ,"15 ,$55 ,!4 54,!
,$ ",1$ $ 0 ,$ ",1$ ,$55 ",1$5 ",0! 3,!
",#0 ",43 1 " ",#0 ",43 ",1$5 ",435 ",3# 11",04
",44 "," " !3 ",44 "," ",435 ","5 ",5 5#,!$
",! ",$1 1" 100 ",! ",$1 ","5 ",$15 ",!0 13#,5#
+&i 100 +&i.)$ $,4
!) Mean
Mean =∑ " i &
∑ " i
Mean =696,46
100
Mean = 6,96
$) %ediani) /etak median
/etak median =1
2 N
/etak median =1
2 100
/etak median = 50
%edian terletak pada nilai ke50 pada Interval ke4
ii) 2ilai median
%edian = %&$()+ [1
2N - *k$()-1
"$()]×I
%edian = 6,715 + [50 - 43
8 ]× 0,24
%edian = 6,715 + [78 ]×0,24
%edian = 6,715 + 0,21
%edian = ,$3
10) %odus
%odus = %&$ + [
"$ - "$-1
( "$- "$-1 )+ ( "$ - "$+1 ) ]× I
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%odus = 6,475 + [22 - 9
(22 - 9 ) + (22 - 8 ) ] × 0,#4
%odus = 6,475 +
[13
13 + 14
] ×
0,#4
%odus = 6,475 + [13
27 ]× 0,#4
%odus = 6,475 + 0,11
%odus = 6,59
11) ersentili) /etak 5
%tk 5 =
I
100 N
%tk 5 =5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada Interval ke1
ii) /etak 50
%tk 50 =I
100 N
%tk 50 =50
100 100
%tk 50 = 50
ersentil 50 terletak pada data ke50 pada Interval ke4
iii) /etak $5
%tk 95 =I
100 N
%tk 95 =
95
100 100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada Interval ke!
i9) 2ilai 5
5 = %&5 + [I
100N - *k5-1
"5]× I
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5 = 5,995 + [5 - 0
12 ] × 0,#4
5 = 5,995 +
[5
12
]×
0,#4
5 = 5,995 + 0,10
5 = 6,10
9) 2ilai 50
50 = %& 50 + [I
100N - *k50-1
"50]×I
50 = 6,715 + [50 - 438 ] × 0,#4
50 = 6,715 + [78 ]×0,24
50 = 6,715 + 0,21
50 = 6,93
9i) 2ilai $5
95 = %&95 +[I
100N - *k95-1
" 95]×I
95 = 7,675 + [95 - 83
17 ]×0,24
95 = 7,675 + [12
17 ]×0,24
95 = 7,675 + 0,17
95 = 7,84
e) angkauan tanganTabel 3.4.51. angkauan tangan
%
n1 2 3 4 5 6 7 8 9 10 ´ x
1 "1,#5 $,3 "4,0 4,3 5," ",4 !,$ $,3 "4,$ !0,! "1,#5
2 "0," 5,0 5,0 "#,! 3," 5,4 "5,3 !0,# ", !0,0 "0,$
3 "4,11 3," 3," "4,4 "!,3 "!,4 "!," ""," "$,0 $,$ "3,3!
4 "0,$4 "1,$ "1,$ "3," ",# ,0 3,4 3,# "!,# "5,! "0,5"
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5 "1,"# 1, 1, !0,1 ",5 $,0 "5,0 !1,0 ,3 5,3 "1,01
6 "0,1 1," 1," "",! 1,$ #,3 "1,! "1,3 !,! ",5 $,1$
7 $,$ $," $," ",5 "1,0 $,3 #,3 "",# "4,0 !,! $,!
8 "1,#3 "4,! "4,! "1, #,3 "#,! ",5 ",5 "3,0 "$,! "1,#"
9 "1,0$ ",$ ",$ $," "3, "0,1 "!," 1,4 "0,$ 5,$ "#,1
10 $,"$ ",1 ",1 !0," $,0 ,1 "1,5 !0,0 5,$ #,# $,#5
+ ´ X "11,4
1) %enghitung rararata subgroup
= ∑ k
= ∑ 711,46
10
=71,15
#) %enghitung standar de9iasi
σ =√∑ ( x1−´ x )2
N
σ = √ (77,9-70,89 )2 + .... + (62,2 – 70,89 )2
100
σ = 5,63
3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup
σ =σ
√ k
σ =5,63
√ 10
σ = 1,78
4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!
Zt = Z[1-[1- β
2 ]]Zt = 1 [1- [1-0,96
2 ]]Zt = 0,98
>t ; #,0
5) U(i keseragaman data
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i) Batas kontrol atas B)
B= x+(Z t x σx )
B = 71,15+(2,06 x 1,"!)
B ; "4,!1ii) Batas kontrol ba-ah BB)
BB= x-(Zt x σ x )
BB = 71,15-(2,06 x 1,78)
BB ; ",4!
) U(i ke&ukupan data
N'=[
Zt
α √N∑ i
2- (∑ i )
2
∑i]
2
N'=[
2,06
0,04 √ 50(5115,83) - (502539,21)
7089 ]2
2C ; 1,0eterangan: 2C 2, data &ukup
1 # 3 4 5 " ! $ 10#.00
4.00
.00
!.00
"0.00
"#.00
"4.00
".00
"1.#5"0."
"4.11
"0.$4"1."#
"0.1$.$
"1.#3 "1.0$
$."$
U(i eseragaman data
B BB bar
+ata subgroup
<ambar 3.4.$. eta kontrol u(i keseragaman data (angkauan tangan
") Distribusi *rekuensii) angkauan +)
+ ; DmaM Dmin
+ ; !1,00 E 1,40+ ; 1$,0
ii) Banyak kelas ; 1 F 3,3 log n ; 1 F 3,3 log 100
; ", K $
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iii) Interval G)
I= R
K
I=19,60
9
I = 2,18
Tabel 3.4.5#. Tabel distribusi *rekuensi T
Interval &i ' ()( *)( ()B *)B )$ &i.)$
1,40 3,5" 13 13 1,40 3,5" 1,3$5 3,5"5 #,4$ !1#,31
3,5! 5,"5 " #0 3,5! 5,"5 3,5"5 5,"55 4," 45#,
5," ",$3 1# 3# 5," ",$3 5,"55 ",$35 ,!5 !0#,14
",$4 "0,11 15 4" ",$4 "0,11 ",$35 "0,115 $,03 1035,3!
"0,1# "#,#$ 10 5" "0,1# "#,#$ "0,115 "#,#$5 "1,#1 "1#,05
"#,30 "4,4" 1# $ "#,30 "4,4" "#,#$5 "4,4"5 "3,3$ !!0,#
"4,4! ",5 10 "$ "4,4! ",5 "4,4"5 ",55 "5,5" "55,5
", "!,!3 1# $1 ", "!,!3 ",55 "!,!35 "","5 $3#,$4
"!,!4 !1,01 $ 100 "!,!4 !1,01 "!,!35 !1,015 "$,$3 "1$,33
+&i 100 +&i.)$ 4$5,15
!) Meanx =
!"i.#$
!"i
x =4695,15
100
x =46,95
$) %ediani) /etak median
/etak median ;
1
2 x N
/etak median ;1
2 x 100
/etak median ; 50
ii) 2ilai median
i=%&$ + [1
2N- (*$-1)
"$] I
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i=70,115+ [1
2 100- 47
10 ] 2,18
i=70,115+ [50-4710 ] 2,18
i=70,115+ [ 0,26] 2,18
i=70,115+ 0,654
i=70,77
10) %odus
)=%& $) + [ ("$) -" $)-1)("$) - "$)+1 ) + ("$) - "$)-1 ) ] I
=67,935+ [15-12
(15-10 )+ (15-12) ] 2,18
=67,935+ [ 3
5+3 ] 2,18
=67,935+0,82
=¿ !,"5
11) ersentili)/etak 5
%tk 5 =I
100 N
%tk 5 =5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada Interval ke1
ii) /etak 50
%tk 50 =I100
N
%tk 50 =50
100 100
%tk 50 = 50
ersentil 50 terletak pada data ke50 pada Interval ke4
iii) /etak $5
%tk 95 =I
100 N
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%tk 95 =95
100 100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada Interval ke!
