Chapter 4, Problem 16E

(a)

To determine

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

Answer to Problem 16E

Set 2 is larger than Set 1

Explanation of Solution

Given:

Set 1

4, 7, 7, 7, 10

Set 2

4, 6, 7, 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:

  $\begin{array}{c}\frac{4+7+7+7+10}{5}\\ =\frac{35}{5}\\ =7\end{array}$

Standard deviation of first set

  $\begin{array}{c}=\sqrt{\frac{\begin{array}{l}{\left(7-4\right)}^2+{\left(7-7\right)}^2+{\left(7-7\right)}^2\\ +{\left(7-7\right)}^2+{\left(10-7\right)}^2\end{array}}{5-1}}\\ =\sqrt{\frac{9+0+0+0+9}{4}}\\ =\sqrt{4.5}\\ =2.12\end{array}$

Mean of second set

  $\begin{array}{c}\frac{4+7+7+7+10}{5}\\ =\frac{35}{5}\\ =7\end{array}$

Standard deviation of second set

  $\begin{array}{c}=\sqrt{\frac{\begin{array}{l}{\left(7-4\right)}^2+{\left(7-6\right)}^2+{\left(7-7\right)}^2\\ +{\left(7-8\right)}^2+{\left(7-10\right)}^2\end{array}}{5-1}}\\ =\sqrt{\frac{9+1+0+1+9}{4}}\\ =\sqrt{5}\\ =2.24\end{array}$

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. In the set 1 7 is continuous coming 3 time where in set 2 the data is at least separated with the 1 digit.

(b)

To determine

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

Answer to Problem 16E

Set 1 is equal to Set 2.

Explanation of Solution

Given:

Set 1

100, 140, 150, 160, 200

Set 2

10, 50, 60, 70, 110

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:

  $\begin{array}{c}\frac{100+140+150+160+200}{5}\\ =\frac{750}{5}\\ =150\end{array}$

Standard deviation of first set

  $\begin{array}{c}=\sqrt{\frac{\begin{array}{l}{\left(150-100\right)}^2+{\left(150-140\right)}^2+{\left(150-150\right)}^2\\ +{\left(150-160\right)}^2+{\left(150-200\right)}^2\end{array}}{5-1}}\\ =\sqrt{\frac{2500+100+0+100+2500}{4}}\\ =\sqrt{\frac{5200}{4}}\\ =36.1\end{array}$

Mean of second set:

  $\begin{array}{c}\frac{10+50+60+70+110}{5}\\ =\frac{300}{5}\\ =60\end{array}$

Standard deviation of second set

  $\begin{array}{c}=\sqrt{\frac{\begin{array}{l}{\left(60-10\right)}^2+{\left(60-50\right)}^2+{\left(60-60\right)}^2\\ +{\left(60-70\right)}^2+{\left(60-110\right)}^2\end{array}}{5-1}}\\ =\sqrt{\frac{2500+100+0+100+2500}{4}}\\ =\sqrt{\frac{5200}{4}}\\ =36.1\end{array}$

Hence, SD (set 1) = 36.1 and SD (set 2) = 36.1

By seeing the set 1 and set 2, it observed that set 2’s standard deviation is equal to the set 1 the reason is that both the data sets are equally separated subtracted with the 90 from set, it will give the set 2 data set.

(c)

To determine

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

Answer to Problem 16E

Set 2 is larger than Set 1

Explanation of Solution

Given:

Set 1

10, 16, 18, 20, 22, 28

Set 2

48, 56, 58, 60, 62, 70

Formula used:

For the standard deviation

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

Calculation:

  $\begin{array}{c}\frac{10+16+18+20+22+28}{6}\\ =\frac{114}{6}\\ =19\end{array}$

Standard deviation of first set

  $\begin{array}{c}=\sqrt{\frac{\begin{array}{l}{\left(19-10\right)}^2+{\left(19-16\right)}^2+{\left(19-18\right)}^2\\ +{\left(19-20\right)}^2+{\left(19-22\right)}^2+{\left(19-28\right)}^2\end{array}}{6-1}}\\ =\sqrt{\frac{81+9+1+1+9+81}{5}}\\ =\sqrt{\frac{150}{5}}\\ =6.0\end{array}$

Mean of second set:

  $\begin{array}{c}\frac{48+56+58+60+62+70}{6}\\ =\frac{354}{6}\\ =59\end{array}$

Standard deviation of second set

  $\begin{array}{c}=\sqrt{\frac{\begin{array}{l}{\left(59-48\right)}^2+{\left(59-56\right)}^2+{\left(59-58\right)}^2\\ +{\left(59-60\right)}^2+{\left(59-62\right)}^2+{\left(59-70\right)}^2\end{array}}{6-1}}\\ =\sqrt{\frac{121+9+1+1+9+121}{5}}\\ =\sqrt{\frac{262}{5}}\\ =7.2\end{array}$

Hence, SD (set 1) = 6.0 and SD (set 2) = 7.2

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