Chapter 12, Problem 25SGR

To determine

The probability for the given condition.

Answer to Problem 25SGR

The probability is 0.9772.

Explanation of Solution

Given:

The normal distribution and the mean is given as:

  $\mu =121,\sigma =9,P\left(x>103\right)$

The z- score for the x:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ \text{where,}\\ x=103\\ \text{Hence,}\\ z=\frac{103-121}{9}\\ =-2\end{array}$

Evaluating the probability that the random variable x has value greater than 103:

  $\begin{array}{l}P\left(x>103\right)=P\left(z>-2\right)\\ =1-P\left(z>-2\right)\\ =1-0.0228\\ =0.9772\end{array}$

Therefore, the probability is 09772.