i9) 2ilai 5
5 = %&5 + [I
100N - *k5-1
"5] I
5 = 61,395 + [5
100100 - 0
10 ] 2,45
5 = #,#3
9) 2ilai 50
50 = %& 50+
I
100N - *k50-1
"50
I
50 = 70,115 + 2,18 [50
100100 - 36
17 ] 2,18
50 = 70,77
9i) 2ilai $5
95 = %&95 + [I
100N - *k95-1
" 95] I
95
= 78,335+
[
95
100100 - 89
11
] 2,18
50 = 79,80
4.2.1.1.3. Pintu
%eran&ang pintu yang sesuai data anthropometri praktikan mata kuliah IJ. DataanthropometrI yang digunakan untuk meran&ang pintu adalah tinggi badan berdiri tbb), tinggi bahu berdiri thb), lebar bahu, lebar (ari #,3,4,5 dan pan(ang (ari 3. /alu perhitungannya adalahsebagai berikut:a, Tinggi badan berdiri tbb)
Tabel 3.4.53. Data antrhopometri tinggi badan berdiri tbb)
-o 1 2 3 4 5 6 7 8 9 10 ×
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1154,00 1!,40 151,0 15,"0 1"#,0 1"1,"0 14!,10 14,!0 1",00 15#,10 11,0
21"$,#0 1#,10 1"!,$0 1"5,0 15#,$0 1"!,40 1"!,!0 10,00 155,10 154,30 1",53
31",#0 1$,30 1!0,"0 1!1,10 1"3,$0 1"0,"0 1$,00 1!0,30 1"3,00 154,!0 1"#,00
4 10,50 1"5,00 153,00 13,30 151,40 14!,#0 14$,$0 1"$,50 14,00 1"5,!0 1#,0
514",10 15,00 15$,#0 155,10 1"#,$0 1!3,50 154,0 154,"0 15!,!0 1#,!0 11,3"
61"5,50 151,"0 10,$0 1"4,0 15!,10 1#,"0 153,0 15#,50 155,0 15,10 10,13
71!#,!0 1!,40 1"3,50 1!0,"0 1"3,30 1"4,00 15$,40 1"$,40 11,#0 14",$0 1"0,0
81"4,40 1#,40 150,0 1"3,00 15,#0 10,"0 1!3,50 1!3,#0 1!3,40 1!0,30 1"1,"
910,#0 1$,$0 1!0,$0 1!1,50 151,00 151,00 1"0,!0 1"1,$0 1"!,00 15,"0 1!,0$
+× 14$4,51
1) %enghitung rararata subgroup
= ∑ k
= ∑ 1494,51
9
= 1,0
#) %enghitung standar de9iasi
σ=
√∑ (x1 - x )
2
N
σ = √ (154,00- 14 9 ,45 )2 + .... + (165,70 – 149,45 )2
90
σ = 10,!5
3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup
σ =σ
√ k
σ = 10,85
√ 9
σ =3,#
4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!
Zt = Z[1-[1- β
2 ]]
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Zt = 1 [1- [1-0,96
2 ]]Zt = 0,98
>t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
B= x + ( Zt x σx )
B =166,09 + (2,06 x 3,#)
B ; 1"3,51ii) Batas kontrol ba-ah BB)
BB = x - ( Zt x σx )
BB = 166,09-(2,06 x 3,62)BB ; 15!,1
) U(i ke&ukupan data
N'=[
Zt
α √N∑ i
2- (∑ i )
2
∑i]
2
N'=
[
2,06
0,04 √ 90(24 .922. 328 ,59) - (223 .356 .014,01)
14 .945,10
]
2
2C ; 11,3
eterangan: 2C 2, data &ukup
1 # 3 4 5 " ! $150.00
155.00
10.00
15.00
1"0.00
1"5.00
11.0
1".53
1"#.00
1#.0 11.3"10.13
1"0.01"1."
1!.0$
U(i keseragaman data
B BB bar
+ata subgroup
<ambar 3.4.10. eta kontrol u(i keseragaman data TBB
") Distribusi *rekuensii) angkauan +)
+ ; DmaM Dmin
+ ; 1!3,50 E 14",10
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+ ; 3,40
ii) Banyak kelas ; 1 F 3,3 log n ; 1 F 3,3 log 100
; ",0 K $
iii) Interval G)
I =R
I=36,40
9
I = 4,05
Tabel 3.4.54. Tabel distrbusi *rekuensi TBB
Interval &i ' ()( *)( ()B *)B )$ 'i.)$
14",10 151,14 ! ! 14",10 151,14 14",0$5 151,145 14$,1# 11$#,$
151,15 155,1$ 15 #3 151,15 155,1$ 151,145 155,1$5 153,1" ##$",55
155,#0 15$,#4 5 #! 155,#0 15$,#4 155,1$5 15$,#45 15",## "!,10
15$,#5 13,#$ 11 3$ 15$,#5 13,#$ 15$,#45 13,#$5 11,#" 1""3,$"
13,30 1",34 $ 4! 13,30 1",34 13,#$5 1",345 15,3# 14!",!!
1",35 1"1,3$ " 55 1",35 1"1,3$ 1",345 1"1,3$5 1$,3" 11!5,5$
1"1,40 1"5,44 13 ! 1"1,40 1"5,44 1"1,3$5 1"5,445 1"3,4# ##54,4
1"5,45 1"$,4$ $ "" 1"5,45 1"$,4$ 1"5,445 1"$,4$5 1"",4" 15$",#3
1"$,50 1!3,54 13 $0 1"$,50 1!3,54 1"$,4$5 1!3,545 1!1,5# #35$,"
+&i $0 +&i.)$ 14$35,50
!) Mean
´ x=∑fi.cm
∑fi
´ x=14935,50
90
´ x= 15,$5
$) %ediani) /etak median
/etak median ;1
2 N
/etak median ;1
2 90
/etak median ; 45
ii) 2ilai median
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Median=%& $+ [1
2N- ( *$-1 )
"$] I
Median=163,295+ [12
90 - 39
9 ] 4,05
Median=163,295 + [0,66 ] 4,05
Median=163,295 + #,"0
Median= 11,$5
10)%odus
)=%& $) + [ (" $) - " $)-1 )("$) - "$)+1) + ("$) -"$)-1 ) ] I
=151,145 + [15 - 8
(15-8 )+ (15-10) ] 4,05
=151,145 + [ 7
7+5 ] 4,05
= 151,145 + #,3
=¿ 15#,!1
11)ersentili) /etak 5
%tk 5 =I
100 N
%tk 5 =5
100 90
%tk 5 =4,5
ersentil 5 terletak pada data ke5 pada Interval ke1
ii) /etak 50
%tk 50 =I
100 N
%tk 50 =50
100 90
%tk 50 = 45
ersentil 50 terletak pada data ke50 pada Interval ke4
iii) /etak $5
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%tk 95 =I
100 N
%tk 95 =95
100
90
%tk 95 = 85,5
ersentil $5 terletak pada data ke$5 pada Interval ke!
i9) 2ilai 5
5 = %&5 + I [I
100N - *k5-1
"5]
5 = 147,095 + 4,05
[5
10090 - 0
8 ]5 =14$,3
9) 2ilai 50
50 = %& 50+ I [I
100N - *k50-1
"50]
50 = 163,295 + 4,05 [50100
90 - 39
9 ]50 =
1!,#5
9i) 2ilai $5
95 = %&95 + I [I
100N - *k95-1
" 95]
95 = 179,495+ 4,05 [95
10090 - 77
13 ]50 =
1!5,10
b) Tinggi bahu berdiri thb)Tabel 3.4.55. data antrhopometri tinggi bahu berdiri thb)
n
1 2 3 4 5 6 7 8 9 10 ´ x
115#,"
0
14#,3
0
14!,#
0
140,5
0
13!,"
0
14$,0
0
14#,3
0
1#$,
0
13,5
0
13",#
0 141,"0
2 143,30
133,"0
135,!0
14,30
1#5,00
14$,"0
150,50
13!,30
13,40
151,40 141,04
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>t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
= + (Zt σ)
= 139,56 + (2,06 3,01 )
= 145,"
ii) Batas kontrol ba-ah BB)
= - (Zt σ)
= 139,56 + (2,06 3,01 )
= 133,35
) U(i ke&ukupan data dengan H ; 0,04
N'=[
Zt
α √N∑ i
2- (∑ i )
2
∑ i]
2
N'=[
2,06
0,04 √ 100(1956693) – (13955,7 )2
13955,7 ]2
N'=
[2,06
0,04 √ 195669261 – 194761562,5
13955,7 ]2
N'=[51,5 952,73
13955,7 ]2
N'= 12,36
eterangan: 2C 2, data &ukup
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1#5
130
135
140
145
150
141." 141.04 141.
13".13.43 13.05
14#.5#
13".5$13!.$"
14#.0"
U(i keseragaman data
B BB bar
+ata subgroup
<ambar 3.4.11. peta kontrol u(i keseragaman data thb
") Distribusi *rekuensi
i) angkauan +)R = $- $i
R = 155,0 - 122,1
R = 32,9
ii) Banyak kelas )
= 1 + 3,3
= 1 + 3,3 100
= 7,6 K !
