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Chapter 5: Q. 5.19 (page 213)

Let X be a normal random variable with mean 12and variance 4. Find the value of csuch that localid="1646649699736" PX>c=0.10.

Short Answer

Expert verified

The value of c is 14.56

Step by step solution

01

Step 1. Given information.

Xis a normal random variable with meanlocalid="1646649744005" 渭=12and variance蟽2=4

02

Step 2. Find value of c. 

As given in the question,

Mean渭=12and Variance 蟽2=4.

Hence, Standard Deviation 蟽=2.

As required,

PX>c=0.101-PX鈮�c=1-0.10PX鈮�c=0.90PX-M蟽鈮�c-M蟽=0.90PZ鈮�c-122=0.90c-122=1.28c=1.28脳2+12c=14.56

Hence, the required value of cis 14.56.

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