Chapter 1, Problem 12PT

(a).

To determine

To Graph: Histogram for given data.

Explanation of Solution

Given:

  $\begin{array}{cccccccccc}7& 20& 24& 25& 25& 28& 28& 30& 32& 35\\ 42& 43& 44& 45& 46& 47& 48& 48& 50& 51\\ 72& 75& 77& 78& 79& 83& 87& 88& 135& 151\end{array}$

Graph:s

Interpretation:

For the given data, the proper histogram is as above.

(b).

To determine

Justification for the outliners.

Answer to Problem 12PT

Outliner is:

Min is $19$ , max is $325$ , Median is $48$ , Q1 is $32$ , Q3 is $80$ .

Explanation of Solution

Given:

  $\begin{array}{cccccccccc}7& 20& 24& 25& 25& 28& 28& 30& 32& 35\\ 42& 43& 44& 45& 46& 47& 48& 48& 50& 51\\ 72& 75& 77& 78& 79& 83& 87& 88& 135& 151\end{array}$

Calculation:

From the above histogram it is observed that,

1. The minimum: $19$
2. (The median of the initial group is the lower or initial quartile) Q1: $32$
3. The median: $48$
4. (The median of the second group is the upper or third quartile) Q3: $80$
5. The maximum: $325$

Conclusion:

Thus, five number summary is: Min is $19$ , max is $325$ , Median is $48$ , Q1(The median of the first group is the lower or first quartile) is $32$ , Q3 (The median of the second group is the upper or third quartile) is $80$

(c).

To determine

To find: Reason to use the mean and standard deviation to describe the cemter and spread of the distribution.

Answer to Problem 12PT

Sample standard deviation is $75.57302930$ .

Explanation of Solution

Given:

  $\begin{array}{cccccccccc}7& 20& 24& 25& 25& 28& 28& 30& 32& 35\\ 42& 43& 44& 45& 46& 47& 48& 48& 50& 51\\ 72& 75& 77& 78& 79& 83& 87& 88& 135& 151\end{array}$

Calculation:

The mean of the given data is,

  $\overline{x}=\frac{1}{n}{\displaystyle \sum _{i=1}^{n}xi}$

Where,

  $n=$ number of values.

  $xi,i=1.n$ are the values.

As here in this example $30$ points are given, then $n=30$

The sum of point is,

  ${\displaystyle \sum _{i=1}^{n}xi}=2382$

Thus, the mean $\overline{x}=\left(\frac{1}{30}\right)2382=\frac{397}{5}=79.4$

The sample standard variation of given data is,

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

Where,

  $n=$ number of values.

  $xi,i=1.n$ are the values.

  $\overline{x}=$ mean of the values.

The mean of data is,

  $\overline{x}=397/5$

As here in this example $30$ points are given, then $n=30$

Sum of,

  $\begin{array}{l}{\left(xi-\overline{x}\right)}^2{\displaystyle \sum _{i=1}^n{\left(xi-\overline{x}\right)}^2}=828136/5\\ {\left(xi-\overline{x}\right)}^2{\displaystyle \sum _{i=1}^n{\left(xi-\overline{x}\right)}^2}=\left[1/\left(30-1\right)\right]\left(828136/5\right)=\left(1/29\right)\left(828136/5\right)=828136/145\end{array}$

Finally,

  $s=\sqrt{828136/145}=\left(2\sqrt{30019930}\right)/145\approx 75.57302930$

Conclusion:

Thus, sample standard deviation is $75.57302930$ .