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iii) Interval G)
I =R
I = 32,98
I = 4,12
Tabel 3.4.5. Tabel distribusi *rekuensi T'B)
Interval &i ' ()( *)( ()B *)B )$ &i.)$
1##,10 1#,#1 14 14 1##,10 1#,#1 1##,0$5 1#,#15 1#4,1 1"3!,1"
1#,## 130,33 " #1 1#,## 130,33 1#,#15 130,335 1#!,#! !$",$3
130,34 134,45 11 3# 130,34 134,45 130,335 134,455 13#,40 145,35
134,4 13!,5"15 4" 134,4 13!,5" 134,455 13!,5"5 13,5# #04","3
13!,5! 14#,$ 10 5" 13!,5! 14#,$ 13!,5"5 14#,$5 140,4 140,35
14#,"0 14,!1 1! "5 14#,"0 14,!1 14#,$5 14,!15 144," #05,5$
14,!# 150,$3 11 ! 14,!# 150,$3 14,!15 150,$35 14!,!! 13",3
150,$4 155,05 14 100 150,$4 155,05 150,$35 155,055 153,00 #141,$3
+&i 100 +&i.)$ 13$31,
!) Mean
Mean =∑ " i &
∑ " i
Mean = 13931,66100
Mean = 13$,3#
$) %ediani) /etak median
/etak median =1
2 N
/etak median =1
2 100
/etak median = 50
%edian terletak pada nilai ke50 pada Interval ke5
ii) 2ilai median
%edian = %&$+ [1
2N - *k$-1
"$]× G
%edian = 138,53+ [50 - 47
10 ]× 4,12
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%edian = 138,53 + [310 ]×4,12
%edian = 138,53 + 1,24
%edian = 13$,""
10)%odus
%odus = %&$ + [
"$ - "$-1
( "$ - "$-1 )+( "$ - "$+1 ) ]× G
%odus = 142,65 + [18 - 10
(18 - 10 ) + (18 - 11 ) ] × 4,1#
%odus = 142,65 +
[8
8 + 7
] × 4,1#
%odus = 142,65 + [815 ]× 4,1#
%odus = 142,65 + 2,20
%odus = 144,85
11)ersentili) /etak 5
%tk 5 =I
100 N
%(tk 5 =5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada Interval ke1
ii) /etak 50
%tk 50 =I
100 N
%(tk 50 =50
100 100
%tk 50 = 50
ersentil 50 terletak pada data ke50 pada Interval ke5
iii) /etak $5
%tk 95 =I
100 N
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%tk 95 =95
100 100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada Interval ke!
i9) 2ilai 5
5 = %&5 + [I
100N - *k5-1
"5]× G
5 = 122,055 + [5 - 0
14 ] × 4,1#
5 = 122,055 +
[5
14
]×
4,1#
5 = 122,055 + 1,47
5 = 123,57
9) 2ilai 50
50 = %& 50 + [I
100N - *k50-1
"50]×I
50 = 138,535 + [50 - 47
10 ] × 4,1#
50 = 138,535 + [310 ]×4,12
50 = 138,535 + 1,24
50 = 139,81
9i) 2ilai $5
95 = %&95 +[I
100N - *k95-1
" 95 ]×I
95 = 150,895 + [95 - 86
14 ]×4,12
95 = 150,895 +[914 ]×4,12
95 = 150,895 + 2,65
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95 = 153,58
&) /ebar bahu lb)Tabel 3.4.5" Data anthropometri lebar bahu lb)
n
1 2 3 4 5 6 7 8 9 10 ´ x
1 40,0 41,40 3$,0 40,0 4,#0 3!,30 35,40 44,"0 3",!0 3,50 40,11
2 45,10 3",#0 3,30 3,#0 40,50 3,00 3!,!0 44,10 41,!0 45,00 40,10
3 44,40 3$,#0 41,"0 3!,10 4#,10 3!,0 35,!0 34,0 41,!0 34,$0 3$,1#
4 3!,30 41,10 3,10 3!,50 3$,10 3,40 3",#0 34,00 41,30 3!,"0 3!,0"
5 34,$0 3,10 41,40 4#,0 45,30 3$,$0 3,10 40,!0 35,30 45,00 3$,"4
6 3,40 3!,!0 3",0 35,0 40,#0 44,30 44,50 3!,"0 3",!0 3$,30 3$,3#
7 45,00 44,"0 3$,40 3,!0 40,00 40,#0 44,!0 34,!0 40,!0 35,00 40,15
8 3!,!0 34,10 43,10 3$,50 40,"0 3,#0 34,"0 4#,$0 3!,00 35,30 3!,33
9 4,#0 34,!0 4,00 3,$0 40,"0 3,#0 4#,"0 43,50 41,"0 3",50 40,#
10 35,#0 35,10 3",10 4#,50 4,00 4#,#0 3$,40 45,!0 44,50 3",50 40,53
L ´ x 3$,0$
1) %enghitung ratarata subgroup
=∑ i
k
=
396,09
10
= 39,61
#) %enghitung standar de9iasi
σ =√∑(i- )
2
N
= √(40,60 – 39,61 )2 + (41,40 – 39,61 )2 + / + (37,50 – 39,61 )2
100
; 3,4"
3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup
σ =σ
√ k
σ =3,47
√ 10
σ = 1,10
4) 'itung nilai > tabel >t), dengan ? ; 0,$
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Zt = Z[1-[1- β
2 ]]
Zt = 1
[1-
[1-0,96
2
]]Zt = 0,98
>t ; #,0
5) U(i keseragaman dataa) Batas kontrol atas B)
= + (Zt σ )
= 39,60 + (2,06 1,10 )
= 41,!"
b) Batas kontrol ba-ah BB)
= - (Zt σ)
= 39,60 + (2,06 1,10 )
= 37,35
) U(i ke&ukupan data dengan H ; 0,04
N
'
=
[
Zt
α √N∑ i
2- (∑ i )
2
∑ i
]
2
N'=[
2,06
0,04 √ 100(158093,8) – (3960,9 )2
3960,9 ]2
N'=[
2,06
0,04 √ 15809383 – 15688729
3960,9 ]2
N' =[51,5 347,353960,9 ]
2
N'= 20,42
eterangan: 2C 2, data &ukup
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1 # 3 4 5 " ! $ 103.00
3".00
3!.00
3$.00
40.00
41.00
4#.00
40.11 40.10
3$.1#
3!.0" 3!.0"
3$."43$.3#
40.15
3!.33
40.#
U(i keseragaman data
B BB bar
+ata subgroup
<ambar 3.4.1#. eta kontrol u(i keseragaman data /B
") Distribusi *rekuensia) angkauan +)
R = $- $I
R = 46,2 - 34
R = 12,2
b) Banyak kelas )
= 1 + 3,3
= 1 + 3,3 100
= 7,6 K !
&) Interval G)
I =R
I =12,2
8
I = 1,53
Tabel 3.4.5!. Tabel distribusi *rekuensi lebar bahu
Interval &i ' ()( *)( ()B *)B )$ &i.)$
34,00 35,5# 14 14 34,00 35,5# 33,$$5 35,5#5 34," 4!,4
35,53 3",05 15 #$ 35,53 3",05 35,5#5 3",055 3,#$ 544,35
3",0 3!,5! 13 4# 3",0 3!,5! 3",055 3!,5!5 3",!# 4$1,
3!,5$ 40,11 15 5" 3!,5$ 40,11 3!,5!5 40,115 3$,35 5$0,#5
40,1# 41,4 13 "0 40,1# 41,4 40,115 41,45 40,!! 531,44
41,5 43,1" 11 !1 41,5 43,1" 41,45 43,1"5 4#,41 4,51
43,1! 44,"0 ! !$ 43,1! 44,"0 43,1"5 44,"05 43,$4 351,5#
44,"1 4,#3 11 100 44,"1 4,#3 44,"05 4,#35 45,4" 500,1"
+&i 100 +&i.)$ 3$#,54
!) Mean
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Mean =∑ " i &
∑ " i
Mean =
3962,54
100
Mean = 39,63
$) %ediani) /etak median
/etak median =1
2 N
/etak median =1
2 100
/etak median = 50
%edian terletak pada nilai ke50 pada Interval ke4
ii) 2ilai median
%edian = %&$()+ [1
2N - *k$()-1
"$()]×I
%edian = 38,545 + [50 - 42
15 ]× 1,53
%edian = 38,545 + [815 ]×1,53
%edian = 38,545 + 0,82
%edian = 3$,3
iii) %odus
%odus = %&$+ [
"$ - "$-1
( "$ - "$-1)+ ( "$ - "$+1 ) ]× I
%odus = 38,545 + [15 - 13
(15 - 13) + (15 - 13 ) ] × 1,53
%odus = 38,545 + [2 2 + 2 ] × 1,53
%odus = 38,545 + [24 ]× 1,53
%odus = 38,545 + 0,77
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%odus = 39,31
10) ersentili) /etak 5
%tk 5 = I100
N
%tk 5 =5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada Interval ke1
ii) /etak 50
%tk 50 =I
100
N
%tk 50 =50
100 100
%tk 50 = 50
ersentil 50 terletak pada data ke50 pada Interval ke4
iii) /etak $5
%tk 95 =I
100 N
%tk 95 =95100
100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada Interval ke!
