Chapter 11.3, Problem 3Q

To determine

To find: The probability.

Answer to Problem 3Q

The required probability is given by $P\left(z\le -1.5\right)=0.0668$

Explanation of Solution

Given:

  $\mu =32.$

  $\sigma =4.$

Calculation:

In this question you have that a normal distribution has a mean of $32$ and standard deviation of $4$ . That is $\mu =32$ and $\sigma =4.$

You have to find the probability $P\left(x\le 26\right).$

So, for that you need to find the $z-score$ corresponding to an $x-valueof26.$

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{26-32}{4}\\ =-1.5\end{array}$

Thus the required probability is given by the value of $P\left(z\le -1.5\right).$

Now, the standard normal table to find $P\left(z\le -1.5\right).$

To find $P\left(z\le -1.5\right)$ , using standard normal table, you have to find the value where $row-1andcolumn.5$ intersects that is $\mathrm{.0668.}$

This value is the probability $P\left(z\le -1.5\right).$

  

The required probability is given by

  $P\left(z\le -1.5\right)=0.0668$

Conclusion:

Hence, the required probability is given by $P\left(z\le -1.5\right)=0.0668$