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3
lim𝑥→−∞
𝐹 𝑥 = 0 lim𝑥→∞
𝐹 𝑥 = 1
𝐹: ℝ → [0,1]
𝐹 𝑥 = 𝑃𝑟 𝑋 ≤ 𝑥 , 𝑥 ∈ ℝ
1
x
F(x) 𝑓 𝑥 =𝑑𝐹(𝑥)
𝑑𝑥
x𝐹 𝑥 =
−∞
𝑥
𝑓 𝑥 𝑑𝑥
1) 𝑓(𝑥) ≥ 0
2)
−∞
+∞
𝑓 𝑥 𝑑𝑥 = 1
𝑃𝑟 𝑎 < 𝑋 < 𝑏 =
𝑎
𝑏
𝑓 𝑥 𝑑𝑥
𝑃𝑟 𝑋 < 𝑏 =
−∞
𝑏
𝑓 𝑥 𝑑𝑥
𝑚𝑘 = 𝐸𝑋𝑘 =
−∞
+∞
𝑥𝑘𝑓 𝑥 𝑑𝑥
𝑚𝑘 = 𝐸𝑋𝑘 =
𝑖=1
𝑛
𝑥𝑖𝑘Pr(𝑋 = 𝑥𝑖)
x
𝜇𝑘 = 𝐸(𝑋 − 𝐸𝑋)𝑘
𝜇2 = 𝐷2𝑋 = 𝐸𝑋2 − (𝐸𝑋)2
𝜎 = 𝜇2
m
s
𝐽𝑒ż𝑒𝑙𝑖 𝑌 = 𝑎𝑋 + 𝑏, 𝑡𝑜
𝐽𝑒ż𝑒𝑙𝑖 𝑋, 𝑌 𝑠ą 𝑛𝑖𝑒𝑧𝑎𝑙𝑒ż𝑛𝑒, 𝑡𝑜
𝐽𝑒ż𝑒𝑙𝑖 𝑍 = 𝑎𝑋 + 𝑏𝑌, 𝑡𝑜
𝐸𝑌 = 𝑎𝐸𝑋 + 𝑏
𝐸(𝑋𝑌) = 𝐸𝑋 ∙ 𝐸𝑌
𝐸𝑍 = 𝑎𝐸𝑋 + 𝑏𝐸𝑌
𝐷2𝑌 = 𝑎2𝐷2𝑋
𝐷2𝑌 = 𝑎2𝐷2𝑋 + 𝑏2𝐷2𝑋
𝐽𝑒ż𝑒𝑙𝑖 𝑌 = 𝜑 𝑋 , 𝑡𝑜 𝐸𝑌 = 𝐸 𝜑 𝑋 =
−∞
+∞
𝜑 𝑥 𝑓 𝑥 𝑑𝑥
x
1 2 3
x
ϵ
𝑜𝑟𝑎𝑧Pr(𝑋 ≤ 𝑥𝑝) ≥ 𝑝 Pr 𝑋 ≥ 𝑥𝑝 ≥ 1 − 𝑝
𝐹 𝑥𝑝 = 𝑝 1
1 xp 6 x
F(x)
p
xp
−∞
𝑥𝑝
𝑓 𝑥 𝑑𝑥 = 𝑝
xX1/2
𝑃𝑟 𝑋 = 𝑥1 = 𝑝 𝑃𝑟 𝑋 = 𝑥1 = 𝑞 𝑔𝑑𝑧𝑖𝑒 𝑝 + 𝑞 = 1
𝑞 = 1 − 𝑝
𝐸𝑋 = 𝑥1 ∙ 𝑝 + 𝑥2 ∙ 𝑞 = 𝑥1 ∙ 𝑝 + 𝑥2 ∙ (1 − 𝑝)
BB: 𝑃𝑟 𝑋 = 𝑥1 ∙ 𝑃𝑟 𝑋 = 𝑥1 = 𝑝2
CC: 𝑃𝑟 𝑋 = 𝑥2 ∙ 𝑃𝑟 𝑋 = 𝑥2 = (1 − 𝑝)2
