Chapter 0.9, Problem 9E

To determine

To find: The values of mean, median, range and standard deviation for the provided data set.

Explanation of Solution

Given:

The data set is:

40
36
29
45
51
36
48
34
36
22
13
42
31
44
32
34

Calculation:

The mean can be calculated as:

  $\begin{array}{c}\text{Mean=}\frac{\text{Sum of observations}}{\text{Number of observations}}\\ =\frac{40+36+\mathrm{....}+34}{16}\\ =35.8125\end{array}$

The median is:

First,we arrange the data in ascending order: $13,22,29,31,32,34,34,36,36,36,40,42,44,45,48,51$ .

  $\begin{array}{c}\text{Median = }\frac{{\left(\frac{n}{2}\right)}^{th}term+{\left(\frac{n}{2}+1\right)}^{th}term}{2}\\ =\frac{{\left(\frac{16}{2}\right)}^{th}term+{\left(\frac{16}{2}+1\right)}^{th}term}{2}\\ =\frac{36+36}{2}\\ =36\end{array}$

The range is:

  $\begin{array}{c}\text{Range = Highest value - Lowest value}\\ \text{=51-13}\\ \text{=38}\end{array}$

The standard deviation is:

  $\begin{array}{c}s=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{{\left(40-35.8125\right)}^2+{\left(36-35.8125\right)}^2+\mathrm{......}+{\left(36-35.8125\right)}^2}{16-1}}\\ =7.715\end{array}$

The mean, median are closely spread, that is, they are very nearer .The range value is used to describe the wide spread of observations but it doesn’t give any information about any of the data points. The standard deviation is very much deviated from the mean.