Chapter 0.9, Problem 5E

To determine

To find:

The range, variance, and standard deviation of each set.

Answer to Problem 5E

  $\sqrt{242}\approx 15.55,\sqrt{1115.44}\approx 33.39$

Explanation of Solution

Given:

Means of sets

  $A=36.66,B=43.77$

Concept used:

The range is the difference between the highest and lowest value.

The variance is the average squared deviation from the mean.

The standard deviation is the average deviation from the mean.

Calculation:

Means of sets

  $A=36.66,B=43.77$

The range is

  $\begin{array}{l}61-14=47\\ 100-8=92\end{array}$

The variance is

Adding the all numbers from each set and the square and divide the number of integers in set

  $\begin{array}{l}A={\left(330\right)}^2=108900\\ B={\left(394\right)}^2=155236\\ A=\frac{108900}{9}=12100\\ B=\frac{155236}{9}=17248.44\end{array}$

  $\begin{array}{l}A={\left(14\right)}^2+{\left(22\right)}^2+{\left(25\right)}^2+\mathrm{....}=14036\\ B={\left(8\right)}^2+{\left(15\right)}^2+{\left(19\right)}^2+\mathrm{....}=26172\end{array}$

  $\begin{array}{l}14036-12100=1936\\ 26172-17248.44=8923.55\end{array}$

Total number of integers in set minus one

  $\begin{array}{l}9-1=8\\ \frac{1936}{8}=242\\ \frac{8923.55}{8}=1115.44\end{array}$

The standard deviation is

  $\begin{array}{l}\sqrt{242}\approx 15.55\\ \sqrt{1115.44}\approx 33.39\end{array}$