Question #252132

If X is binomially distributed with mean 3.20 and variance 1.152, find the complete binomial probability distribution


1
Expert's answer
2021-10-19T02:44:47-0400

Let u - mean, d - variance

Since we have binpmial distribution, then u = np, d = np(1-p), where n - number of experiments, p - probability of succes in experiment

Then we have system of two equations {np=3.2np(1p)=1.152}\begin{Bmatrix} np = 3.2 \\ np(1-p) = 1.152 \end{Bmatrix} . After solving it, we receive n = 5 and p = 0.64

So, given distribution is Bin(5, 0.64)

Complete binomial probability distribution is the sum of P(x≤n) for every n from 0 to 5, where x is the number of successful experiments. Let's mark the complete binomial probability distribution as P(n) for n from 0 to 5.

P(0)=0P(0) = 0

P(1)=0+P(1)=(51)(0.64)1(0.36)4=0.0537P(1) = 0 + P(1) = {5 \choose 1}*(0.64)^{1}*(0.36)^{4} = 0.0537

P(2)=0.0537+P(2)=0.0537(52)(0.64)2(0.36)3=0.245P(2) = 0.0537 + P(2) =0.0537 {5 \choose 2}*(0.64)^{2}*(0.36)^{3}=0.245

P(3)=0.245+P(3)=0.245+(53)(0.64)3(0.36)2=0.585P(3) = 0.245 + P(3) = 0.245+ {5 \choose 3}*(0.64)^{3}*(0.36)^{2}=0.585

P(4)=0.585+P(4)=0.585+(54)(0.64)4(0.36)1=0.887P(4) = 0.585 + P(4) =0.585+ {5 \choose 4}*(0.64)^{4}*(0.36)^{1} = 0.887

P(5)=1P(5) = 1



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