Find Var(X) ifP(X=a)=p=1P(X=b)

Short Answer

Expert verified

We have found to beVar(X)=(a-b)2(1-p)p

Step by step solution

01

Given information

Given in the question is,

P(X=a)=p=1-P(X=b)

02

Substitution

The random variable Xcan consider only two values, awith probability pand bwith the probability 1-p.

Using the meaning of the mean, we have that

E(X)=a×p+b×(1-p)=:μ

03

Calculation

Currently, operating the definition of the variance, we have that

Var(X)=E(X-μ)2=(a-μ)2·p+(b-μ)2·(1-p)

=(a-a×p-b×(1-p))2p+(b-a×p+b×(1-p))2(1-p)

Substitute the given expression,

=(a-b)2(1-p)2p+(b-a)2p2(1-p)

role="math" localid="1646669147074" =(a-b)2(1-p)p[1-p+p]

We get,

=(a-b)2(1-p)p.

04

Final answer

The solution of theVar(X)is found to be(a-b)2(1-p)p.

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