BC: 𝑃𝑟 𝑋 = 𝑥1 ∙ 𝑃𝑟 𝑋 = 𝑥2 = 𝑝(1 − 𝑝)
CB: 𝑃𝑟 𝑋 = 𝑥2 ∙ 𝑃𝑟 𝑋 = 𝑥1 = 1 − 𝑝 𝑝
BB: 𝑃𝑟 𝑋 = 𝑥1 ∙ 𝑃𝑟 𝑋 = 𝑥1 = 𝑝2
CC: 𝑃𝑟 𝑋 = 𝑥2 ∙ 𝑃𝑟 𝑋 = 𝑥2 = (1 − 𝑝)2
BC: 𝑃𝑟 𝑋 = 𝑥1 ∙ 𝑃𝑟 𝑋 = 𝑥2 = 𝑝(1 − 𝑝)
CB: 𝑃𝑟 𝑋 = 𝑥2 ∙ 𝑃𝑟 𝑋 = 𝑥1 = 1 − 𝑝 𝑝
𝑝2 + 1 − 2𝑝 + 𝑝2 + 𝑝 − 𝑝2 + 𝑝 − 𝑝2 = 1
𝑃𝑟 𝑋 = 𝑘 =𝑛
𝑘𝑝𝑘(1 − 𝑝)𝑛−𝑘
𝐸 𝑥𝑖 = 1 ∙ 𝑝 + 0 ∙ 1 − 𝑝 = 𝑝
𝑥 =
𝑖
𝑥𝑖
𝐸 𝑥 =
𝑖=1
𝑛
𝐸(𝑥𝑖) = 𝑖=1
𝑛
𝑝 = 𝑛𝑝
𝐷2 𝑥𝑖 = 𝐸(𝑥𝑖)2 − (𝐸 𝑥𝑖 )
2
𝐸(𝑥𝑖)2 = 12 ∙ 𝑝 + 02 ∙ (1 − 𝑝) = 𝑝
𝐷2 𝑥𝑖 = 𝑝 − 𝑝2 = 𝑝(1 − 𝑝)
𝐷2 𝑥 = 𝑛𝑝(1 − 𝑝) = 𝑛𝑝𝑞
𝑘=1
𝑛
𝐷2(𝑥𝑘) =𝑛𝐷2(𝑥𝑖) =
𝑃𝑟 𝑋 = 𝑘 =𝜆𝑘𝑒−𝜆
𝑘!
𝐸𝑋 = 𝜆 𝐷2𝑋 = 𝜆 𝑀𝑒 ≈ 𝜆 +1
3+
0,02
𝜆
1
𝑏 − 𝑎
0 a bx
f(x)
−∞
+∞
𝑓 𝑥 𝑑𝑥 = 1
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 = 1
𝑓 𝑥
𝑎
𝑏
𝑑𝑥 = 𝑓 𝑥 𝑏 − 𝑎 = 1
𝑓 𝑥 =
1
𝑏 − 𝑎𝑑𝑙𝑎 𝑥 ∈ [𝑎, 𝑏]
0 𝑑𝑙𝑎 𝑝𝑜𝑧𝑜𝑠𝑡𝑎ł𝑦𝑐ℎ 𝑥
𝐹 𝑥 =
−∞
𝑥
𝑓 𝑥 𝑑𝑥
𝑎
𝑥1
𝑏 − 𝑎𝑑𝑥 =
1
𝑏 − 𝑎
𝑎
𝑥
𝑑𝑥 = 𝑥
𝑏 − 𝑎
𝑥𝑎
=𝑥 − 𝑎
𝑏 − 𝑎
𝐹 𝑥 =
