Chapter 2.5, Problem 2.33E

a.

To determine

To find: The range of the data.

Answer to Problem 2.33E

The range of the datais $42$ .

Explanation of Solution

Given information: The data is,

56	89	51	79	58	82	52	88	52	78	69	75	77	72	71
55	87	53	85	61	93	54	76	80	81	59	86	78	71	77

Calculation:

The range of the data is,

  $\begin{array}{c}Range=93-51\\ =42\end{array}$

Thus, the range of the data is $42$ .

b.

To determine

To find:The standard deviation using range approximation.

Answer to Problem 2.33E

The standard deviation using range approximation is $10.5$ .

Explanation of Solution

Given information: The data is,

56	89	51	79	58	82	52	88	52	78	69	75	77	72	71
55	87	53	85	61	93	54	76	80	81	59	86	78	71	77

Calculation:

The range of the data is,

  $\begin{array}{c}Range=93-51\\ =42\end{array}$

The standard deviation is,

  $\begin{array}{c}s=\frac{Range}{4}\\ =\frac{42}{2.5}\\ =10.5\end{array}$

Thus, the standard deviation using range approximation is $10.5$ .

c.

To determine

To find: The standard deviation.

Answer to Problem 2.33E

The standard deviation is $13.1$ .

Explanation of Solution

Given information: The data is,

56	89	51	79	58	82	52	88	52	78	69	75	77	72	71
55	87	53	85	61	93	54	76	80	81	59	86	78	71	77

Calculation:

The data is shown below.

$x$	$x^2$
56	3136
89	7921
51	2601
9	81
58	3364
82	6724
52	2704
78	6084
69	4761
75	5625
77	5929
72	5184
71	5041
87	7569
53	2809
85	7225
61	3721
93	8649
54	2916
76	5776
80	6400
81	6561
59	3481
86	7396
78	6084
71	5041
77	5929
${\displaystyle \sum x}=2145$	${\displaystyle \sum x^2}=158345$

The variance is,

  $\begin{array}{c}{s}^2=\frac{1}{n-1}\left[{\displaystyle \sum x_i^2}-\frac{{\left({\displaystyle \sum x}\right)}^2}{n}\right]\\ =\frac{1}{30-1}\left[158345-\frac{4601025}{30}\right]\\ =172.5689\end{array}$

The standard deviation is,

  $\begin{array}{c}s=\sqrt{172.5689}\\ =13.1\end{array}$

Thus, the standard deviation is $13.1$ .

d.

To determine

To find: The proportion of the measurement lies within the two standard deviation.

Answer to Problem 2.33E

The proportion of the measurement lies within the two standard deviationis $95\%$ and $99.7\%$

Explanation of Solution

Given information: The data is,

56	89	51	79	58	82	52	88	52	78	69	75	77	72	71
55	87	53	85	61	93	54	76	80	81	59	86	78	71	77

Calculation:

The interval is,

  $\begin{array}{c}\left(\overline{x}\pm s\right)=71.5\pm 13.1\\ =\left(58.4,84.6\right)\end{array}$

The interval is,

  $\begin{array}{c}\left(\overline{x}\pm 2s\right)=71.5\pm 2\times 13.1\\ =\left(45.3,97.7\right)\end{array}$

The interval is,

  $\begin{array}{c}\left(\overline{x}\pm 3s\right)=71.5\pm 3\times 13.1\\ =\left(32.2,110.8\right)\end{array}$

According to the Empirical formula the percentage of values falling into the intervals are $95\%$ and $99.7\%$

Thus, the proportion of the measurement lies within the two standard deviation is $95\%$ and $99.7\%$