Chapter 4, Problem 15E

(a)

To determine

To find: the larger standard deviation without doing any calculation and also by hand.

Answer to Problem 15E

Set 2 is larger than Set 1

Explanation of Solution

Given:

Set 1

3, 5, 6, 7, 9

Set 2

2, 4, 6, 8, 10

Formula used:

For the standard deviation

  $\sqrt{\frac{1}{n}{\displaystyle {\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}}$

Calculation:

Mean of first set:

  $\begin{array}{c}\frac{3+5+6+7+9}{5}\\ =\frac{30}{5}\\ =6\end{array}$

Standard deviation of first set

  $\begin{array}{c}=\sqrt{\frac{\begin{array}{l}{\left(6-3\right)}^2+{\left(6-5\right)}^2+{\left(6-6\right)}^2\\ +{\left(6-7\right)}^2+{\left(6-9\right)}^2\end{array}}{5-1}}\\ =\sqrt{\frac{9+1+0+1+9}{4}}\\ =\sqrt{5}\\ =2.2\end{array}$

Mean of second set:

  $\begin{array}{c}\frac{3+5+6+7+9}{5}\\ =\frac{30}{5}\\ =6\end{array}$

Standard deviation of second set

  $\begin{array}{c}=\sqrt{\frac{\begin{array}{l}{\left(6-2\right)}^2+{\left(6-4\right)}^2+{\left(6-6\right)}^2\\ +{\left(6-8\right)}^2+{\left(6-10\right)}^2\end{array}}{5-1}}\\ =\sqrt{\frac{16+4+0+4+16}{4}}\\ =\sqrt{10}\\ =3.2\end{array}$

Hence, SD (set 1) = 2.2 and SD (set 2) = 3.2

By seeing the set 1 and set 2, it observed that set 2’s standard deviation is greater than set 1 the reason is that the set 2’s data is more spread out than set 1. For example set 2 is evenly separated by 1 where set 1 is also separated by 1 but 6 and 7 continuous.

(b)

To determine

To find: the larger standard deviation without doing any calculation and also by hand.

Answer to Problem 15E

Set 2 is larger than Set 1

Explanation of Solution

Given:

Set 1

10, 14, 15, 16, 20

Set 2

10, 11, 15, 19, 20

Formula used:

For the standard deviation

  $\sqrt{\frac{1}{n}{\displaystyle {\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}}$

Calculation:

Mean of first set:

  $\begin{array}{c}\frac{10+14+15+16+20}{5}\\ =\frac{75}{5}\\ =15\end{array}$

Standard deviation of first set

  $\begin{array}{c}=\sqrt{\frac{\left[\begin{array}{l}{\left(15-10\right)}^2+{\left(15-14\right)}^2+{\left(15-15\right)}^2\\ +{\left(15-16\right)}^2+{\left(15-20\right)}^2\end{array}\right]}{5-1}}\\ =\sqrt{\frac{25+1+0+1+25}{4}}\\ =\sqrt{\frac{52}{4}}\\ =3.6\end{array}$

Mean of second set:

  $\begin{array}{c}\frac{10+14+15+16+20}{5}\\ =\frac{75}{5}\\ =15\end{array}$

Standard deviation of second set

  $\begin{array}{c}=\sqrt{\frac{\left[\begin{array}{l}{\left(15-10\right)}^2+{\left(15-11\right)}^2+{\left(15-15\right)}^2\\ +{\left(15-19\right)}^2+{\left(15-20\right)}^2\end{array}\right]}{5-1}}\\ =\sqrt{\frac{25+16+0+16+25}{4}}\\ =\sqrt{\frac{82}{4}}\\ =4.6\end{array}$

Hence, SD (set 1) = 3.6 and SD (set 2) = 4.5

By seeing the set 1 and set 2, it observed that set 2’s standard deviation is greater than set 1 the reason is that the set 2’s data is more spread out than set 1

(c)

To determine

To find: the larger standard deviation without doing any calculation and also by hand.

Answer to Problem 15E

Set 1 is equal to Set 2

Explanation of Solution

Given:

Set 1

2, 6, 6, 9, 11, 14

Set 2

82, 86, 86, 91, 94

Formula used:

For the standard deviation

  $\sqrt{\frac{1}{n}{\displaystyle {\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}}$

Calculation:

Mean of first set:

  $\begin{array}{c}\frac{2+6+6+9+11+14}{6}\\ =\frac{48}{6}\\ =8\end{array}$

Standard deviation of first set

  $\begin{array}{c}=\sqrt{\frac{\left[\begin{array}{l}{\left(8-2\right)}^2+{\left(8-6\right)}^2+{\left(8-6\right)}^2\\ +{\left(8-9\right)}^2+{\left(8-11\right)}^2+{\left(8-14\right)}^2\end{array}\right]}{6-1}}\\ =\sqrt{\frac{36+4+4+1+9+36}{5}}\\ =\sqrt{\frac{90}{5}}\\ =4.2\end{array}$

Mean of second set:

  $\begin{array}{c}\frac{82+86+86+89+91+94}{6}\\ =\frac{528}{6}\\ =88\end{array}$

Standard deviation of second set

  $\begin{array}{c}=\sqrt{\frac{\left[\begin{array}{l}{\left(88-82\right)}^2+{\left(88-86\right)}^2+{\left(88-86\right)}^2\\ +{\left(88-89\right)}^2+{\left(88-91\right)}^2+{\left(88-94\right)}^2\end{array}\right]}{6-1}}\\ =\sqrt{\frac{36+4+4+1+9+36}{5}}\\ =\sqrt{\frac{90}{5}}\\ =4.2\end{array}$

Hence, SD (set 1) =4.2 and SD (set 2) = 4.2

By seeing the set 1 and set 2, it observed that set 1’s standard deviation is equal to the set 1 the reason is that there is equal spread out in both the set.