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Suppose that X is a normal random variable with mean 5. If PX>9=.2, approximately what is Var(X)?

Short Answer

Expert verified

The required variance is 22.66.

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.

Here,

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

Hence, VarX=22.66.

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