Chapter 10, Problem 19SGR

To determine

To calculate and compare the mean and standard deviation

Answer to Problem 19SGR

Day customer mean is 9.58 standard deviation is 12.58

Night customer mean is 11.67 and standard deviation is 2.46

Explanation of Solution

Given information:

The data of day and night customer,

Day customers	Night customers
10,3,12,15,7,8,4,12,9,14,12,9	15,12,13,9,11,12,14,12,8,16,9,9

Formula used:

Mean is,

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

Standard deviation is,

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

Calculation:

For the day customer

For mean, substituting the value,

  $\overline{x}=\frac{10+3+12+15+7+8+4+12+9+14+12+9}{12}$

So,

  $\overline{x}=\frac{115}{12}$

On solving,

  $\overline{x}=9.58$

Hence, mean is 9.58

For standard deviation, substituting the values,

  $sd=\sqrt{\frac{\begin{array}{}$ {(10-9.58)}^2+{(3-9.58)}^2+{(12-9.58)}^2+{(15-9.58)}^2+{(7-9.58)}^2+{(8-9.58)}^2+{(4-9.58)}^2+$$ {(12-9.58)}^2+{(9-9.58)}^2+{(14-9.58)}^2+{(12-9.58)}^2+{(9-9.58)}^2$\end{array}}{12}}$

So,

  $sd=\sqrt{\frac{\begin{array}{l}\text{0}\text{.1736+ 43}\text{.3403+ 5}\text{.8403+ 29}\text{.3403+6}\text{.6736+2}\text{.5069+31}\text{.1736+5}\text{.8403++0}\text{.3403+19}\text{.5069+}\\ \text{5}\text{.8403+0}\text{.3403}\end{array}}{12}}$

so,

  $sd=\sqrt{12.57638889}$

On solving,

  $sd=12.58$

Hence, standard deviation is 12.58

For the night customer

For mean, substituting the value,

  $\overline{x}=\frac{15+13+12+6+11+12+14+12+8+16+9+9}{12}$

So,

  $\overline{x}=\frac{140}{12}$

On solving,

  $\overline{x}=11.67$

Hence, mean is 9.58

For standard deviation, substituting the values,

  $sd=\sqrt{\frac{\begin{array}{}$ {(15-11.67)}^2+{(13-11.67)}^2+{(12-11.67)}^2+{(6-11.67)}^2+{(11-11.67)}^2+{(12-11.67)}^2+$$ {(14-11.67)}^2+{(12-11.67)}^2+{(8-11.67)}^2+{(16-11.67)}^2+{(9-11.67)}^2+{(9-11.67)}^2$\end{array}}{12}}$

So,

  $sd=\sqrt{\frac{\begin{array}{l}\text{11}\text{.1111+ 0}\text{.1111+1}\text{.7778+7}\text{.1111+0}\text{.4444+0}\text{.1111+5}\text{.4444+0}\text{.1111+13}\text{.4444+18}\text{.7778+}\\ \text{7}\text{.1111+7}\text{.1111}\end{array}}{12}}$

so,

  $sd=\sqrt{6.055555556}$

On solving,

  $sd=2.46$

Hence, standard deviation is 2.46

Comparing the day and night customer mean, the mean of night time customer is larger.

Comparing the day and night customer standard deviation, the standard deviation of day time customer is larger.