Answer to Question #320018 in Statistics and Probability for Mirylle Anne

Question

A normal random variable X has unknown mean and standard deviation. The probability that X exceeds 4 is 0.9722, and the probability that X exceeds 5 is 0.9332. Find mean and standard deviation.

Answer 1

P(X>4)=0.9772

P(X>5)=0.9332

1-0.9772=0.0228

1-0.9332=0.0668

From the z table the corresponding z values for the areas 0.0228 and 0.0668 are as follows :

z1 =-2

z2=-1.5

Z=(x-$\mu$)/$s$

-2=(4-$\mu$)/s

-2s=4-$\mu$

-1.5=(5-$\mu$)/s

-1.5s=5-$\mu$

Solving the equations simultaneously

-2s=4-$\mu$

-1.5s=5-$\mu$

Using substitution method

$\mu$ =4+2s

-1.5s=5-4-2s

0.5s=1

s=2

μ=8

Therefore the mean is 8 and the standard deviation is 2