Question

The random variable x has a normal distribution with standard deviation 21. It is known that the probability that x exceeds 160 is .90. Find the mean mu of the probability distribution.

Answer 1

Answer:

Mean, $\mu$ = 133.09

Step-by-step explanation:

We are given that the random variable x has a normal distribution with standard deviation 21,i.e;

X ~ N($\mu,\sigma = 21$)

The Z probability is given by;

              Z = $\frac{X-\mu}{\sigma}$ ~ (0,1)

Also, it is known that the probability that x exceeds 160 is 0.90 ,i.e;

 P(X > 160) = 0.90

 P( $\frac{X-\mu}{\sigma}$ > $\frac{160-\mu}{21}$ ) = 0.90

From the z% table we find that at value of x = -1.2816, the value of

P(X > 160) is 90%

which means;     $\frac{160-\mu}{21}$ = -1.2816

                           160 - $\mu$ = 21*(-1.2816)

                             $\mu$ = 160 - 26.914 = 133.09

Therefore, mean $\mu$ of the probability distribution is 133.09.