Exercise :

Tugas statistik ke-3

Bismillahirrahmaanirrahiin

Keterangan:

N=distribusi populasi ; cth ada 500 orang

n=distribusi sampling ; cth misal dari 500 org diambil 17 org, nah 17 org adl distribusi

samplingnya

σ=”sigma”// (adl standar deviasi populasi)

μ=”Mu”// (adl mean populasi)

π=Pi (adl proporsi dari populasi)

SE=adalah standart deviasi dari distribusi sampling mean

7.5) refer to the population of exercise 6.8.

a. what is the standart error of the mean for n=16?

b. what is the standart error of the mean for n=64?

c. what is true about the relationship between n and SE(X ) ?

jawab :

di soal 6.8 (halk= nilai mean adl 50 tahun / Standar deviasinya adl 12

a. SE= σ√n

= 12√16

= 3

b. SE= σ√n

= 12√64

= 1.5

c. Hubungannya semakin besar nilai “n” maka nilai “SE(X )” semakin kecil

Dan semakin kecil nilai “n” maka nilai “SE(X 9)” semakin besar

7.6) suppose heights of 20-year-old men are approximetly normally distributed with a

mean of 71 in . and a population standart deviation of 5 in. a random sample of

fifteen 20-year-old men is selected and measured. Find the probability that the

sample mean (X )

a. is at least 77 in.

b. lies between 65 and 75 in

c. is not more than 63 in.

diketahui:

distribusi normal adl mean populasi

distribusi sample adl

μ=71 (mean populasi)

σ=5

Jawaban

a. SE= σ√n

= 5√15

= 1.29

Z= x−μSE

=77−711,29

=4.65

b. Z= x−μSE

= 65−711.29

= -4.65

Z= x−μSE

= 75−711.29

= 3.10 +

c. Z= x−μSE

= 63−711.29

= -6.20

7.7) if the mean length of newborn infants is 52.5 cm and the standart deviation is

4.5 cm, what is the probability that the mean of a sample of (a) size 10 and (b) size

15 is greater than 56 cm ?

Jawab :

μ=52.5

σ=4.5

P(X > 56)

n= 10

SE= 4.5√10

= 1.42

Z= 56−52.51.42

=2.46 P= 0.4931

P(X > 56) = 0.5-0.4931=0.0069

ATAU RUMUSnya

Z= 56 – 52.5 = 2.46 P = 4.5√ 10

7.8) suppose that the mean weight of infant born in a community is µ = 3360 g and σ

=490 g.

a. find P(2300< x < 4300)

b. find P(x ≤ 2500)

c. find P (x≥ 5000)

what must you assume about the distribution of birth weights to make the answer to

(a),(b), and (c) valid ?

jawab :

μ=3360

σ=490

a. Antara 2300 dan 4300

Z1= 2300 – 3360 = -2.16 P = .4846 490 Z2 = 4300 – 3360 = 1.92 P = .4726 + 490P (2300 < X < 4300) = 0.9572

b. Z= 2500 – 3360 = -1.76 P = .4608 490

c. Z= 5000 – 3360 = -3.35 P = .99960 490

find P (x≥ 5000) =

7.9) suppose you select a sample of 49 infants from the population describe in

exercise 7.8.

a. what are the mean and standart error of this sampling distribution ?

(apa kesalahan mean dan standar eror dari distribusi sampling)

b.find P(3100<X<3600)

c.find P(X<2500)

d.find P(X>3540)

what must you assume about the distribution of birth weight to make the answers to

(b),(c) and (d) valid?

Jawab:\

a. b. Antara 3100 dan 3600

Z1= 3100 – 3360 = - 3.71 P = . 490√ 49 Z2 = 3600 – 3360 = 3.43 P = . + 490√ 49

P (3100< X < 3600)

c. Z= 2500 – 3360 = - 12.29 P = . 490√ 49 P(X<2500) <0.001

d. . Z= 3540 – 3360 = 0.05 P = . 490√ 49

P(X>3540)

7.10) if the mean number of cigarettes smoked by pregnant women is 16 and the

standard deviation is 8 , find the probability that in a random sample of 100 pregnant

women yhe mean number of cigarettes smoked will be greater than 24.

7.11) a. describe the three main points of the central limit theorem

b. what conditions must be met for the central limit theorem to be applicable ?

c. explain why the central limit theorem plays such an important role in

inferential statistics

7.12)

a) describe the difference between the distribution of observations from a population

and a distribution of its sample means.

b) what are the differences between the standard deviation and the standart error?

c) when would we want to use the standart deviation, and when the standart error?

7.14) if the cholesterol level of men in community is normally distributed with a mean

of 200 and a standart deviation of 25, what is the probability that a randomly selected

sample of 49 men will have a mean a between 190 and 205?

7.15) compare the critical value (Z=±1.96)that corresponds to 5% of the tail area of

the normal distribution with the critical values of the t distribution for df=9,19, 29, and

∞ . As the degree of freedom increases (which means that the sample size

increases), what happens to the value of t compared with the value of Z? explain

why this is occurring?

7.16) if the forced vital capacity of 11-year-old white males isnormally distributed with

a mean of 2400 cc and σ=400, find the probability that a sample of size n=64 will

provide a mean

a. greater than 2500

b. between 2300 and 2500

c. less than 2350

7.17)

a. find the standart error in exercise 7.16.

b. if you want SE(X ) to be one-half its size in exercise 7.16 ho large a sample would

you need to have ?

7.18) suppose systolic blood pressure of 17-year-old females is approximately

normally distributed with a mean of 118 mmHg and a standart deviation of 12 mmHg.

a. what proportion of girls would you expect to have blood pressures between 112

mmHg and 124 mmHg?

b. if you were to select a sample of 16 girls and obtain their mean systolic blood

pressure, what proportion of such samples would you expect between 112 mmHg

and 124 mmHg?

c. compare the results of (a) and (b), and explain the reasons for the difference.

7.19) for data that are normally distributed, how much area is included under the

normal curve

a. within ±1σ

b. with ±SE(X )for a distribution of sample means

compare (a) and (b) , and state why the results do or do not surprise you.