Practice 1 Solutions

1. $z=–2.36$

$\mathrm{P}(z<–2.36)=0.0091=0.91\%$

2. $z=0.48$

$\mathrm{P}(z<0.48)=0.6844=68.44\%$

3. 0.9642

$z=1.80$

4. 0.1207

$z=–1.17$

5. $z=–1.35, \mu =13.45, X=5.1$

$\begin{array}{rcl}–1.35& =& \frac{5.1–13.45}{\sigma }\\ –1.35\sigma & =& –8.35\end{array}$

$\sigma =6.19$

6. $\mathrm{P}=0.624, \mu =24.74, \sigma =4.32$

$\begin{array}{rcl}0.624& \approx & 0.6255\\ z& =& 0.32\\ 0.32& =& \frac{X–24.74}{4.32}\\ 1.3824& =& X–24.74\end{array}$

$X=26.12$

7. What proportion of fish have a length smaller than 10 centimeters?

$\begin{array}{rcl}z& =& \frac{X–\mu }{\sigma }\\ z& =& \frac{10–10.12}{1.57}\\ z& =& –0.0764\approx –0.08\end{array}$

$\mathrm{P}\left(\mathrm{length}<10 \mathrm{cm}\right)=0.4681$

8. What proportion of fish were between 7.5 centimeters and 11 centimeters in length? 

$$\begin{gathered} z=\frac{X–\mu }{\sigma } \\ \begin{array}{cc}z=\frac{7.5–10.12}{1.57}\begin{array}{c}\end{array}& z=\frac{11–10.12}{1.57}\\ z=–1.67& z=0.56\end{array} \\ \mathrm{P}\left(\mathrm{length} <7.5\right)=0.0475 \mathrm{P}\left(\mathrm{length}<11\right)=0.7123 \\ 0.7123–0.0475=0.6648 \end{gathered}$$

$\mathrm{P}\left(7.5<\mathrm{length}<11\right)=0.6648$

9. What proportion of fish were longer than 11.5 centimeters?

$$\begin{gathered} z=\frac{X–\mu }{\sigma } \\ \begin{array}{l}X=11.5\\ z=\frac{11.5–10.12}{1.57}\\ z=0.88\\ 1–\mathrm{P}\left(\mathrm{length} <11.5\right)=1–0.8106\\ \mathrm{P}\left(\mathrm{length}>11.5\right)=0.1894\end{array} \end{gathered}$$

$\mathrm{P}\left(\mathrm{length}>11.5\right)=0.1894$

10. What length would represent 93% of the fish in the lake?

$\begin{array}{rcl}0.93& \approx & 0.9306\\ z& =& 1.48\\ 1.48& =& \frac{X–10.12}{1.57}\\ 2.3236& =& X–10.12\\ X& =& 12.4436\end{array}$

The length of 93% of fish will be 12.44 centimeters or less.

11. How long is the commute for 90% of Time Inc. employees?

$\begin{array}{rcl}0.90& \approx & 0.8997\\ z& =& 1.28\\ 1.28& =& \frac{X–20}{5}\\ 6.4& =& X–20\\ X& =& 26.4 \end{array}$

The commute is about 26 minutes or less for 90% of employees.

12. What proportion of employees have a commute of 10 minutes or less?

$\begin{array}{rcl}X& =& 10\\ z& =& \frac{10–20}{5}\\ z& =& –2\end{array}$

$\mathrm{P}\left(\mathrm{commute}<10\right)=0.0228$

13. What is the maximum expected commute time for 6% of employees?

$\begin{array}{rcl}0.06& \approx & 0.0606\\ z& =& –1.55\\ –1.55& =& \frac{X–20}{5}\\ –7.75& =& X–20\\ X& =& 12.25\end{array}$

The maximum commute time is about 12 minutes for 6% of employees.

14. Determine the proportion of employees who have a commute between 15 and 20 minutes at Time Inc.

$$\begin{gathered} z=\frac{X–\mu }{\sigma } \\ \begin{array}{ll}X=15& X=20\\ z=\frac{15–20}{5}& z=\frac{20–20}{5}\\ z=–1.000& z=0\\ \mathrm{P}\left(\mathrm{commute} <15\right)=0.1587& \mathrm{P}\left(\mathrm{commute}<20\right)=0.5000\\ 0.5000–0.1587=0.3413& \end{array} \end{gathered}$$

$\mathrm{P}\left(15<\mathrm{commute}<20\right)=0.3413$

15. Complete the table to determine if the sample group of students represents a normal data set.

16. Is the sample normally distributed? Explain.

Sample: The sample is normally distributed because the predicted number of data points and the actual number of data points are nearly identical.