i9) 2ilai 5
5 = %&5 + [I
100N - *k5-1
"5]× I
5 = 33,955 + [5 - 014 ] × 1,53
5 = 33,955 + [514 ]× 1,53
5 = 33,955 + 0,55
5 = 34,50
9) 2ilai 50
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50 = %& 50 + [I
100N - *k50-1
"50]×I
50 = 38,545 + [50 - 4215 ] × 1,53
50 = 38,545 + [815 ]×1,53
50 = 38,545 + 0,82
50 = 39,36
9i) 2ilai $5
95 = %&95 +[I100
N - *k95-1
" 95 ]×I
95 = 44,665 + [95 - 89
11 ]×1,53
95 = 44,665 + [611 ]×1,53
95 = 44,665 + 0,83
95 = 45,49
d) an(ang (ari 3Tabel 3.4.5$. Data antrhopometri pan(ang (ari 3
n
1 2 3 4 5 6 7 8 9 10 ´ x
1 !,#0 !,#0 !,$0 ",10 !,0 !,10 ",30 $,00 !,30 ",0 !,13
2 ",10 ",!0 !,$0 ",10 ",30 ",40 !,00 $,00 ",$0 !,!0 ",$3
3 ",#0 ",10 !,30 !,50 !,40 ",50 !,40 ",10 ",#0 ",#0 ",$
4 !,30 !,!0 !,30 !,!0 !,0 $,00 ",30 ",10 ",00 ",#0 !,04
5 ",$0 ","0 !,#0 ",$0 $,00 ",!0 ",0 !,40 ",10 ",#0 ",!!
6 !,00 !,0 ",50 ",!0 ",!0 !,0 !,30 !,!0 !,10 !,40 !,1$
7 ",30 ",00 !,"0 ",!0 !,0 ","0 ",$0 ",#0 ",10 ",30 ",
8 ",40 ",!0 ",0 !,0 !,50 ",#0 !,30 ",$0 !,50 ",40 ",$#
9 !,30 ",0 ",0 ",50 !,40 !,#0 ",#0 ",!0 ",10 !,$0 ",!
10 !,10 !,$0 ",$0 ","0 !,10 !,00 ",00 ","0 !,10 ",30 ",!!
∑ ´ "$,1!
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1) %enghitung ratarata subgroup
=∑ i
k
=79,18
10
= 7,92
#) %enghitung standar de9iasi
σ =√∑(i- )
2
N
=
√
(8,20 – 7,92 )2 + (8,20 – 7,92 )2 + / + (7,30 – 7,92 )2
100
; 0,5$
3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup
σ =σ
√ k
σ =0,59
√ 10
σ = 0,19
4) 'itung nilai > tabel >t), dengan ? ; 0,$
Zt = Z[1-[1- β
2 ]]Zt = 1 [1- [1-0,96
2 ]]Zt = 0,98
>t ; #,0
5) U(i keseragaman dataa) Batas kontrol atas B)
= + (Zt σ )
= 7,92 + (2,06 0,19 )
= !,31
b) Batas kontrol ba-ah BB)
= - (Zt σ)
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= 7,92 - (2,06 0,19 )
= 7,53
) U(i ke&ukupan data dengan H ; 0,04
N'=[
Zt
α √N∑ i
2- (∑ i )
2
∑ i]
2
N'=[
2,06
0,04 √ 100(6305,32) – (791,80)2
791,80 ]2
N' =
[2,06
0,04
√ 630532 – 626947,24
791,80 ]2
N'=[51,5 59,87
791,80 ]2
N'= 15,16
eterangan: 2C 2, data &ukup
1 # 3 4 5 " ! $ 10".00
".#0
".40
".0
".!0
!.00
!.#0
!.40
!.13
".$3
".$
!.04 !.04
".!!
!.1$
".
".$#".!
U(i keseragaman data
B BB bar +ata ,bgrup
<ambar 3.4.13. eta kontrol u(i keseragaman data 3
") Distribusi *rekuensii) angkauan +)
R = $- $I
R = 9 - 7
R = 2
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ii) Banyak kelas )
= 1 + 3,3
= 1 + 3,3 100
= 7,6≈7
iii) Interval G)
I =R
I =2
7
I = 0,29
Tabel 3.4.0. Tabel distribusi pan(ang (ari 3 Interval &i ' ()( *)( ()B *)B )$ &i.)$
",00 ",#! #0 #0 ",00 ",#! ,$$5 ",#!5 ",14 14#,!0
",#$ ",5" 1# 3# ",#$ ",5" ",#!5 ",5"5 ",43 !$,1
",5! ",! 1 4! ",5! ",! ",5"5 ",!5 ","# 1#3,5#
",!" !,15 14 # ",!" !,15 ",!5 !,155 !,01 11#,14
!,1 !,44 1 "! !,1 !,44 !,155 !,445 !,30 13#,!0
!,45 !,"3 10 !! !,45 !,"3 !,445 !,"35 !,5$ !5,$0
!,"4 $,0# 1# 100 !,"4 $,0# !,"35 $,0#5 !,!! 10,5
+&i 100 +&i.)$ "$#,!!
8) Mean
Mean = ∑ " i &
∑ " i
Mean =792,88
100
Mean = 7,93
$) %ediani) /etak median
/etak median =1
2 N
/etak median =1
2 100
/etak median = 50
%edian terletak pada nilai ke50 pada inter9al ke4
ii) 2ilai median
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%edian = %&$()+ [1
2N - *k$()-1
"$()]×I
%edian = 7,87 + [50 - 4814 ]× 0,29
%edian = 7,87 + [214 ]×0,29
%edian = 7,87 + 0,04
%edian = ",$1
10) %odus
%odus = %&$ +
["$ - "$-1
( "$ - "$-1 )+( "$ - "$+1 ) ]× I
%odus = 6,99 + [20 - 0
(20 - 0 ) + (20 - 12) ] × 0,#$
%odus = 6,99 + [20
20 + 8 ] × 0,#$
%odus = 6,99 + [20
28 ]× 0,#$
%odus = 6,99 + 0,20
%odus = 7,20
11) ersentili) /etak 5
%tk 5 =I
100 N
%tk 5 =
5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada inter9al ke1
ii) /etak 50
%tk 50 =I
100 N
%tk 50 =50
100 100
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%tk 50 = 50
ersentil 50 terletak pada data ke50 pada inter9al ke4
iii) /etak $5
%tk 95 = I100
N
%tk 95 =95
100 100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada inter9al ke"
i9) 2ilai 5
5 = %&5 +
I
100N - *k5-1
"5
× I
5 = 6,99 + [5 - 0
20 ] × 0,#$
5 = 6,99 + [520 ]× 0,#$
5 = 6,99 + 0,07
5
= 7,06
9) 2ilai 50
50 = %& 50 + [I
100N - *k50-1
"50]×I
50 = 7,87 + [50 - 48
14 ] × 0,#$
50 = 7,87+
[2
14
]×0,29
50 = 7,87 + (0.04)
50 = 7,91
9i) 2ilai $5
95 = %&95 +[I
100N - *k95-1
" 95]×I
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95 = 8,74 + [95 - 88
12 ]×0,29
95 = 8,74 +
[7
12
]×0,29
95 = 8,74 + 0,16
95 = 8,90
e) /ebar (ari #,3,4 dan 5Tabel 3.4.1. Data anthropometri lebar (ari #,3,4 dan 5
n
1 2 3 4 5 6 7 8 9 10 ´ x
1 ,4 , ,! ",1 ",! ",$ ,4 ,3 ,3 ,3 ,"$
2 ",5 ,5 ,# ",3 ",1 ,0 ", ",$ ", ,$ ",0
3 ",5 , ",4 ",1 ",# ,# ",4 "," ",3 ,5 ",0$
4 ", ,0 ",3 ",! ",0 ",3 , "," ,# ",1 ",0
5 "," ",# ",0 ," , ",$ ,$ ,1 ,! ", ",05
6 ",3 "," ",1 ",# ",$ ",3 ",0 ",4 ,$ ,0 ",1!
7 ," ,# ",# ,4 ,0 ",! ,$ ",# ,! ," ,"$
8 ",$ ,! ,5 ,0 ," , ",1 ," , ",! ,!"
