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