Chapter 7.3, Problem 8Q

(a)

To determine

To find:

The mean, median, mode, range, and standard deviation of the prices .

Answer to Problem 8Q

Mean $=122$ , median $=109.2$ , mode $=98$ , range $=116$ , and standard deviation $=36$

Explanation of Solution

Given:

The table

  $x,\overline{x},x-\overline{x},{\left(x-\overline{x}\right)}^2$

Concept used:

The mean of the squared deviation or variance

  $=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}$

The square root of the variance

  $\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}$

Calculation:

The mean is

  $\begin{array}{l}\overline{x}=\frac{98+119+95+211+130+98+100+125}{8}=\frac{976}{8}\\ \overline{x}=122\end{array}$

Median

  $95,98,98,100,119,125,130,211$

  $\begin{array}{l}=\frac{100+119}{2}\\ =109.5\end{array}$

Mode

  $\begin{array}{l}95,98,98,100,119,125,130,211\\ \to 98\end{array}$

Range:-difference of largest and smallest value

  $\begin{array}{l}=211-95\\ =116\end{array}$

The mean of the squared deviation or variance

  $\begin{array}{l}=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}\\ =\frac{729+576+\mathrm{...........}+7921}{8}\\ =1296\end{array}$

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
$95$	$122$	$-27$	$729$
$98$	$122$	$-24$	$576$
$98$	$122$	$-24$	$576$
$100$	$122$	$-22$	$484$
$119$	$122$	$-3$	$9$
$125$	$122$	$3$	$9$
$130$	$122$	$8$	$64$
$211$	$122$	$89$	$7921$

The square root of the variance

  $\begin{array}{l}\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}\\ \sigma =\sqrt{\frac{729+576+\mathrm{...........}+7921}{8}}\\ \sigma =\sqrt{1296}\\ \sigma \approx 36\end{array}$

(b)

To determine

To find:

The outlier affect the mean median and mode of the prices .

Answer to Problem 8Q

This is increases the mean and median and this is not affect the mode.

Explanation of Solution

Given:

The table

  $x,\overline{x},x-\overline{x},{\left(x-\overline{x}\right)}^2$

Concept used:

The mean of the squared deviation or variance

  $=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}$

The square root of the variance

  $\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}$

Calculation:

The mean is

  $\begin{array}{l}\overline{x}=\frac{98+119+95+211+130+98+100+125}{8}=\frac{976}{8}\\ \overline{x}=122\end{array}$

Median

  $95,98,98,100,119,125,130,211$

  $\begin{array}{l}=\frac{100+119}{2}\\ =109.5\end{array}$

Mode

  $\begin{array}{l}95,98,98,100,119,125,130,211\\ \to 98\end{array}$

Range:-difference of largest and smallest value

  $\begin{array}{l}=211-95\\ =116\end{array}$

The mean of the squared deviation or variance

  $\begin{array}{l}=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}\\ =\frac{729+576+\mathrm{...........}+7921}{8}\\ =1296\end{array}$

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
$95$	$122$	$-27$	$729$
$98$	$122$	$-24$	$576$
$98$	$122$	$-24$	$576$
$100$	$122$	$-22$	$484$
$119$	$122$	$-3$	$9$
$125$	$122$	$3$	$9$
$130$	$122$	$8$	$64$
$211$	$122$	$89$	$7921$

The square root of the variance

  $\begin{array}{l}\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}\\ \sigma =\sqrt{\frac{729+576+\mathrm{...........}+7921}{8}}\\ \sigma =\sqrt{1296}\\ \sigma \approx 36\end{array}$

The outlier is $211$ .

Since this is greater than most of the data.

This is increases the mean and median and this is not affect the mode.

(c)

To determine

To find:

The interquartile range of the data and the shape of the distribution of the prices .

Answer to Problem 8Q

The distribution is skewed right.

Explanation of Solution

Given:

The table

  $x,\overline{x},x-\overline{x},{\left(x-\overline{x}\right)}^2$

Concept used:

The mean of the squared deviation or variance

  $=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}$

The square root of the variance

  $\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}$

Calculation:

Order the data

  $95,98,98,100,119,125,130,211$

Least value $=95$

First quartile: median of lower half $\left(95,98,98,100\right)$

  $\begin{array}{l}=\frac{98+98}{2}\\ =98\end{array}$

Median

  $95,98,98,100,119,125,130,211$

  $\begin{array}{l}=\frac{100+119}{2}\\ =109.5\end{array}$

Third quartile: median of upper half $\left(119,125,130,211\right)$

  $\begin{array}{l}=\frac{125+130}{2}\\ =127.5\end{array}$

Greatest value $=211$

The right whisker is longer than the left whisker and most of the data are on the left side of the plot

Therefore,

The distribution is skewed right.

(d)

To determine

To find:

The mean, median, mode, range, and standard deviation of the prices.

Answer to Problem 8Q

Mean $=115.9$ , median $=104.025$ , mode $=93.1$ , range $=110.2$ , and standard deviation $=34.2$

Explanation of Solution

Given:

The table

  $x,\overline{x},x-\overline{x},{\left(x-\overline{x}\right)}^2$

Concept used:

The mean of the squared deviation or variance

  $=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}$

The square root of the variance

  $\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}$

Calculation:

In this case the data set is multiplied by $0.95\%\left(5\%\text{discount}\right)$

Each value in a numerical data set is multiplied by a real number $k$

Where $k>0$

The measures of center and variation is founded by multiplying the original measures by $k$

Mean

  $\begin{array}{l}=122\left(0.95\right)\\ =115.9\end{array}$

Median

  $\begin{array}{l}=109\left(0.95\right)\\ =104.025\end{array}$

Mode

  $\begin{array}{l}95,98,98,100,119,125,130,211\\ =98\left(0.95\right)\\ =93.1\end{array}$

Range:-difference of largest and smallest value

  $\begin{array}{l}=116\left(0.95\right)\\ =110.2\end{array}$

The mean of the squared deviation or variance

  $\begin{array}{l}=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}\\ =\frac{729+576+\mathrm{...........}+7921}{8}\\ =1296\end{array}$

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
$95$	$122$	$-27$	$729$
$98$	$122$	$-24$	$576$
$98$	$122$	$-24$	$576$
$100$	$122$	$-22$	$484$
$119$	$122$	$-3$	$9$
$125$	$122$	$3$	$9$
$130$	$122$	$8$	$64$
$211$	$122$	$89$	$7921$

The square root of the variance

  $\begin{array}{l}\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}\\ \sigma =\sqrt{\frac{729+576+\mathrm{...........}+7921}{8}}\\ \sigma =\sqrt{1296}\\ \sigma =36\left(0.95\right)\\ \sigma =34.2\end{array}$