𝑑𝑙𝑎 𝑥 ∈ (−∞, 𝑎)
=
−∞
𝑥
0𝑑𝑥 = 0
𝑑𝑙𝑎 𝑥 ∈ [𝑎, 𝑏)
𝐹 𝑥 =
𝑑𝑙𝑎 𝑥 ∈ [𝑏, +∞)
𝑎
𝑏1
𝑏 − 𝑎𝑑𝑥 +
𝑏
+∞
0𝑑𝑥 = 𝑥
𝑏 − 𝑎𝑏𝑎
=𝑏 − 𝑎
𝑏 − 𝑎= 1
1
𝑏 − 𝑎𝐹 𝑥 =
0 𝑑𝑙𝑎 𝑥 < 𝑎𝑥 − 𝑎
𝑏 − 𝑎𝑑𝑙𝑎 𝑎 ≤ 𝑥 < 𝑏
1 𝑑𝑙𝑎 𝑥 ≥ 𝑏
0 a bx
f(x)
0 a bx
F(x)
1
EX
𝐸𝑋 =𝑏 + 𝑎
2
𝐷2𝑋 =(𝑏 − 𝑎)2
12
𝑓 𝑥 = 0 𝑑𝑙𝑎 𝑥 < 0
𝜆𝑒−𝜆𝑥 𝑑𝑙𝑎 𝑥 ≥ 0, 𝜆 > 0
x
f(x)
𝐸𝑋 =1
𝜆𝐷2𝑋 =
1
𝜆2
𝐹 𝑥 = 0 𝑑𝑙𝑎 𝑥 < 0
1 − 𝑒−𝜆𝑥 𝑑𝑙𝑎 𝑥 ≥ 0, 𝜆 > 0
F(x)
x
𝑓 𝑥 =
0 𝑑𝑙𝑎 𝑥 < 0
𝑘
𝜆
𝑥
𝜆
𝑘−1
𝑒−( 𝑥
𝜆)𝑘
𝑑𝑙𝑎 𝑥 ≥ 0, 𝜆 > 0
𝐹 𝑥 = 0 𝑑𝑙𝑎 𝑥 < 0
1 − 𝑒−( 𝑥
𝜆)𝑘
𝑑𝑙𝑎 𝑥 ≥ 0, 𝜆 > 0
𝜙 𝑥 =1
𝜎 2𝜋𝑒
(−𝑥−𝑚 2
2𝜎2)
( , s)
(0, 1)
𝜙 𝑥 =1
2𝜋𝑒(−
𝑥2
2 )
𝑌 = 𝑋1 + 𝑋2 + ⋯ + 𝑋𝑛𝑡𝑜 𝑌 𝑚𝑎 𝑟𝑜𝑧𝑘ł𝑎𝑑 𝑁 0,1 𝑜𝑟𝑎𝑧
𝐸𝑌 = 𝐸𝑋1 + 𝐸𝑋2 + ⋯ + 𝐸𝑋𝑛
𝐷2𝑌 = 𝐷2𝑋1 + 𝐷2𝑋2 + ⋯ + 𝐷
2𝑋𝑛
Φ 𝑥 =
−∞
𝑥1
2𝜋𝑒(−
𝑥2
2 )𝑑𝑥
Φ 𝑥 =1
21 + 𝑒𝑟𝑓
𝑧
2
𝑓 𝑥 =
0 𝑑𝑙𝑎 𝑥 < 0𝜃𝑘
Γ(𝑘)𝑥𝑘−1𝑒(−𝜃𝑥) 𝑑𝑙𝑎 𝑥 ≥ 0, 𝑏 > 0
Γ 𝑘 =
0
∞
𝑥𝑘−1𝑒−𝑥𝑑𝑥
𝐹 𝑥 =𝜃𝑘
Γ(𝑘)
0
𝑥
𝑥𝑝−1 𝑒(−𝜃𝑥)𝑑𝑥
𝐸𝑋 =𝑘
𝜃𝐸𝑋2 =
𝑘(𝑘 + 1)
𝜃2𝐷2𝑋 =
𝑘
𝜃2
3