Chapter 10.3, Problem 11PPS

To determine

To calculate standard deviation.

Answer to Problem 11PPS

Wildcat’s standard deviation is 2.23 and mean is 11.88.

Trojan’s standard deviation is 2.07 and mean is 10.06.

Explanation of Solution

Given information: Wildcat’s fouls are 15,12,13,9,11,12,14,12,8,16,9,9,11,13,12,14

Trojan’s fouls are 9,10,14,13,7,8,10,10,9,7,11,9,12,11,13,8

Formula used: Mean is,

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

Standard deviation is,

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

Where, SD= standard deviation,

x= value

Calculation:

For Wildcats,

Substituting the values in mean formula,

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

So,

  $\overline{x}=\frac{190}{16}$

On solving,

  $\overline{x}=11.88$

Hence, Mean is 11.88

For standard deviation substituting the values in the formula,

  $SD=\sqrt{\frac{\begin{array}{}$ {(15-11.88)}^2+{(12-11.88)}^2+{(13-11.88)}^2+{(9-11.88)}^2+{(11-11.88)}^2+{(12-11.88)}^2+$$ {(14-11.88)}^2+{(12-11.88)}^2+{(8-11.88)}^2+{(16-11.88)}^2+{(9-11.88)}^2+{(9-11.88)}^2+$$ {(11-11.88)}^2+{(13-11.88)}^2+{(12-11.88)}^2+{(14-11.88)}^2$\end{array}}{16}}$

So,

  $SD=\sqrt{\frac{\begin{array}{}$ {(3.12)}^2+{(0.12)}^2+{(1.12)}^2+{(-2.88)}^2+{(-0.88)}^2+{(0.12)}^2+{(2.12)}^2+{(0.12)}^2+{(-3.88)}^2+$$ {(4.12)}^2+{(-2.88)}^2+{(-2.88)}^2+{(-0.88)}^2+{(1.12)}^2+{(0.12)}^2+{(2.12)}^2$\end{array}}{16}}$

On solving,

  $SD=\sqrt{\frac{79.75}{16}}$

So,

  $SD=\sqrt{4.98}$

On solving,

  $SD=2.23$

Standard deviation 2.23

For Trojans,

For mean substituting the values in mean formula,

  $\overline{x}=\frac{9+10+14+13+7+8+10+10+9+7+11+9+12+11+13+8}{16}$

So,

  $\overline{x}=\frac{161}{16}$

On solving,

  $\overline{x}=10.06$

Mean is 10.06

Substituting the values in standard deviation formula,

  $SD=\sqrt{\frac{\begin{array}{}$ {(9-10.06)}^2+{(10-10.06)}^2+{(14-10.06)}^2+{(13-10.06)}^2+{(7-10.06)}^2+{(8-10.06)}^2+$$ {(10-10.06)}^2+{(10-10.06)}^2+{(9-10.06)}^2+{(7-10.06)}^2+{(11-10.06)}^2+{(9-10.06)}^2+$$ {(12-10.06)}^2+{(11-10.06)}^2+{(13-10.06)}^2+{(8-10.06)}^2$\end{array}}{16}}$

So,

  $SD=\sqrt{\frac{\begin{array}{}$ {(-1.06)}^2+{(-0.06)}^2+{(3.94)}^2+{(2.94)}^2+{(-3.06)}^2+{(-2.06)}^2+{(-0.06)}^2+{(-0.06)}^2+{(-1.06)}^2+$$ {(-3.06)}^2+{(0.84)}^2+{(-1.06)}^2+{(1.94)}^2+{(0.94)}^2+{(2.94)}^2+{(-2.06)}^2$\end{array}}{16}}$

On solving,

  $SD=\sqrt{\frac{68.76}{16}}$

So,

  $SD=\sqrt{4.30}$

On solving,

  $SD=2.07$

Hence, Standard deviation is 2.07

Comparing the means of wildcats and Trojan, the average fouls of wildcats are more than that of Trojan. Standard deviation of wildcats is more, hence, the foul points of wildcats are spread over a large number of values.