Chapter 2.7, Problem 2.40E

(a)

To determine

To find the five number summary and the IQR.

Answer to Problem 2.40E

The five number summary consists $0,3.25,5.5,6.75,8$ .

IQR= $3.5$ .

Explanation of Solution

The data set given in the question is as:- $\{8,7,1,4,6,6,4,5,7,6,3,0\}$ .

Let us arrange it in ascending order as:- $\{0,1,3,4,4,5,6,6,6,7,7,8\}$ .

The five number summary consists of the minimum, the first quartile, the median, the third quartile and the maximum.

The minimum number is $0$ .

The median is the middle value of the sorted data set. Since the data values are even, the median is the average of the two middle values, as,

  $M=Q_2=\frac{5+6}{2}=\frac{11}{2}=5.5$

The lower quartile range is calculated as,

  $\begin{array}{l}\text{Position of }{Q}_1=0.25(n+1)\\ =0.25(12+1)\\ =0.25(13)\\ =3.25\end{array}$

Thus, $\begin{array}{l}{Q}_1=3+0.25(4-3)\\ =3+0.25\\ =3.25\end{array}$

Now, the upper quartile range is calculated as,

  $\begin{array}{l}\text{Position of }{Q}_3=0.75(n+1)\\ =0.75(12+1)\\ =0.75(13)\\ =9.75\end{array}$

Thus, $\begin{array}{l}{Q}_3=6+0.75(7-6)\\ =6+0.75\\ =6.75\end{array}$

And the maximum value of the data values is $8$ .

Thus, the five number summary consists $0,3.25,5.5,6.75,8$ .

The Interquartile range IQR is the difference between the lower and upper quartile.

This implies, IQR is equal to

  $\begin{array}{l}=Q_3-Q_1=6.75-3.25\\ =3.5\end{array}$

(b)

To determine

To calculate the sample mean and standard deviation for the data given in the question.

Answer to Problem 2.40E

  $s=2.454$ and $\overline{x}=4.75$ .

Explanation of Solution

The data set given in the question is as:- $\{8,7,1,4,6,6,4,5,7,6,3,0\}$ .

The mean of the data is calculated by the formula,

  $\overline{x}=\frac{{\displaystyle \sum x_i}}{n}$

Here, we have, ${\displaystyle \sum x=\text{ Sum of all observations =}}57$

  ${\displaystyle \sum x_i{}^2=\text{ Sum of all squares of the observation }=}337$

  $({\displaystyle \sum x_i{)}^2}={\text{ Square of the sum of observation =57}}^2=3249$

And $n=12$ .

Putting in the values from above, we have,

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_i}}{n}\\ \overline{x}=\frac{57}{12}\\ \overline{x}=4.75\end{array}$

The actual value of standard deviation can be calculated by the formula of variance given below as:-

  $s^2=\frac{{\displaystyle \sum x_i{}^2-\frac{({\displaystyle \sum x_i{)}^2}}{n}}}{n-1}$

Putting in the values from above, we have,

  $\begin{array}{l}{s}^2=\frac{{\displaystyle \sum x_i{}^2-\frac{({\displaystyle \sum x_i{)}^2}}{n}}}{n-1}\\ s^2=\frac{337-\frac{3249}{12}}{12-1}\\ s^2=\frac{337-270.75}{11}\\ s^2=\frac{66.25}{11}\\ s^2=6.023\\ s=\sqrt{6.023}\\ s=2.454\end{array}$

(c)

To determine

To calculate the z-score for the smallest and the largest observations and find are there any of the observations unusually large or unusually small.

Answer to Problem 2.40E

The z-score is $-1.94$ and $1.32$ .

Explanation of Solution

The data set given in the question is as:- $\{8,7,1,4,6,6,4,5,7,6,3,0\}$ .

Also from part (b) $s=2.454$ and $\overline{x}=4.75$ .

The z-score value is decreased by the mean and divided by the standard deviation as:

The smallest observation can be calculated as,

  $z=\frac{x-\overline{x}}{s}=\frac{0-4.75}{2.545}\approx -1.94$

The largest observation can be calculated as,

  $z=\frac{x-\overline{x}}{s}=\frac{8-4.75}{2.545}\approx 1.32$

Unusual observations have a z-score below $-2$ or above $2$ .

Since both the observations are between them, neither of the observations are unusual.