Chapter 2, Problem 2.65SE

a.

To determine

Find the range for the given data set.

Answer to Problem 2.65SE

The range is $6.8$ .

Explanation of Solution

Given:

The data shows that the miles per gallon for each of twenty medium sized cars.

  

Calculation:

Arrange the data values from smallest to largest.

  $\begin{array}{l}20.2,21.3,22.2,22.7,22.9,23.1,23.2,23.6,23.7,24.2,24.4,24.4,24.6,24.7\\ 24.7,24.9,25.3,25.9,26.2,27\end{array}$

The minimum value $=$

  $20.2$

The maximum value $=27$

Range $=$ max-min

Range $=27-20.2=6.8$

Hence the range is $6.8$ .

b.

To determine

Draw a relative frequency histogram.

Answer to Problem 2.65SE

The standard deviation $s=8.025$ .

Explanation of Solution

Given:

The data shows that the miles per gallon for each of twenty medium sized cars.

  

Calculation:

Arrange the data values from smallest to largest.

  $\begin{array}{l}20.2,21.3,22.2,22.7,22.9,23.1,23.2,23.6,23.7,24.2,24.4,24.4,24.6,24.7\\ 24.7,24.9,25.3,25.9,26.2,27\end{array}$

Draw a relative frequency table.

Class	Frequency	Relative frequency $=\frac{\text{frequency}}{\text{total}\text{frequency}}$
$\begin{array}{l}20-<21\\ 21-<22\\ 22-<23\\ 23-<24\\ 24-<25\\ 25-<26\\ 26-<27\end{array}$	$\begin{array}{l}1\\ 1\\ 3\\ 4\\ 7\\ 2\\ 2\end{array}$	$\begin{array}{l}\frac{1}{20}=0.05\\ \frac{1}{20}=0.05\\ \frac{3}{20}=0.15\\ \frac{4}{20}=0.20\\ \frac{7}{20}=0.35\\ \frac{2}{20}=0.10\\ \frac{2}{20}=0.10\end{array}$

The histogram is

  

The width of the bars has to be the same.

The height of the bars has to be equal to the relative frequency.

The distribution is slightly skewed left, because highest bar is right and tail of smaller bars to the left in histogram.

c.

To determine

Find the mean and the standard deviation for the given data.

Answer to Problem 2.65SE

The value of mean $\overline{x}=23.96$ and the standard deviation $s=1.64$ .

Explanation of Solution

Given:

The data shows that the miles per gallon for each of twenty medium sized cars.

  

Calculation:

Arrange the data values from smallest to largest.

  $\begin{array}{l}20.2,21.3,22.2,22.7,22.9,23.1,23.2,23.6,23.7,24.2,24.4,24.4,24.6,24.7\\ 24.7,24.9,25.3,25.9,26.2,27\end{array}$

  $\text{Mean}=\frac{\text{sum}\text{of}\text{all}\text{values}}{\text{total}\text{number}\text{of}\text{values}}$

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =\frac{\begin{array}{l}20.2+21.3+22.2+22.7+22.9+23.1+23.2+23.6+23.7+24.2+24.4+24.4+24.6+24.7+\\ 24.7+24.9+25.3+25.9+26.2+27\end{array}}{20}\\ =\frac{479.2}{20}\\ =23.96\end{array}$

find the sample variance $s^2=\frac{{\displaystyle \sum x_i{}^2-\frac{{\left({\displaystyle \sum x_i}\right)}^2}{n}}}{n-1}$

  $n=$ total number of data values.

