X ∼ B i n ( n , p ) X\sim Bin(n, p) X ∼ B in ( n , p )
Given n = 6 , p = 0.25 n=6, p=0.25 n = 6 , p = 0.25
1.
μ = n p = 6 ( 0.25 ) = 1.5 \mu=np=6(0.25)=1.5 μ = n p = 6 ( 0.25 ) = 1.5 The mean u = 1.5 u=1.5 u = 1.5
2.
V a r ( X ) = σ 2 = n p ( 1 − p ) = 6 ( 0.25 ) ( 1 − 0.25 ) = 1.125 Var(X)=\sigma^2 =np(1-p)=6(0.25)(1-0.25)=1.125 Va r ( X ) = σ 2 = n p ( 1 − p ) = 6 ( 0.25 ) ( 1 − 0.25 ) = 1.125
standard deviation = σ = 1.125 = 0.75 2 ≈ 1.06066 \text{standard deviation}=\sigma=\sqrt{1.125}=0.75\sqrt{2}\approx1.06066 standard deviation = σ = 1.125 = 0.75 2 ≈ 1.06066 The standard deviation a = 0.75 2 ≈ 1.06066 ≠ 1.125 a=0.75\sqrt{2}\approx1.06066\not=1.125 a = 0.75 2 ≈ 1.06066 = 1.125
3.
P ( X = 3 ) = ( 6 3 ) ( 0.25 ) 3 ( 1 − 0.25 ) 6 − 3 = P(X=3)=\dbinom{6}{3}(0.25)^3(1-0.25)^{6 -3}= P ( X = 3 ) = ( 3 6 ) ( 0.25 ) 3 ( 1 − 0.25 ) 6 − 3 =
= 6 ! 3 ! ( 6 − 3 ) ! ( 1 64 ) ( 27 64 ) = 135 1024 = 0.1318359375 =\dfrac{6!}{3!(6-3)!}(\dfrac{1}{64})(\dfrac{27}{64})=\dfrac{135}{1024}=0.1318359375 = 3 ! ( 6 − 3 )! 6 ! ( 64 1 ) ( 64 27 ) = 1024 135 = 0.1318359375 P ( X = 3 ) = 0.1318359375 P(X=3)=0.1318359375 P ( X = 3 ) = 0.1318359375
4.
P ( X = 4 ) = ( 6 4 ) ( 0.25 ) 4 ( 1 − 0.25 ) 6 − 4 = P(X=4)=\dbinom{6}{4}(0.25)^4(1-0.25)^{6 -4}= P ( X = 4 ) = ( 4 6 ) ( 0.25 ) 4 ( 1 − 0.25 ) 6 − 4 =
= 6 ! 4 ! ( 6 − 4 ) ! ( 1 256 ) ( 9 16 ) = 135 4096 = 0.032958984375 =\dfrac{6!}{4!(6-4)!}(\dfrac{1}{256})(\dfrac{9}{16})=\dfrac{135}{4096}=0.032958984375 = 4 ! ( 6 − 4 )! 6 ! ( 256 1 ) ( 16 9 ) = 4096 135 = 0.032958984375
P ( X = 5 ) = ( 6 5 ) ( 0.25 ) 5 ( 1 − 0.25 ) 6 − 5 = P(X=5)=\dbinom{6}{5}(0.25)^5(1-0.25)^{6 -5}= P ( X = 5 ) = ( 5 6 ) ( 0.25 ) 5 ( 1 − 0.25 ) 6 − 5 =
= 6 ! 5 ! ( 6 − 5 ) ! ( 1 1024 ) ( 3 4 ) = 9 2048 = 0.00439453125 =\dfrac{6!}{5!(6-5)!}(\dfrac{1}{1024})(\dfrac{3}{4})=\dfrac{9}{2048}=0.00439453125 = 5 ! ( 6 − 5 )! 6 ! ( 1024 1 ) ( 4 3 ) = 2048 9 = 0.00439453125
P ( X = 6 ) = ( 6 6 ) ( 0.25 ) 6 ( 1 − 0.25 ) 6 − 6 = P(X=6)=\dbinom{6}{6}(0.25)^6(1-0.25)^{6 -6}= P ( X = 6 ) = ( 6 6 ) ( 0.25 ) 6 ( 1 − 0.25 ) 6 − 6 =
= 1 4096 = 0.000244140625 =\dfrac{1}{4096}=0.000244140625 = 4096 1 = 0.000244140625
P ( X > 4 ) = P ( X = 5 ) + P ( X = 6 ) = P(X>4)=P(X=5)+P(X=6)= P ( X > 4 ) = P ( X = 5 ) + P ( X = 6 ) = = 0.004638671875 =0.004638671875 = 0.004638671875
