Chapter 10, Problem 24SGR

To determine

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

Answer to Problem 24SGR

Mean is 22.5, median is 83, mode is 93, range is 21, and standard deviation is 6.78

Explanation of Solution

Given information:

The data set is,

72,56,71,63,68,59,77,74,76,66

Constant to be added is 16

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,

56,59,63,66,68,71,72,74,76,77

Adding the constant,

72,75,79,82,84,87,88,90,92,93

The minimum is 72

The maximum is 93

For range, substituting the values,

  $range=93-72$

On solving,

  $range=21$

Hence, range is 21.

For mean, substituting the values,

  $z=93$

Hence, mode is 93

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,

  $M=\frac{82+84}{2}$

On solving,

  $M=83$

Hence, median is 83

For mean, substituting the values,

  $\overline{x}=\frac{\text{72+75+79+82+84+87+88+90+92+93}}{10}$

So,

  $\overline{x}=\frac{824}{10}$

On solving,

  $\overline{x}=82.4$

Hence, mean is 82.4

For standard deviation, substituting the values,

  $sd=\sqrt{\frac{\begin{array}{}$ {(72-82.4)}^2+{(75-82.4)}^2+{(79-82.4)}^2+{(82-82.4)}^2+{(84-82.4)}^2+{(87-82.4)}^2+$$ {(88-82.4)}^2+{(90-82.4)}^2+{(92-82.4)}^2+{(93-82.4)}^2$\end{array}}{10}}$ so,

  $sd=\sqrt{\frac{\text{148}\text{.84+84}\text{.64+27}\text{.04+4}\text{.84+0}\text{.04+7}\text{.84+14}\text{.44+33}\text{.64+60}\text{.84+77}\text{.44}}{10}}$

So,

  $sd=\sqrt{45.96}$

On solving,

  $sd=6.78$

Hence, standard deviation is 6.78