Chapter 10, Problem 23SGR

To determine

To calculate mean, median, mode, range and standard deviation

Answer to Problem 23SGR

Mean is 22.5, median is 21.5, mode is 38, range is 24, and standard deviation is 7.37

Explanation of Solution

Given information:

The data set is,

27,21,34,42,20,19,18,26,25,33

Formula used:

Mean is,

  $\overline{x}=\frac{{\displaystyle \sum x}}{n}$

Median is,

  $M=\frac{{(\frac{n}{2})}^{th}+{(\frac{n}{2}+1)}^{th}}{2}$

Mode is,

  $z=\max value$

Range is,

  $range=\max -\min$

Standard deviation is,

  $sd=\sqrt{\frac{{\displaystyle \sum {(x-\overline{x})}^2}}{n}}$

Calculation:

Re-arranging the data in ascending order,

18, 19, 20, 21, 25, 26, 27, 33, 34, 42

Adding the constant,

14, 15, 16, 17, 21, 22, 23, 29, 30, 38

The minimum is 14

The maximum is 38

For range, substituting the values,

  $range=38-14$

On solving,

  $range=24$

Hence, range is 24.

For mean, substituting the values,

  $z=38$

Hence, mode is 38

For median, substituting the values,

  $M=\frac{{(\frac{10}{2})}^{th}+{(\frac{10}{2}+1)}^{th}}{2}$

On solving,

  $M=\frac{5^{th}+6^{th}}{2}$

Substituting the values,

  $M=\frac{21+22}{2}$

On solving,

  $M=21.5$

Hence, median is 21.5

For mean, substituting the values,

  $\overline{x}=\frac{14+15+16+17+21+22+23+29+30+38}{10}$

So,

  $\overline{x}=\frac{225}{10}$

On solving,

  $\overline{x}=22.5$

Hence, mean is 22.5

For standard deviation, substituting the values,

  $sd=\sqrt{\frac{\begin{array}{}$ {(14-22.5)}^2+{(15-22.5)}^2+{(16-22.5)}^2+{(17-22.5)}^2+{(21-22.5)}^2+{(22-22.5)}^2+{(23-22.5)}^2+$$ {(29-22.5)}^2+{(30-22.5)}^2+{(38-22.5)}^2$\end{array}}{10}}$ so,

  $sd=\sqrt{\frac{\text{72}\text{.25+56}\text{.25+42}\text{.25+30}\text{.25+2}\text{.25+0}\text{.25+0}\text{.25+42}\text{.25+56}\text{.25+240}\text{.25}}{10}}$

So,

  $sd=\sqrt{54.25}$

On solving,

  $sd=7.37$

Hence, standard deviation is 7.37