P ( X ≥ 4 ) = P ( X = 4 ) + P ( X = 5 ) + P ( X = 6 ) = P(X\geq4)=P(X=4)+P(X=5)+P(X=6)= P ( X ≥ 4 ) = P ( X = 4 ) + P ( X = 5 ) + P ( X = 6 ) = = 0.03759765625 =0.03759765625 = 0.03759765625
P ( X ≤ 4 ) = 1 − P ( X = 5 ) − P ( X = 6 ) = P(X\leq4)=1-P(X=5)-P(X=6)= P ( X ≤ 4 ) = 1 − P ( X = 5 ) − P ( X = 6 ) = = 0.995361328125 =0.995361328125 = 0.995361328125
P ( X < 4 ) = 1 − P ( X ≥ 4 ) = P(X<4)=1-P(X\geq4)= P ( X < 4 ) = 1 − P ( X ≥ 4 ) = = 0.96240234375 =0.96240234375 = 0.96240234375
5.
P ( X = 2 ) = ( 6 2 ) ( 0.25 ) 2 ( 1 − 0.25 ) 6 − 2 = P(X=2)=\dbinom{6}{2}(0.25)^2(1-0.25)^{6 -2}= P ( X = 2 ) = ( 2 6 ) ( 0.25 ) 2 ( 1 − 0.25 ) 6 − 2 =
= 6 ! 2 ! ( 6 − 2 ) ! ( 1 16 ) ( 81 256 ) = 1215 4096 = 0.296630859375 =\dfrac{6!}{2!(6-2)!}(\dfrac{1}{16})(\dfrac{81}{256})=\dfrac{1215}{4096}=0.296630859375 = 2 ! ( 6 − 2 )! 6 ! ( 16 1 ) ( 256 81 ) = 4096 1215 = 0.296630859375
P ( X = 0 ) = ( 6 0 ) ( 0.25 ) 0 ( 1 − 0.25 ) 6 − 0 = P(X=0)=\dbinom{6}{0}(0.25)^0(1-0.25)^{6 -0}= P ( X = 0 ) = ( 0 6 ) ( 0.25 ) 0 ( 1 − 0.25 ) 6 − 0 =
= 729 4096 = 0.177978515625 =\dfrac{729}{4096}=0.177978515625 = 4096 729 = 0.177978515625
P ( X = 1 ) = ( 6 1 ) ( 0.25 ) 1 ( 1 − 0.25 ) 6 − 1 = P(X=1)=\dbinom{6}{1}(0.25)^1(1-0.25)^{6 -1}= P ( X = 1 ) = ( 1 6 ) ( 0.25 ) 1 ( 1 − 0.25 ) 6 − 1 =
= 6 ( 1 4 ) ( 243 1024 ) = 729 2048 = 0.35595703125 =6(\dfrac{1}{4})(\dfrac{243}{1024})=\dfrac{729}{2048}=0.35595703125 = 6 ( 4 1 ) ( 1024 243 ) = 2048 729 = 0.35595703125
P ( X < 2 ) = P ( X = 0 ) + P ( X = 1 ) = P(X<2)=P(X=0)+P(X=1)= P ( X < 2 ) = P ( X = 0 ) + P ( X = 1 ) = = 0.533935546875 =0.533935546875 = 0.533935546875
P ( X ≤ 2 ) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) = P(X\leq2)=P(X=0)+P(X=1)+P(X=2)= P ( X ≤ 2 ) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) = = 0.83056640625 =0.83056640625 = 0.83056640625
P ( X > 2 ) = 1 − P ( X ≤ 2 ) = P(X>2)=1-P(X\leq2)= P ( X > 2 ) = 1 − P ( X ≤ 2 ) = = 0.16943359375 =0.16943359375 = 0.16943359375
P ( X ≥ 2 ) = 1 − P ( X < 2 ) = P(X\geq2)=1-P(X<2)= P ( X ≥ 2 ) = 1 − P ( X < 2 ) = = 0.466064453125 =0.466064453125 = 0.466064453125
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