Chapter 4, Problem 4.1P

Compute the range and standard deviation of the following 10 scores. (HINT: It will be helpful to organize your computations as in Table 4.4.)

10, 12, 15, 20, 25, 30, 32, 35, 40, 50

To determine

To find:

The range and standard deviation of the given scores

Answer to Problem 4.1P

Solution:

The range and standard deviation of the given scores is 40 and 12.28 respectively.

Explanation of Solution

The given scores are,

$10,12,15,20,25,30,32,35,40,50$

Formula used:

The formula to calculate the range is,

$\text{Range}=\text{High Score}-\text{Low score}$

Let the data values be $X_i$’s.

The formula to calculate the mean is,

$\overline{X}=\frac{{\displaystyle \sum _{i=1}^N{X}_i}}{N}$

The formula to calculate the standard deviation is,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X_i-\overline{X}\right)}^2}}{N}}$

Where, $N$ is the population size.

Calculation:

Arrange the given data in increasing order.

The data in increasing order is given by,

$10,12,15,20,25,30,32,35,40,50$

The highest score is 50 and the lowest score is 10.

The range is given by,

$\text{Range}=\text{High Score}-\text{Low score}$

Substitute 50 for high score and 10 for low score in the above mentioned formula,

$\begin{array}{rll}\text{Range}& =& 50-10\\ & =& 40\end{array}$

The range of population is 40.

The size of the population is 10.

The mean is given by,

$\overline{X}=\frac{{\displaystyle \sum _{i=1}^N{X}_i}}{N}$

Substitute 10 for $N$, 10 for $X_1$, 12 for $X_2$ and so on in the above mentioned formula,

$\begin{array}{rll}\overline{X}& =& \frac{10+12+15+20+25+30+32+35+40+50}{10}\\ & =& \frac{269}{10}\\ & =& 26.9\end{array}$ ……$\left(1\right)$

Consider the following table of sum of squares,

Scores $\left(X_i\right)$	$\left(X_i-\overline{X}\right)$	${\left(X_i-\overline{X}\right)}^2$
10	$-16.9$	285.61
12	$-14.9$	222.01
15	$-11.9$	141.61
20	$-6.9$	47.61
25	$-1.9$	3.61
30	3.1	9.61
32	5.1	26.01
35	8.1	65.61
40	13.1	171.61
50	23.1	533.61
	${\displaystyle \sum \left(X_i-\overline{X}\right)=0}$	${\displaystyle \sum {\left(X_i-\overline{X}\right)}^2=1506.9}$

From equation $\left(1\right)$, substitute $10$ for $X_1$ and $26.9$ for $\overline{X}$ in $\left(X_1-\overline{X}\right)$.

$\begin{array}{rll}\left(X_1-\overline{X}\right)& =& \left(10-26.9\right)\\ \left(X_1-\overline{X}\right)& =& -16.9\end{array}$

Square both sides of the equation.

$\begin{array}{rll}{\left(X_1-\overline{X}\right)}^2& =& {\left(-16.9\right)}^2\\ {\left(X_1-\overline{X}\right)}^2& =& 285.61\end{array}$

Proceed in the same manner to calculate ${\left(X_i-\overline{X}\right)}^2$ for all the $1\le i\le N$ for the rest of the values and refer table for the rest of the ${\left(X_i-\overline{X}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2& =& 285.61+222.01+\mathrm{..................}+171.61+533.61\\ & =& 1506.9\end{array}$ ……$\left(2\right)$

The formula to calculate the standard deviation is given by,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X_i-\overline{X}\right)}^2}}{N}}$

From equation $\left(2\right)$, substitute 1506.9 for ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ and 10 for $N$ in the above mentioned formula,

$\begin{array}{rll}s& =& \sqrt{\frac{1506.9}{10}}\\ & =& \sqrt{150.69}\\ & \approx & 12.28\end{array}$

Thus, range is 40 and standard deviation is 12.28.

Conclusion:

Therefore, the range and standard deviation of the given scores is 40 and 12.28 respectively.