9 ", ",! ,5 ,4 ,5 ",3 ,5 ",! ,3 ,3 ,$0
10 ",! , ," ,# ",4 ,1 , ," "," ," ,!5
∑ $.4
1) %enghitung ratarata subgroup
=∑ i
k
=69,64
10
= 6,97
#) %enghitung standar de9iasi
σ =√∑(i- )
2
N
= √(6,40 – 6,97 )2 + (6,60 – 6,97 )2 + / + (6,70 – 6,97 )2
100
; 0,5"
3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup
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σ =σ
√ k
σ =0,57
√ 10σ = 0,18
4) 'itung nilai > tabel >t), dengan ? ; 0,$
Zt = Z[1-[1- β
2 ]]Zt = 1 [1- [1-0,96
2 ]]Zt = 0,98
>t ; #,0
5) %enentukan B dan BB untuk u(i keseragaman datai) Batas kontrol atas B)
= + (Zt σ)
= 6,97 + (2,06 0,18 )
= ",33
ii) Batas kontrol ba-ah BB) = - (Zt σ)
= 6,97 + (2,06 0,18 )
= 6,56
) U(i ke&ukupan data dengan H ; 0,04
N'=[
Zt
α √N∑ i
2- (∑ i )
2
∑ i]
2
N'=[
2,06
0,04 √ 100(4881,82) – (696,4 )2
696,4 ]2
N'=[
2,06
0,04 √ 488182 – 484973
696,4 ]2
N'=
[
51,5 56,65
696,4
]
2
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N'= 17,55
eterangan: 2C 2, data &ukup
1 # 3 4 5 " ! $ 10
.#0
.40
.0
.!0
".00
".#0
".40
."$
".0 ".0$ ".0 ".05
".1!
."$.!" .$0
.!5
U(i keseragaman data
B BB bar
+ata subgroup
<ambar 3.4.14. eta kontrol u(i keseragaman data lebar (ari #, 3, 4 dan 5
") Distribusi *rekuensii) angkauan +)
R = $- $I
R = 7,90 - 6,00
R = 1,$0
ii) Banyak kelas )
= 1 + 3,3
= 1 + 3,3 100
; ", 8
iii) Interval G)
I =R
I =1,9
8
I = 0,24
Tabel 3.4.#. Tabel distribusi *rekuensi lebar (ari #, 3, 4 dan 5
Interval &i ' ()( *)( ()B *)B )$ 'i.)$
,00 ,#3 1# 1# ,00 ,#3 5,$$5 ,#35 ,1# "3,3!
,#4 ,4" $ #1 ,#4 ,4" ,#35 ,4"5 ,3 5",#0
,4! ,"1 ## 43 ,4! ,"1 ,4"5 ,"15 ,0 145,0$
,"# ,$5 ! 51 ,"# ,$5 ,"15 ,$55 ,!4 54,!
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,$ ",1$ $ 0 ,$ ",1$ ,$55 ",1$5 ",0! 3,!
",#0 ",43 1 " ",#0 ",43 ",1$5 ",435 ",3# 11",04
",44 "," " !3 ",44 "," ",435 ","5 ",5 5#,!$
",! ",$1 1" 100 ",! ",$1 ","5 ",$15 ",!0 13#,5#
+&i 100 +&i.)$ $,4
!) Mean
Mean =∑ " i &
∑ " i
Mean =
696,46
100
Mean = 6,96
$) %ediani) /etak median
/etak median =1
2 N
/etak median =1
2 100
/etak median = 50
%edian terletak pada nilai ke50 pada Interval ke4
ii) 2ilai median
%edian = %&$()+ [1
2N - *k$()-1
"$()]×I
%edian = 6,715 + [50 - 43
8 ]× 0,24
%edian = 6,715 + [78 ]×0,24
%edian = 6,715 + 0,21
%edian = ,$3
10) %odus
%odus = %&$ + [
"$ - "$-1
( "$ - "$-1 )+ ( "$ - "$+1 ) ]× I
%odus = 6,475 + [22 - 9
(22 - 9 ) + (22 - 8 ) ] × 0,#4
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%odus = 6,475 + [13
13 + 14 ] × 0,#4
%odus
= 6,475 +
[13
27
]×
0,#4
%odus = 6,475 + 0,11
%odus = 6,59
11) ersentil
i) /etak 5
%tk 5 =I
100 N
%tk 5
=5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada Interval ke1
ii) /etak 50
%tk 50 =I
100 N
%tk 50 =50
100 100
%tk 50 = 50
ersentil 50 terletak pada data ke50 pada Interval ke4
iii) /etak $5
%tk 95 =I
100 N
%tk 95 =95
100 100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada Interval ke!
i9) 2ilai 5
5 = %&5 + [I
100N - *k5-1
"5]× I
5 = 5,995 + [5 - 0
12 ] × 0,#4
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5 = 5,995 + [512 ]× 0,#4
5 = 5,995 + 0,10
5 = 6,10
9) 2ilai 50
50 = %& 50 + [I
100N - *k50-1
"50]×I
50 = 6,715 + [50 - 43
8 ] × 0,#4
50 = 6,715 + [78 ]×0,24
50 = 6,715 + 0,21
50 = 6,93
9i) 2ilai $5
95 = %&95 +
[
I
100N - *k95-1
"95
]×I
95 = 7,675 + [95 - 83
17 ]×0,24
95 = 7,675 + [12
17 ]×0,24
95 = 7,675 + 0,17
95 = 7,84
4.2.1.1.4. endela
%eran&ang (endela yang sesuai data anthropometri praktikan mata kuliah IJ. Dataanthropometri yang digunakan untuk meran&ang (endela adalah rentang tangan, (angkauan tangandan tinggi badan berdiri. /alu perhitungannya adalah sebagai berikut:
a) Tinggi badan berdiri tbb)Tabel 3.4.3. data anthropometri tinggi badan berdiri tbb)
-o 1 2 3 4 5 6 7 8 9 10 ×
1154,00 1!,40 151,0 15,"0 1"#,0 1"1,"0 14!,10 14,!0 1",00 15#,10 11,0
21"$,#0 1#,10 1"!,$0 1"5,0 15#,$0 1"!,40 1"!,!0 10,00 155,10 154,30 1",53
31",#0 1$,30 1!0,"0 1!1,10 1"3,$0 1"0,"0 1$,00 1!0,30 1"3,00 154,!0 1"#,00
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410,50 1"5,00 153,00 13,30 151,40 14!,#0 14$,$0 1"$,50 14,00 1"5,!0 1#,0
514",10 15,00 15$,#0 155,10 1"#,$0 1!3,50 154,0 154,"0 15!,!0 1#,!0 11,3"
61"5,50 151,"0 10,$0 1"4,0 15!,10 1#,"0 153,0 15#,50 155,0 15,10 10,13
7 1!#,!0 1!,40 1"3,50 1!0,"0 1"3,30 1"4,00 15$,40 1"$,40 11,#0 14",$0 1"0,0
81"4,40 1#,40 150,0 1"3,00 15,#0 10,"0 1!3,50 1!3,#0 1!3,40 1!0,30 1"1,"
910,#0 1$,$0 1!0,$0 1!1,50 151,00 151,00 1"0,!0 1"1,$0 1"!,00 15,"0 1!,0$
+× 14$4,51
1) %enghitung rararata subgroup
= ∑ k
= ∑1494,51
9
= 1,0
#) %enghitung standar de9iasi
σ =√∑ ( x1−´ x )2
N
σ = √ (154,00- 14 9 ,45 )2 + .... + (165,70 – 149,45 )2
90
σ = 10,!5
3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup
σ =σ
√ k
σ =10,85
√ 9
σ =3,#
4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!
Zt = Z[1-[1- β
2 ]]Zt = 1 [1- [1-0,96
2 ]]Zt = 0,98
>t ; #,0
5) U(i keseragaman data
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i) Batas kontrol atas B)
B ¿ ´ x+(Z t x σ x)
B2,06 x
¿71,15+¿ 1,"!)
B ; "4,!1ii) Batas kontrol ba-ah BB)
BB ¿ ´ x−(Z t x σ x)
BB ¿71,15−(2,06 x1,78)
BB ; ",4!
) U(i ke&ukupan data
N
'
=
[
Zt
α √N∑ i
2- (∑ i )
2
∑i
]
2
N'=[
2,06
0,04 √ 50(5115,83) - (502539,21)
7089 ]2
2C ; 1,0eterangan: 2C 2, data &ukup
1 # 3 4 5 " ! $150.00
155.00
10.00
15.00
1"0.00
1"5.00
11.0
1".53
1"#.00
1#.0 11.3"10.13
1"0.01"1."
1!.0$
U(i keseragaman data
B BB bar
+ata subgroup
<ambar 3.4.14. eta kontrol u(i keseragaman data TBB
") Distribusi *rekuensi
i) angkauan +)+ ; DmaM Dmin
+ ; 1!3,50 E 14",10+ ; 3,40
ii) Banyak kelas
; 1 F 3,3 log n ; 1 F 3,3 log 100 ; ",0 K $
iii) Interval G)
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I =R
I=36,40
9I = 4,05
Tabel 3.4.4. Tabel distrbusi *rekuensi TBB
Interval &i & ()( *)( ()B *)B )$ 'i.)$
14",10 151,14 ! ! 14",10 151,14 14",0$5 151,145 14$,1# 11$#,$
151,15 155,1$ 15 #3 151,15 155,1$ 151,145 155,1$5 153,1" ##$",55
155,#0 15$,#4 5 #! 155,#0 15$,#4 155,1$5 15$,#45 15",## "!,10
15$,#5 13,#$ 11 3$ 15$,#5 13,#$ 15$,#45 13,#$5 11,#" 1""3,$"
13,30 1",34 $ 4! 13,30 1",34 13,#$5 1",345 15,3# 14!",!!
