Chapter 3, Problem 23CQ

a.

To determine

The mean of the given data.

Answer to Problem 23CQ

The mean of the given data is $15.3$.

Explanation of Solution

Given Info:

Total numbers in the data set are 10.

The data represent the number of highway miles per gallon of the 10 worst vehicles.

Calculation:

Formula to calculate the sample mean is,

$\text{Mean}=\frac{{\displaystyle \sum X}}{n}$

Substitute the value of $X_1,X_2\cdots X_{10}$ and 10 for $n$ in the above formula,

$\begin{array}{c}\text{Mean}=\frac{{\displaystyle \sum X}}{n}\\ =\frac{12+15+13+14+15+16+17+16+17+18}{10}\\ =\frac{153}{10}\\ =15.3\end{array}$

Thus, mean of the data is 15.3.

b.

To determine

The median of given data.

Answer to Problem 23CQ

The median of given data is $15.5$.

Explanation of Solution

Calculation:

Arrange the data in ascending order.

12

13

14

15

15

16

16

17

17

18

There is an even number of data values. That is 10.

The middle two data values are $15$ and $16$.

Determine the average of two middle values.

$\begin{array}{c}\text{Median}=\frac{15+16}{2}\\ =\frac{31}{2}\\ =15.5\end{array}$

Thus the median of the data is 15.5.

c.

To determine

The mode of given data.

Answer to Problem 23CQ

The mode of given data are 15, 16 and 17.

Explanation of Solution

Calculation:

Since the values 15, 16 and 17 occurs two times in given data, the mode is $15,16,\text{and}17$.

d.

To determine

The midrange of given data.

Answer to Problem 23CQ

The midrange of given data is $15$.

Explanation of Solution

Calculation:

Formula to calculate the mid range is,

$\text{MR=}\frac{\text{lowest}\text{value}\text{+}\text{highest}\text{value}}{\text{2}}$

Substitute 12 for lowest value and 18 for maximum value in the above formula,

$\begin{array}{c}\text{MR}=\frac{12+18}{2}\\ =\frac{30}{2}\\ =15\end{array}$

Thus the mid range of the data is 15.

e.

To determine

The range of given data.

Answer to Problem 23CQ

The range of given data 6.

Explanation of Solution

Calculation:

Formula to calculate the range of the data is,

$\text{R=}\text{highest}\text{value-}\text{lowest}\text{value}$

Substitute 12 for lowest value and 18 for highest value in the above formula to get range,

$\begin{array}{c}\text{R=}\text{highest}\text{value}-\text{lowest}\text{value}\\ \text{=18-12}\\ \text{=6}\end{array}$

Thus, the range of the data is 6.

f.

To determine

The variance of given data.

Answer to Problem 23CQ

The variance of given data is 3.57.

Explanation of Solution

Calculation:

Formula to calculate the variance of the data is,

${\sigma }^2=\frac{{\displaystyle \sum {(X-\overline{X})}^2}}{n-1}$

Determine the deviation for each data value and square each of the deviation,

X	$\mu$	$(X-\overline{X})$	${\left(X-\overline{X}\right)}^2$
12	$15.3$	$12-15.3=-3.3$	10.89
15	$15.3$	$15-15.3=-0.3$	0.09
13	$15.3$	$13-15.3=-2.3$	5.29
14	$15.3$	$14-15.3=-1.3$	1.69
15	$15.3$	$15-15.3=-0.3$	0.09
16	$15.3$	$16-15.3=0.7$	0.49
17	$15.3$	$17-15.3=1.7$	2.89
16	$15.3$	$16-15.3=0.7$	0.49
17	$15.3$	$17-15.3=1.7$	2.89
18	$15.3$	$18-15.3=2.7$	7.29

Determine the sum of the squares,

$\begin{array}{c}{{\displaystyle \sum \left(X-\overline{X}\right)}}^2=10.89+0.09+5.29+1.69+0.09+0.49+2.89+0.49+2.89+7.29\\ =32.1\end{array}$

Substitute 32.1 for ${{\displaystyle \sum \left(X-\mu \right)}}^2$ and 10 for n in the above formula to get population variance,

$\begin{array}{c}{\sigma }^2=\frac{{\displaystyle \sum {(X-\overline{X})}^2}}{n-1}\\ =\frac{32.1}{10-1}\\ =3.57\end{array}$

Thus, the population variance is 3.57.

g.

To determine

The standard deviation of the given data.

Answer to Problem 23CQ

The standard deviation of the given data is 1.9.

Explanation of Solution

Calculation:

Formula to calculate the sample standard deviation is,

$\sigma =\sqrt{\frac{{\displaystyle \sum {(X-\overline{X})}^2}}{n-1}}$

Take the square root of 3.57 to get population standard deviation.

$\begin{array}{c}\sigma =\sqrt{3.57}\\ =1.9\end{array}$

Thus the population standard deviation is 1.9