Chapter B.2, Problem 38E

(a)

To determine

To find: The mean, variance and standard deviation of the revised data.

Answer to Problem 38E

The explanation is given below.

Explanation of Solution

Given:

Multiply each number in exercise 28 by 3.

The data set of exercise 28 is given as:

  $1,2,3,4,5,6,7$ .

The new data is given as:

  $3,6,9,12,15,18,21$ .

The mean, variance and standard deviation of new data is calculated as:

  $\begin{array}{c}\text{Mean}=\frac{3+6+9+12+15+18+21}{7}\\ =12\\ \text{Variance}=\frac{{\left(3-12\right)}^2+{\left(6-12\right)}^2+{\left(9-12\right)}^2+{\left(12-12\right)}^2+{\left(15-12\right)}^2+{\left(18-12\right)}^2+{\left(21-12\right)}^2}{7}\\ =36\end{array}$

Further simplified as:

  $\begin{array}{c}\text{Standard deviation}=\sqrt{\frac{{\left(3-12\right)}^2+{\left(6-12\right)}^2+{\left(9-12\right)}^2+{\left(12-12\right)}^2+{\left(15-12\right)}^2+{\left(18-12\right)}^2+{\left(21-12\right)}^2}{7}}\\ =\sqrt{36}\\ =6\end{array}$

The mean, variance and standard deviation of data of exercise 28 is calculated as:

  $\begin{array}{c}\text{Mean}=\frac{1+2+3+4+5+6+7}{7}\\ =3\\ \text{Variance}=\frac{{\left(1-3\right)}^2+{\left(2-3\right)}^2+{\left(3-3\right)}^2+{\left(4-3\right)}^2+{\left(5-3\right)}^2+{\left(6-3\right)}^2+{\left(7-3\right)}^2}{7}\\ =5\end{array}$

Further simplified as:

  $\begin{array}{c}\text{Standard deviation}=\sqrt{\frac{{\left(1-3\right)}^2+{\left(2-3\right)}^2+{\left(3-3\right)}^2+{\left(4-3\right)}^2+{\left(5-3\right)}^2+{\left(6-3\right)}^2+{\left(7-3\right)}^2}{7}}\\ =\sqrt{5}\\ \approx 2.236\end{array}$

Therefore, the mean, variance and standard deviation is increased after multiplying the data of exercise 28 by 3.

(b)

To determine

To find: The change in data of exercise 28 if each number is multiplied by a constant $k$ .

Answer to Problem 38E

The explanation is given below.

Explanation of Solution

Case 1: If constant is greater than1.

Then after multiplying by the constant $k$ of each number of the data, the mean, variance, and standard deviation will increase as compare to original data.

Case 1: If constant is less than $-1$ .

Then after multiplying by the constant $k$ of each number of the data, the mean will decrease but variance, and standard deviation will increase as compare to original data.

Case 1: If constant is between $-1\&1$ .

Then after multiplying by the constant $k$ of each number of the data, the mean variance, and standard deviation will decrease as compare to original data.

Therefore, the explanation is given above.