1",35 1"1,3$ " 55 1",35 1"1,3$ 1",345 1"1,3$5 1$,3" 11!5,5$
1"1,40 1"5,44 13 ! 1"1,40 1"5,44 1"1,3$5 1"5,445 1"3,4# ##54,4
1"5,45 1"$,4$ $ "" 1"5,45 1"$,4$ 1"5,445 1"$,4$5 1"",4" 15$",#3
1"$,50 1!3,54 13 $0 1"$,50 1!3,54 1"$,4$5 1!3,545 1!1,5# #35$,"
+&i $0 +&i.)$ 14$35,50
!) Mean
´ x=∑fi.cm
∑fi
´ x=14935,50
90
´ x= 15,$5
$) %ediani) /etak median
/etak median ;1
2 N
/etak median ;1
2 90
/etak median ; 45
ii) 2ilai median
Median=%& $()+ [1
2N- ( *$()-1 )
"$()] I
Median=163,295+ [1
2 90 - 39
9 ] 4,05
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Median=163,295 + [0,66 ] 4,05
Median=163,295 + #,"0
Median= 11,$5
10) %odus
)=%& $) + [ (" $) - " $)-1 )("$) - "$)+1) + ("$) -"$)-1 ) ] I
=151,145 + [15 - 8
(15-8 )+ (15-10) ] 4,05
=151,145 + [ 7
7+5 ] 4,05
= 151,145 + #,3
=¿ 15#,!1
11) ersentili) /etak 5
%tk 5 =I
100 N
%tk 5
=5
100 90
%tk 5 =4,5
ersentil 5 terletak pada data ke5 pada Interval ke1
ii) /etak 50
%tk 50 =I
100 N
%tk 50 =50
100 90
%tk 50 = 45
ersentil 50 terletak pada data ke50 pada Interval ke4
iii) /etak $5
%tk 95 =I
100 N
%tk 95 =95
100 90
%tk 95 = 85,5
ersentil $5 terletak pada data ke$5 pada Interval ke!
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i9) 2ilai 5
5 = %&5 + I
I
100N - *k5-1
"5
5 = 147,095 + 4,05 [5
10090 - 0
8 ]5 =
14$,3
9) 2ilai 50
50 = %& 50+ I
[I
100
N - *k50-1
"50 ]50 = 163,295 + 4,05 [
50
10090 - 39
9 ]50 =
1!,#5
9i) 2ilai $5
95 = %&95 + I [I
100 N - *k95-1
" 95 ]95 = 179,495+ 4,05 [
95
10090 - 77
13 ]50 =
1!5,10
b) +entang tangan rt)Tabel 3.4.5. +entang tangan
%
n1 2 3 4 5 6 7 8 9 10 ´ x
1 1!5," 153, 1!, 11," 14$, 1$#, 1", 15!, 150, 14," 1,1
2 1!",1 15$, 10, 10,4 10 153, 154," 1!1, 11 1$0,$ 1,$3
3 1"! 1$3, 1!", 1!4,$ 1!5 1"#, 1!4,$ 1"#, 15", 1$," 1"!,3
4 1!#, 154, 1$#, 1$4 1"3, 1", 1, 1"4 1"$, 15"," 1"5,05
5 1"4,3 10, 1$0, 10,$ 1$#, 15$, 1"4, 1!, 15, 1"1,$ 1"#,!
6 1"3," 153, 14", 1"3,$ 1!3 15, 1"", 11, 1", 1$1, 1"0,5
714",# 14!, 1!3, 1"5,# 155, 14!, 1!, 1!", 154 1",5 1,33
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8 1"$ 15 1, 15$,$ 1!$, 13, 1!1 1"3 1$#, 1!3,4 1"4,4$
9 153,4 1$3, 1", 1!$,3 1!0, 1" 1!0,# 1!3 1!, 15$ 1"4,1!
10 1"!,1 1!4, 1!0, 10,3 1!1, 1", 1$3,4 15$, 1$0, 1$# 1"$,
∑ ´ x 1."#4,"3
1) %enghitung ratarata subgroup
=∑ i
k
=1726,73
10
= 172,67
#) %enghitung standar de9iasi
σ =√∑(i- )
2
N
= √(185,7 – 172,67 )2 + (153,6 – 172,67 )2 + / + (192,0 – 172,67 )2
100
; 13,4"
3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup
σ =σ√ k
σ =13,47
√ 10
σ = 4,26
4) 'itung nilai > tabel >t), dengan ? ; 0,$
Zt = Z[1-[1- β
2 ]]Zt = 1 [1- [1-0,96
2 ]]Zt = 0,98
>t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
= + (Zt σ )
= 172,67+ (2,06 4,26 )
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= 1!1,45
ii) Batas kontrol ba-ah BB)
= - (Zt σ)
= 172,67- (2,06 4,26 )
= 163,90
1 # 3 4 5 " ! $ 10155.00
10.00
15.00
1"0.00
1"5.00
1!0.00
1!5.00
1.11".$3
1"!.31"5.05
1"5.05 1"#.!01"0.50
1.33
1"4.4$1"5.1!
U(i keseragaman data
B BB bar
+ata sbgrup
<ambar 3.4.15. eta kontrol u(i keseragaman data +T
) U(i ke&ukupan data dengan H ; 0,04
N'=[
Zt
α √N∑ i
2- (∑ i )
2
∑ i]
2
N'=[
2,06
0,04 √ 100(2999751,95) – (1726,73 )2
1726,73 ]2
N'=[
2,06
0,04 √ 299975195 – 298159649,29
1726,73 ]2
N'=[51,5 1347,40
1726,73 ]2
N'= 40,20
eterangan: 2C 2, data &ukup
") Distribusi *rekuensi
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i) angkauan +)
R = $- $I
R = 194 - 147,20
R = 46,80
ii) Banyak kelas )
= 1 + 3,3
= 1 + 3,3 90
= 7,4 7
iii) Interval G)
I =R
I =46,80
7
I = 6,69
Tabel 3.4.. Tabel distribusi *rekuensi rentang tangan +T)
Interval &i & ()( *)( ()B *)B )$ &i.)$
14",#0 153,!! ! ! 14",#0 153,!! 14",1$5 153,!!5 150,54 1#04,3#
153,!$ 10,5" 1$ #" 153,!$ 10,5" 153,!!5 10,5"5 15",#3 #$!",3"
10,5! 1",# 11 3! 10,5! 1",# 10,5"5 1",#5 13,$# 1!03,1#
1",#" 1"3,$5 1# 50 1",#" 1"3,$5 1",#5 1"3,$55 1"0,1 #04",3#
1"3,$ 1!0,4 1 1"3,$ 1!0,4 1"3,$55 1!0,45 1"",30 #!3,!0
1!0,5 1!",33 1" !3 1!0,5 1!",33 1!0,45 1!",335 1!3,$$ 31#",!3
1!",34 1$4,0# 1" 100 1!",34 1$4,0# 1!",335 1$4,0#5 1$0,! 3#41,5
+&i 100 +&i.)$ 1"#4!,3#
!) Mean
Mean =∑ " i &
∑ " i
Mean =17248,32
100
Mean = 172,48
$) %ediani) /etak median
/etak median =1
2 N
/etak median =1
2 100
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/etak median = 50
%edian terletak pada nilai ke50 pada Interval ke4
ii) 2ilai median
%edian = %&$+ [12
N - *k$-1
"$ ]×I
%edian = 167,26 + [50 - 38
12 ]× 6,69
%edian = 167,26 + [12
12 ]×6,69
%edian = 167,26 + 6,69
%edian = 1"3,$5
10) %odus
%odus = %&$) + [
"$) - "$)-1
( "$) - "$)-1 )+ ( "$) - "$)+1 ) ]× I
%odus = 167,26 + [19 - 8
(19 - 8 ) + (19 - 11) ] × ,$
%odus = 167,26 +
[11
11 + 8 ] × ,$
%odus = 167,26 + [11
19 ]× ,$
%odus = 167,26 + 3,87
%odus = 171,14
11) ersentil
i) /etak 5
%tk 5 =I
100 N
%(tk 5 =5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada Interval ke1
ii) /etak 50
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%tk 50 =I
100 N
%(tk 50 =50
100
100
%tk 50 = 50
ersentil 50 terletak pada data ke50 pada Interval ke4
iii) /etak $5
%tk 95 =I
100 N
%tk 95 =95
100 100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada Interval ke"
i9) 2ilai 5
5 = %&5 + [I
100N - *k5-1
"5]× I
5 = 147,195 + [5 - 0
8 ] × ,$
5 = 147,195 + [5
8 ]× ,$
5 = 147,195 + 4,18
5 = 151,38
9) 2ilai 50
50 = 167,26 +
[
I
100N - *k50-1
" 50
]×I
50 = 167,26 + [50 - 38
12 ] × ,$
50 = 167,26 + [12
12 ]×6,69
50 = 167,26 + 6,69
50 = 173,96
9i) 2ilai $5
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95 = %&95 +[I
100N - *k95-1
" 95]×I
95 = 187,33 + [95 - 10017 ]×6,69
95 = 187,33 + [-517 ]×6,69
95 = 187,33 + (-1,97)
95 = 185,37
&) angkauan tangan (t)Tabel 3.4." angkauan tangan
% n
1 2 3 4 5 6 7 8 9 10 ´ x
1 "",$ $,3 "4 4,3 5," ",4 !,$ $,3 "4,$ !0,! "1,#5
2 5, 5 5 "#,! 3," 5,4 "5,3 !0,# ", !0 "0,$
3 "0 3," 3," "4,4 "!,3 "!,4 "!," ""," "$ $,$ "3,3!
