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

Answer:

P(X > 112) = 0.21186.

Step-by-step 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  =  $\frac{X-\mu}{\sigma}$  ~ 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( $\frac{X-\mu}{\sigma}$ > $\frac{112-100}{15}$ ) = 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.