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

Mean,
`\mu` = 133.09

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 =
`(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(
`(X-\mu)/(\sigma)` >
`(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;
`(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.