4 "4,4 "1,$ "1,$ "3," ",# 3,4 3,# "!,# "5,! "0,5"
5 "3," 1, 1, !0,1 ",5 $ "5 !1 ,3 5,3 "1,01
6 "!,1 1," 1," "",! 1,$ #,3 "1,! "1,3 !,! ",5 $,1$
7 $,1 $," $," ",5 "1 $,3 #,3 "",# "4 !,! $,!
8 !, "4,! "4,! "1, #,3 "#,! ",5 ",5 "3 "$,! "1,#"
9 "",5 ",$ ",$ $," "3, "0,1 "!," 1,4 "0,$ 5,$ "#,1
10 #,$ ",1 ",1 !0," $ ,1 "1,5 !0 5,$ #,# $,#5
∑ ´ x "0!,$
1) %enghitung rararata subgroup
x = ∑ xk
x = ∑ 708,9
10x =70,89
#) %enghitung standar de9iasi
σ =√∑ ( x1−´ x )2
N
σ = √ (77,9-70,89 )2 + .... + (62,2 – 70,89 )2
100
σ = 5,79
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3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup
σx =σ
√ k
σx =5,79
√ 10
σx = 1,83
4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!
Z = 1 – [1 - β
2 ]
Z = 1 –
[1 – 0, 98
2 ]Z = 1 – 0, 01
Z = 0,99
Berdasarkan tabel distribusi normal, nilai yang paling mendekati untuk harga > ; 0,$$
adalah #,0 maka >t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
B
= x+(Z t x σx )
B =70,89+(2,06 x 1,!3)
B ; "4,ii) Batas kontrol ba-ah BB)
BB= x-(Zt x σ x )
BB = x-(2,06 x 1,83)
BB ; ",1#
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1 # 3 4 5 " ! $ 10".1#
!.1#
$.1#
"0.1#
"1.1#
"#.1#
"3.1#
"4.1#
"1.#5"0.$
"3.3!
"0.5""1.01
$.1$
$.!
"1.#"
"#.1
$.#5
U(i keseragaman data
A bar B BB # bar
<ambar 3.4.1". U(i keseragaman data
) U(i ke&ukupan data
N'=[
Zt
α √N∑ x i
2- (∑ x i )
2
∑ xi]2
N'=
[
2,06
0,04 √ 505115,83 - 502539,21
7089
]
2
2C ; 13,0eterangan: 2C 2, data &ukup
") Distribusi *rekuensii) angkauan +)
+ ; DmaA Dmin
+ ; !1,00 E 1,40+ ; 1$,0
ii) Banyak kelas ; 1 F 3,3 log n
; 1 F 3,3 log 100 ; ",Dibulatkan ke atas, sehingga ; $
iii) Gnter9al G)
I=R
I=19,60
9
I=2,18
Tabel 3.4.!. Distribusi *rekuensi
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!) %ean
x =!"i.#$
!"i
x =7079,080100
x =70,79
$) osisi median
osisi median ;1
2 x N
osisi median ;1
2 x 100
osisi median ; 50
10) %edian
x =%&$+ [1
2N- (*$-1 )
"$] x I
~x =67,935+[1
2 x 100- (35 )
51 ] x 2,18
x =67,935+[50-35
51 ]x 2,18
x = 70,05
11) %odus
x =%&$) +[ ( "$) - "$)-1 )( "$)- "$)+1 )+ (" $) - "$)-1) ] x I
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x = 67,935+[17-9
(17-10) + (17-9) ] x 2,18
x =67,935+
[8
7+8
] x 2,18
x =69,30
1#) ersentili) /etak 5
%tk 5 =I
100 N
%(tk 5 =5
100 100
%tk 5 = 5
ersentil 5 terletak pada data ke5 pada interval ke1
ii) /etak 50
%tk 50 =I
100 N
%(tk 50 =50
100 100
%tk 50 = 50
ersentil 50 terletak pada data ke50 pada interval ke4
iii) /etak $5
%tk 95 =I
100 N
%tk 95 =95
100 100
%tk 95 = 95
ersentil $5 terletak pada data ke$5 pada interval ke$
i9) 2ilai 5
5 = %&5 + I [I
100N - *k5-1
"5]
5 = 61,395, + 2,18 [5
100100 - 0
13 ]5 = 61,89
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9) 2ilai 50
50 = %&50 + I
I
100N - *k50-1
" 50
50 = 67,935 + 2,18 [50
100100 - 35
16 ]50 = 70,05
9i) 2ilai $5
95 = %&95 + I
[I
100
N - *k95-1
" 95 ]95 = 78,835 + 2,18 [
95
100100 - 91
9 ]50 = 80,01
4.2.3 Peran#angan Produ
4.2.3.1. Peran#angan Produ dan /uangan seretariat !urusan
eran&angan usulan tata letak ruangan merupakan pengembangan atau perbaikan dari ran&anganyang sudah ada dan masih bersi*at usulan, hal ini bertu(uan untuk membuat tata letak ruanganyang lebih e*ekti* dan nyaman sehingga segala ma&am kegiatan yang ada didalamnya bisadilakukan dengan lebih baik kedepannya. Berikut adalah desain usulan tata letak ruang sekretariat (urusan:
<ambar 3.4.#3. Usulan ruangan sekretariat (urusan
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Gambar 3.4.24. Desain produk meja
Tabel 3.4.69. Data anthropometri meja
Data antropometri Persentil 5 Persentil 50 Persentil 95
angkauan tangan #,30 "0,$5 !0,01
+entang tangan 150,!# 1"3,$5 1$#,$0
Tinggi siku duduk 1",#0 ##,00 #,#1
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<ambar 3.4.#5 Desain produk lemari kabinet
Tabel 3.4."0. Desain produk lemari kabinet
Data antropometri Persentil 5 Persentil 50 Persentil 95
Tinggi bahu berdiri 1#3,5" 13$,!1 153,5!
/ebar bahu 34,50 3$,3 45,4$
an(ang (ari 3 ",0 ",$1 !,$0
/ebar (ari #,3,4 dan 5 ,10 ,$3 ",!4
angkauan tangan #,30 "0,$5 !0,01
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<ambar 3.4.#. Desain produk pintu
Tabel 3.4.71. Data anthropometri meja
Data antropometri Persentil 5 Persentil 50 Persentil 95
Tinggi badan berdiri 14$,3 1!,#5 1!5,10
Tinggi bahu berdiri 1#3,5" 13$,!1 153,5!
/ebar bahu 34,50 3$,3 45,4$
an(ang (ari 3 ",0 ",$1 !,$0
/ebar (ari #,3,4 dan 5 ,10 ,$3 ",!4
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P95 (LB)=45,75
P95 (TBB)=182,44
P5 (JT)=80,72
P50 (TSB)= 122,172
<ambar 3.4.#". Desain produk (endela
Tabel 3.4.72. Data anthropometri meja
Data antropometri Persentil 5 Persentil 50 Persentil 95Tinggi badan berdiri 14$,3 1!,#5 1!5,10
+entang tangan 151,3! 1"3,$ 1!3,3"
angkauan tangan #,30 "0,$5 !0,01
4.2.4. ata pen#ahaaan dan ebisingan
4.2.4.1. Perhitungan data pen#ahaaan
a) 'ari pertamaTabel 3.4."3. lux meter hari pertama
n
1 2 3 4 5 6 7 8 9 10 x
1 0,#$ 0,#! 0,30 0,## 0,30 0,#3 0,#$ 0,#3 0,30 0,30 0,#"
2 0,30 0,# 0,30 0,30 0,30 0,#! 0,30 0,30 0,#3 0,#$ 0,#$
3 0,30 0,30 0,30 0,## 0,# 0,#! 0,30 0,30 0,## 0,3# 0,#!
