Chapter 14.4, Problem 22E

(a)

To determine

To find:

Themedian of the given data.

Answer to Problem 22E

Median $=148.5$

Explanation of Solution

Given:

The table

  $110,110,330,200,88,110,88,110,165,390,150,440,536,200,110,165,88,147,110,165$

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:

Median

  $110,110,330,200,88,110,88,110,165,390,150,440,536,200,110,165,88,147,110,165$

Rearrange the term

  $88,88,88,110,110,110,110,110,110,147,150,165,165,165,200,200,330,390,440,536$

  $\begin{array}{l}=\frac{147+150}{2}\\ =148.5\end{array}$

Range:-difference of largest and smallest value

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

(b)

To determine

To find:

The firstquartile range and third quartile point.

Answer to Problem 22E

First quartile $=110$ ,Third quartile $=200$

Explanation of Solution

Given:

The table

  $110,110,330,200,88,110,88,110,165,390,150,440,536,200,110,165,88,147,110,165$

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

  $88,88,88,110,110,110,110,110,110,147,150,165,165,165,200,200,330,390,440,536$

Least value $=88$

First quartile: median of lower half $\left(88,88,88,110,110,110,110,110,110,147\right)$

  $\begin{array}{l}=\frac{110+110}{2}\\ =110\end{array}$

Third quartile: median of upper half $\left(150,165,165,165,200,200,330,390,440,536\right)$

  $\begin{array}{l}=\frac{200+200}{2}\\ =200\end{array}$

Greatest value $=536$

(c)

To determine

To find:

The interquartile range of the given data.

Answer to Problem 22E

The interquartile range is $90$

Explanation of Solution

Given:

The table

  $110,110,330,200,88,110,88,110,165,390,150,440,536,200,110,165,88,147,110,165$

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:

Median

  $110,110,330,200,88,110,88,110,165,390,150,440,536,200,110,165,88,147,110,165$

Rearrange the term

  $88,88,88,110,110,110,110,110,110,147,150,165,165,165,200,200,330,390,440,536$

  $\begin{array}{l}=\frac{147+150}{2}\\ =148.5\end{array}$

Range:-difference of largest and smallest value

  $\begin{array}{l}=200-110\\ =90\end{array}$

(d)

To determine

To find:

The semi-quartile range of the given data.

Answer to Problem 22E

The semi-quartile range is $45$

Explanation of Solution

Given:

The table

  $110,110,330,200,88,110,88,110,165,390,150,440,536,200,110,165,88,147,110,165$

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:

Median

  $110,110,330,200,88,110,88,110,165,390,150,440,536,200,110,165,88,147,110,165$

Rearrange the term

  $88,88,88,110,110,110,110,110,110,147,150,165,165,165,200,200,330,390,440,536$

  $\begin{array}{l}=\frac{147+150}{2}\\ =148.5\end{array}$

Semi-quartile Range:-

  $\begin{array}{l}=\frac{\text{interquartile}\text{range}}{2}\\ =\frac{90}{2}\\ =45\end{array}$

(e)

To determine

To draw:

A box and whisker plotof the given data.

Explanation of Solution

Given:

The table

  $110,110,330,200,88,110,88,110,165,390,150,440,536,200,110,165,88,147,110,165$

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:

Median

  $110,110,330,200,88,110,88,110,165,390,150,440,536,200,110,165,88,147,110,165$

Rearrange the term

  $88,88,88,110,110,110,110,110,110,147,150,165,165,165,200,200,330,390,440,536$

  $\begin{array}{l}=\frac{147+150}{2}\\ =148.5\end{array}$

Semi-quartile Range:-

  $\begin{array}{l}=\frac{\text{interquartile}\text{range}}{2}\\ =\frac{90}{2}\\ =45\end{array}$

Draw a box and whisker plot of the data