  $n=20$

  $\begin{array}{l}{\displaystyle \sum x_i}=20.2+21.3+22.2+22.7+22.9+23.1+23.2+23.6+23.7+24.2+24.4+24.4+24.6+24.7+\\ 24.7+24.9+25.3+25.9+26.2+27=479.2\\ {\displaystyle \sum x_i{}^2}={20.2}^2+{21.3}^2+{22.2}^2+{22.7}^2+{22.9}^2+{23.1}^2+{23.2}^2+{23.6}^2+{23.7}^2+{24.2}^2+{24.4}^2+{24.4}^2+{24.6}^2+{24.7}^2+\\ {24.7}^2+{24.9}^2+{25.3}^2+{25.9}^2+{26.2}^2+{27}^2=11532.8\end{array}$

  $\begin{array}{l}{s}^2=\frac{11532.8-\frac{{479.2}^2}{20}}{20-1}\\ =\frac{11532.8-\frac{{479.2}^2}{20}}{19}\\ \approx 2.69\end{array}$

  $s=\sqrt{2.69}=1.64$

Hence the value of mean $\overline{x}=23.96$ and the standard deviation $s=1.64$ .

d.

To determine

Find the median for the given data set.

Answer to Problem 2.65SE

The median is $m=24.3$ .

Explanation of Solution

Given:

The data shows that the miles per gallon for each of twenty medium sized cars.

  

Calculation:

Arrange the data values from smallest to largest.

  $\begin{array}{l}20.2,21.3,22.2,22.7,22.9,23.1,23.2,23.6,23.7,24.2,24.4,24.4,24.6,24.7\\ 24.7,24.9,25.3,25.9,26.2,27\end{array}$

The median $=$ average of the two middle values (number of data values is even)$n=20$

  $n=$ total number of data values.

  $\begin{array}{l}m=\frac{\text{10th}\text{datavalue+11th}\text{data}\text{value}}{\text{2}}\\ =\frac{24.2+24.4}{2}\\ =\frac{48.6}{2}\\ =24.3\end{array}$

Hence the median is $m=24.3$ .

e.

To determine

Find the lower and upper quartiles.

Answer to Problem 2.65SE

The lower quartile is ${\text{Q}}_L=23.4$ and upper quartile is ${\text{Q}}_U=24.85$ .

Explanation of Solution

Given:

The data shows that the miles per gallon for each of twenty medium sized cars.

  

Calculation:

Arrange the data values from smallest to largest.

  $\begin{array}{l}20.2,21.3,22.2,22.7,22.9,23.1,23.2,23.6,23.7,24.2,24.4,24.4,24.6,24.7\\ 24.7,24.9,25.3,25.9,26.2,27\end{array}$

Use five-number summary $\to$ it consists of the minimum, the lower quartile, the median, the upper quartile and the maximum.

The minimum $=0.53$

The median $=$ average of the two middle values (number of data values is even)$n=20$

  $n=$ total number of data values.

  $\begin{array}{l}m=\frac{\text{10th}\text{datavalue+11th}\text{data}\text{value}}{\text{2}}\\ =\frac{24.2+24.4}{2}\\ =\frac{48.6}{2}\\ =24.3\end{array}$

For lower quartiles.

  $0.25\left(n+1\right)=0.25\left(20+1\right)=5.25\to \text{not an integer}$

The lower quartile is the $\text{5th}$ data value increased by $\text{0}\text{.25}$ times the difference between the $\text{5th}$ and the $\text{6th}$ data value.

  ${\text{Q}}_L\text{=22}\text{.9+0}\text{.25}\left(23.1-22.9\right)=22.9+0.05=23.4$

For upper quartiles.

  $0.75\left(n+1\right)=0.75\left(20+1\right)=15.75\to \text{not an integer}$

The lower quartile is the $\text{9th}$ data value increased by $\text{0}\text{.75}$ times the difference between the $\text{9th}$ and the $\text{10th}$ data value.

  ${\text{Q}}_U\text{=24}\text{.7+0}\text{.75}\left(24.9-24.7\right)=\text{24}\text{.7}+0.15=24.85$

Hence the lower quartile is ${\text{Q}}_L=23.4$ and upper quartile is ${\text{Q}}_U=24.85$ .