4 0,30 0,#" 0,#3 0,30 0,31 0,3# 0,## 0,30 0,30 0,#4 0,#!
5 0,# 0,#! 0,30 0,#$ 0,30 0,30 0,3# 0,#5 0,30 0,## 0,#!
∑ : 140
1) %enghitung rararata subgroup
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x =∑ x
k
x = 1,405
x =0,28
#) %enghitung standar de9iasi
σ =√∑ (x1- x )2
N
σ =
√
(0,29-0,28 )2 + .... + (0,28-0,28 )
2
50
σ = 0,03
3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup
σx =σ
√ k
σx =0,03
√ 5
σx = 0,01
4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!
Z = 1 – [1 - β
2 ]Z = 1 – [1 – 0,98
2 ]Z = 1 – 0,01
Z = 0,99
Berdasarkan tabel distribusi normal, nilai yang paling mendekati untuk harga > ; 0,$$adalah #,0 maka >t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
B= x+(Z t x σx¿
B =0,28+ (2,06 x 0,01)
B ;0,31
ii) Batas kontrol ba-ah BB)
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BB= x-(Zt x σ x )
BB = 0,28-(2,06 x 0,01)
BB ; 0,#5
1 # 3 4 50.#5
0.#"
0.#$
0.31
0.27
0.290.28 0.28 0.28
U(i eseragaman Data
bka bkb bar
rata sub!roup
<ambar 3.4.#!. eta kontrol u(i keseragaman data
) U(i ke&ukupan data
N'=[
Zt
α √N∑ x i
2- (∑ x i )
2
∑ x i]2
N'=[
2,06
0,04 √ 50 x 3,97 - 196,28
14,01 ]2
2C ; 31,!eterangan: 2C 2, data &ukup
b) 'ari kedua
Tabel 3.4."4. lux meter hari kedua
1 2 3 4 5 6 7 8 9 10 x
1 0,34 0,34 0,34 0,3# 0,35 0,34 0,35 0,35 0,35 0,35 0,34
2 0,35 0,35 0,34 0,35 0,35 0,35 0,3# 0,35 0,35 0,35 0,35
3 0,35 0,3 0,35 0,34 0,34 0,35 0,35 0,35 0,3# 0,3 0,35
4 0,3 0,35 0,35 0,35 0,3" 0,3 0,3 0,3" 0,35 0,3# 0,35
5 0,35 0,35 0,35 0,35 0,31 0,35 0,3" 0,34 0,35 0,3$ 0,35
∑ : 174
1) %enghitung rararata subgroup
x =∑ x
k
x =1,74
5
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x =0,35
#) %enghitung standar de9iasi
σ =
√∑ (
x1
- x
)
2
N
σ = √ (0,34-0,35)2 + .... + (0,39-0,35 )2
50
σ = 0,01
3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup
σx =σ
√ k
σx =0,01
√ 5
σx = 0,004
4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!
Z = 1 – [1 - β
2 ]Z = 1 –
[
1 – 0,98
2
]Z = 1 – 0,01
Z = 0,99
Berdasarkan tabel distribusi normal, nilai yang paling mendekati untuk harga > ; 0,$$adalah #,0 maka >t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
B= x+(Z t x σx¿
B =0,01+ (2,06 x 0,004)B ;0,3
ii) Batas kontrol ba-ah BB)
BB= x-(Zt x σ x )
BB = 0,01-(2,06 x 0,00 4)
BB ; 0,34
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1 2 3 4 "0.33
0.34
0.34
0.3"0.3"
0.36
0.36
0.34300.340 0.34"0
0.35400.3510
U(i eseragaman Data
bka bkb bar
rata sub!roup
<ambar 3.4.#$. eta kontrol u(i keseragaman data
) U(i ke&ukupan data
N'=[
Zt
α √N∑ x i
2- (∑ x i )
2
∑ x i]2
N'=[
2,06
0,04 √ 50 x 6,07 - 303,11
17,41 ]2
2C ; 4,00eterangan: 2C 2, data &ukup
4.2.4.2 Perhitungan data ebisingan
a) 'ari pertamaTabel 3.4."5. sound level hari pertama
1 2 3 4 5 6 7 8 9 10 x
1 ",00 3,30 0,#0 1,0 1,$0 ","0 4,10 3,0 3,50 0,0 3,35
2 4,50 3,!0 ,50 5,!0 0,30 1,!0 3,10 0,40 ","0 1,$0 3,5!
3 0,#0 3,50 ,50 5,!0 4,50 3,10 3,40 ","0 3,0 1,!0 4,01
4 0,40 5,!0 ,50 0,#0 1,!0 3,0 4,50 5,!0 ,50 1,!0 3,$
5 5,!0 0,30 3,50 1,!0 ,50 0,30 5,!0 4,50 5,!0 4,"0 3,$0
∑ :
31853
1) %enghitung rararata subgroup
x =∑ x
k
x =318,53
5
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x =63,71
#) %enghitung standar de9iasi
σ =
√∑ (
x1
- x
)
2
N
σ = √ (67,00-63,71 )2 + .... + (64,70-63,71)2
50
σ = 63,47
3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup
σx =σ
√ k
σx =63,47
√ 5
σx = 28,38
4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!
Z = 1 – [1 - β
2 ]Z = 1 –
[
1 – 0,98
2
]Z = 1 – 0,01
Z = 0,99
Berdasarkan tabel distribusi normal, nilai yang paling mendekati untuk harga > ; 0,$$adalah #,0 maka >t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
B= x+(Z t x σx¿
B =63,47+ (2,06 x #!,3!)B ;1##,1!
ii) Batas kontrol ba-ah BB)
BB= x-(Zt x σ x )
BB = 63,47-(2,06 x 28,38)
BB ; 0,#5
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1 # 3 4 55$.00
0.00
1.00
#.00
3.004.00
5.00
.00
".00
3.35 3.5! 4.01 3.$ 3.$0
U(i eseragaman Data
bka bkb bar
rata sub!roup
<ambar 3.4.#$. eta kontrol u(i keseragaman data
) U(i ke&ukupan data
N'=[
Zt
α √N∑ x i
2- (∑ x i )
2
∑ x i ]2
N'=[
2,06
0,04 √ 50 x 203188,29- 10146136,09
3185,30 ]2
2C ; 3,4"eterangan: 2C 2, data &ukup
b) 'ari keduaTabel 3.4.". sound level hari kedua
1 2 3 4 5 6 7 8 9 10 x
1 ","0 $,#0 5,10 "3,#0 !,30 4,0 #,50 4,50 5,40 4,!0 ,53
2 ,10 ,!0 4,#0 #,30 3,00 ",30 4,00 4,40 ,50 !,$0 5,35
3 !,$0 3,30 ",50 5,30 ,50 4,50 !,"0 ,"0 5,40 3,50 ,03
4 ,!0 !,"0 "#,10 4,0 $,30 4,50 #,30 4,10 1,!0 ",!0 ,#0
5 3,"0 !,40 5,50 4,30 ,#0 ,$0 $,!0 !,30 5,10 #,$0 ,11
∑ : 33022
1) %enghitung rararata subgroup
x =∑ x
k
x =330,22
5
x =66,04
#) %enghitung standar de9iasi
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σ =√∑ (x1- x )2
N
σ =
√ (67,70-66,04 )
2 + .... + (62,90-66,04 )
2
50
σ = 2,52
3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup
σx =σ
√ k
σx =2,52
√ 5
σx = 1,13
4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!
Z = 1 – [1 - β
2 ]Z = 1 – [1 – 0,98
2 ]Z = 1 – 0,01
Z = 0,99Berdasarkan tabel distribusi normal, nilai yang paling mendekati untuk harga > ; 0,$$adalah #,0 maka >t ; #,0
5) U(i keseragaman datai) Batas kontrol atas B)
B= x+(Z t x σx¿
B =66,04+ (2,06 x 1,13)
B ;!,3"
ii) Batas kontrol ba-ah BB)
BB= x-(Zt x σ x )
BB = 66,04-(2,06 x 1,13)
BB ; 3,"#
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1 # 3 4 51.00
#.00
3.00
4.00
5.00
.00
".00
!.00
$.00
.53
5.35
.03 .#0 .11
U(i eseragaman Data
bka bkb bar
rata sub!roup
<ambar 3.4.30. eta kontrol u(i keseragaman data
) U(i ke&ukupan data
N'=[
Zt
α √N∑ x i
2- (∑ x i )
2
∑ x i]2
N' =
[2,06
0,04
√ 50 x 218402,16 - 10904524,84
3302,20
]
2
2C ; 3,"$ eterangan: 2C 2, data &ukup