Question

Assume that the random variable X is normally distributed, with mean p = 100 and standard deviation o = 15. Compute the

probability P(X > 112).

Answer 1

P(X > 112) = 0.21186.

Explanation:

We are given that the random variable X is normally distributed, with mean
`\mu` = 100 and standard deviation
`\sigma` = 15.

Let X = a random variable

The z-score probability distribution for the normal distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = population mean = 100

`\sigma` = standard deviaton = 15

Now, the probability that the random variable X is greater than 112 is given by = P(X > 112)

P(X > 112) = P(
`(X-\mu)/(\sigma)` >
`(112-100)/(15)` ) = P(Z > 0.80) = 1- P(Z
`\leq` 0.80)

= 1 - 0.78814 = 0.21186

The above probability is calculated by looking at the value of x = 0.80 in the z table which has an area of 0.78814.