Chapter 14.5, Problem 26E

a.

To determine

Find the mean of the data.

Answer to Problem 26E

  $12.375$

Explanation of Solution

Given information:

Given: The distribution shows the random sample of the weekend shoppers at the store.

  $\begin{array}{llllllllll}NumberofMinutes\hfill & 3-5\hfill & 5-7\hfill & 7-9\hfill & 9-11\hfill & 11-13\hfill & 13-15\hfill & 15-17\hfill & 17-19\hfill & 19-21\hfill \\ Frequency\hfill & 1\hfill & 3\hfill & 5\hfill & 12\hfill & 17\hfill & 13\hfill & 7\hfill & 4\hfill & 2\hfill \end{array}$

Calculation:

Here, we have to find the mean for the given data:

  $\begin{array}{l}Mean\left(\overline{X}\right)=\frac{{\displaystyle \sum _1^n{f}_i{x}_i}}{{\displaystyle \sum _1^n{f}_i}}\\ Mean\left(\overline{X}\right)=\frac{1.4+3.6+8.5+10.12+12.17+14.13+16.7+18.4+20.2}{64}\\ Mean\left(\overline{X}\right)=12.375\end{array}$

Hence, the mean is $12.375$ .

b.

To determine

Find the standard deviation of the data.

Answer to Problem 26E

  $3.37$

Explanation of Solution

Given information:

Given: The distribution shows the random sample of the weekend shoppers at the store.

  $\begin{array}{llllllllll}NumberofMinutes\hfill & 3-5\hfill & 5-7\hfill & 7-9\hfill & 9-11\hfill & 11-13\hfill & 13-15\hfill & 15-17\hfill & 17-19\hfill & 19-21\hfill \\ Frequency\hfill & 1\hfill & 3\hfill & 5\hfill & 12\hfill & 17\hfill & 13\hfill & 7\hfill & 4\hfill & 2\hfill \end{array}$

Calculation:

Here, we have to find the standard deviation for the given data:

  $\begin{array}{l}\sigma =\sqrt{\frac{\begin{array}{l}{\left(4-12.375\right)}^2+{\left(6-12.375\right)}^2+{\left(8-12.375\right)}^2+{\left(10-12.375\right)}^2+{\left(12-12.375\right)}^2\\ +{\left(14-12.375\right)}^2+{\left(16-12.375\right)}^2+{\left(18-12.375\right)}^2+{\left(20-12.375\right)}^2\end{array}}{64}}\\ \sigma =3.37\end{array}$

Hence, the standard deviation is $3.37$ .

c.

To determine

Find the standard error of the mean.

Answer to Problem 26E

  $0.42$

Explanation of Solution

Given information:

Given: The distribution shows the random sample of the weekend shoppers at the store.

  $\begin{array}{llllllllll}NumberofMinutes\hfill & 3-5\hfill & 5-7\hfill & 7-9\hfill & 9-11\hfill & 11-13\hfill & 13-15\hfill & 15-17\hfill & 17-19\hfill & 19-21\hfill \\ Frequency\hfill & 1\hfill & 3\hfill & 5\hfill & 12\hfill & 17\hfill & 13\hfill & 7\hfill & 4\hfill & 2\hfill \end{array}$

Calculation:

Here, we have to find the standard error of the mean:

  $\begin{array}{l}{\sigma }_{\overline{X}}=\frac{standarddeviation}{\sqrt{N}}\\ {\sigma }_{\overline{X}}=\frac{3.37}{\sqrt{64}}=0.42\end{array}$

Hence, the standard error is $0.42$ .

d.

To determine

Find the interval about the sample mean having $0.95$ chance that the true mean lies within the interval.

Answer to Problem 26E

  $\left[11.55,13.20\right]$

Explanation of Solution

Given information:

Given: The distribution shows the random sample of the weekend shoppers at the store.

  $\begin{array}{llllllllll}NumberofMinutes\hfill & 3-5\hfill & 5-7\hfill & 7-9\hfill & 9-11\hfill & 11-13\hfill & 13-15\hfill & 15-17\hfill & 17-19\hfill & 19-21\hfill \\ Frequency\hfill & 1\hfill & 3\hfill & 5\hfill & 12\hfill & 17\hfill & 13\hfill & 7\hfill & 4\hfill & 2\hfill \end{array}$

Calculation:

Here, we have to find the interval:

  $\begin{array}{l}P=95\%correspondstot=1.96\\ \left[\overline{X}-t{\sigma }_X,\overline{X}+t{\sigma }_X\right]\\ \left[11.55,13.20\right]\end{array}$

Hence, the interval is $\left[11.55,13.20\right]$ .

e.

To determine

Find the probability that the mean of population will be less than one minute from the mean of sample.

Answer to Problem 26E

  $P=98.4\%$

Explanation of Solution

Given information:

Given: The distribution shows the random sample of the weekend shoppers at the store.

  $\begin{array}{llllllllll}NumberofMinutes\hfill & 3-5\hfill & 5-7\hfill & 7-9\hfill & 9-11\hfill & 11-13\hfill & 13-15\hfill & 15-17\hfill & 17-19\hfill & 19-21\hfill \\ Frequency\hfill & 1\hfill & 3\hfill & 5\hfill & 12\hfill & 17\hfill & 13\hfill & 7\hfill & 4\hfill & 2\hfill \end{array}$

Calculation:

Here, we have to find the probability that the mean of population will be less than one minute from the mean of sample.

Now, it belongs to the interval:

  $\begin{array}{l}\left[\overline{X}-1,\overline{X}+1\right]\\ Findthevalueoft:\\ t{\sigma }_{\overline{X}}=1\\ t\left(0.42\right)=1\\ t=\frac{1}{0.42}=2.38\\ t=2.38correspondstoP=98.4\%\\ P=98.4\%\end{array}$

Hence, the probability is $P=98.4\%$ .