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Suppose that X is a normal random variable with

mean 5. If P{X > 9} = .2, approximately what is Var(X)?

Short Answer

Expert verified

The required variance is22.66.

Step by step solution

01

Step 1. Given Information.

Xis a random variable with mean 5.

Also,PX>9=0.2.

02

Step 2. Find VarX.

It is given that,

PX>9=0.2Then,PX-μσ>9-μσ=0.21-φ9-μσ=0.89-μσ=0.84σ=9-μ0.84σ=9-50.84σ=4.76σ2=4.762=22.66

Hence,VarX=22.66

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