Question

Let x be a normal random variable with mean 20 and standard deviation 4. find p(16<X<22).

A) 0.4672

B) 0.5328

C) 0.1498

D) 0.0328

Answer 1

Hello

YOUR REQUIRED ANSWER IS OPTION B : 0.5328

Given :

X is a normal random variable with $\mu$ = 20 and $\sigma$ = 4.

We can find the z-score using the formula:

$z = \frac{x - \mu}{\sigma}$

Hence, z score for 16 is :

16-20

and z score for 22 is :

$z = \frac{22 - 20}{4} = 0.5$

Hence, the required probability p(16<x<22) = p(-1 < z < 0.5) = p(z < 0.5) - p(z < -1)

= 0.691462 - 0.158655 = 0.532807

I hope this solves your query.

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