Chapter 10, Problem 8PT

To determine

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

Answer to Problem 8PT

Range is 22, mean is 16.3, standard deviation is 7.06, median is 15 and mode is 9

Explanation of Solution

Given information:

The data set is,

9, 17, 31, 21, 17, 25, 13, 9, 12, 9

Constant that is multiplied with each data value is 3

Formula used:

The mean is,

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

Standard deviation is,

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

Median is,

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

Range is,

  $range=\max -\min$

Mode is,

z = most repeated value.

Calculation:

Re-arranging the data in an ascending order,

9, 9, 9, 12, 13, 17, 17, 21, 25, 31

The maximum value is 31.

The minimum value id 9.

For range, substituting the values,

  $range=31-9$

On solving,

  $range=22$

For the mean, substituting the value

  $\overline{x}=\frac{9+9+9+12+13+17+17+21+25+31}{10}$

So,

  $\overline{x}=\frac{163}{10}$

On solving,

  $\overline{x}=16.3$

Hence, mean is 16.3

For standard deviation, substituting the values,

  $sd=\sqrt{\frac{\begin{array}{l}{(}^9+{(}^9+{(}^9+{(}^{12}+{(}^{13}+{(}^{17}+{(}^{17}+\\ {(}^{21}+{(}^{25}+{(}^{31}\end{array}}{10}}$ So,

  $sd=\sqrt{\frac{{(-7.3)}^2+{(-7.3)}^2+{(-7.3)}^2+{(-4.3)}^2+{(-3.3)}^2+{(0.7)}^2+{(0.7)}^2+{(4.7)}^2+{(8.7)}^2+{(14.7)}^2}{10}}$ On solving,

  $sd=\sqrt{49.80}$

On solving,

  $sd=7.06$

Hence, standard deviation is 7.06

For median, substituting the value,

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

On solving,

  $M=\frac{5^{th}+6^{th}}{2}$

Substituting the value,

  $M=\frac{13+17}{2}$

On solving,

  $M=15$

Hence, median is 15.

For mode, substituting the value,

  $z=9$

Hence, mode is 9