Chapter 6.3, Problem 16E

Solution

To find: the probability that a randomly selected x-value from the distribution is in the given interval

  $P\left(x\le 37\right)=0.84$

Given information:

The given interval is: at most 37

A normal distribution has a mean of 33 and a standard deviation of 4.

Concept Used:

Areas under a Normal Curve

A normal distribution with mean $\overline{x}$ and standard deviation $\sigma$ has the following properties:

• The total area under the related normal curve is 1.
• About 68% of the area lies within 1 standard deviation of the mean.
• About 95% of the area lies within 2 standard deviation of the mean.
• About 99.7% of the area lies within 3 standard deviation of the mean.

  

Explanation:

The normal distribution curve for the given distribution with mean $\overline{x}=33$ and standard deviation $\sigma =4$ is:

  

The required interval of at most 37 represents the shaded area under the normal curve (it is the area lies before $\overline{x}+\sigma$ ) as shown below:

  

So, the probability that a randomly selected x-value from the distribution is in the given interval is:

  $\begin{array}{c}P\left(x\le 37\right)=0.34+0.34+0.135+0.0235+0.0015\\ =0.84\end{array}$