Chapter 11, Problem 5CT

To determine

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

Answer to Problem 5CT

Mean is 15, median is 13.86, mode is 23.50, range is 15.7 and standard deviation is 5.64

Explanation of Solution

Given information:

  $\$15.50,\$18.90,\$10.60,\$12.25,\$7.80,\$23.50,\$9.75,\$21.70$

Formula used:

Mean is,

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

Median is,

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

Mode is,

  $Mode=MaxValue$

Range is,

  $range=\max -\min$

Standard deviation is,

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

Calculation:

For mean, substituting the values,

  $\overline{X}=\frac{15.50+18.90+10.60+12.25+7.80+23.50+9.75+21.70}{8}$

On solving,

  $\overline{X}=\frac{120}{8}$

So,

  $\overline{X}=15$

Hence, mean is 15.

Re-arranging the data

  $7.80,9.75,10.60,12.25,15.50,18.90,21.70,23.50$

For median, substituting the values,

  $M=\frac{{(\frac{8}{2})}^{th}term+{(\frac{8}{2}+1)}^{th}term}{2}$

So,

  $M=\frac{4^{th}+5^{th}term}{2}$

Substituting the values,

  $M=\frac{12.25+15.50}{2}$

So,

  $M=\frac{27.725}{2}$

On solving,

  $M=13.86$

Hence, median is 13.86

For mode, substituting the values,

  $Mode=23.50$

Hence, mode is 23.50

  $7.80,9.75,10.60,12.25,15.50,18.90,21.70,23.50$

Minimum is 7.80 and maximum is 23.50

For range, substituting the values,

  $range=23.50-7.80$

On solving,

  $range=15.7$

Hence, range is 15.7

For standard deviation, substituting the values,

  $SD=\sqrt{\frac{\begin{array}{l}(7.80-15)+(9.75-15)+(10.60-15)+(12.25-15)+(15.50-15)+(18.90-15)+(21.70-15)\\ +(23.50-15)\end{array}}{8}}$

On solving,

  $SD=\sqrt{\frac{238.925}{8}}$

So,

  $SD=\sqrt{29.87}$

On solving,

  $SD=5.64$

Hence, standard deviation is 5.64