Chapter 11.5, Problem 11PPS

(a)

To determine

To find: The number of members went to the gym at least $50$ times.

Answer to Problem 11PPS

The number of members went to the gym at least $50$ times are $509$ .

Explanation of Solution

Given:

The sample are $522$ .

The mean is $88$ .

The standard deviation is $19$ .

Calculation:

Find the z-value for $X=50$ .

  $\begin{array}{c}z=\frac{X-\mu }{\sigma }\\ =\frac{50-88}{19}\\ =-2\end{array}$

Find the z-value for $X=42.7$ .

  $\begin{array}{c}z=\frac{X-\mu }{\sigma }\\ =\frac{42.7-29.3}{6.7}\\ =2\end{array}$

So, $50$ is $2\sigma$ away from the mean.

The $42.7$ is $2\sigma$ , more than the mean, and $22.6$ is $1\sigma$ less than the mean.

According to Empirical rule, the percent of data more than 50 is,

  $13.5\%+34\%+34\%+13.5\%+2.35\%+0.15\%=97.5\%$

The number of members went to the gym at least 50 times are as follows.

  $\begin{array}{c}97.5\%\text{of}522=\frac{97.5}{100}\left(522\right)\\ =508.9\\ \approx 509\end{array}$

Therefore, the number of members went to the gym at least $50$ times are $509$ .

(b)

To determine

To find: The probability that a member selected at random went to the gym more than $145$ times.

Answer to Problem 11PPS

The probability that a member selected at random went to the gym more than $145$ times is $0.15\%$ .

Explanation of Solution

Given:

The sample are $522$ .

The mean is $88$ .

The standard deviation is $19$ .

Calculation:

Find the z-value for $X=145$ .

  $\begin{array}{c}z=\frac{X-\mu }{\sigma }\\ =\frac{145-88}{19}\\ =3\end{array}$

So, the value more than $145$ are the values more than $\mu +3\sigma$ . According to Empirical rule, $0.15\%$ of the values are more than $3\sigma$ .

Therefore, the probability that a member selected at random went to the gym more than $145$ times is $0.15\%$ .