Chapter 4, Problem 29CAP

To determine

Find the value of SS by assuming 16 students constitute a population.

Find the variance by assuming 16 students constitute a population.

Find the standard deviation by assuming 16 students constitute a population.

Answer to Problem 29CAP

The value of SS by assuming 16 students constitute a population is 1,755.

The variance by assuming 16 students constitute a population is 109.69.

The standard deviation by assuming 16 students constitute a population is 10.47.

Explanation of Solution

The data is 16 students records the time to complete the exam is considered.

The data is 23, 32, 44, 20, 25, 14, 29, 41, 43, 21, 39, 33, 48, 38, 50, and 40.

Calculation:

Follow the below steps to find the variance for given data.

• Obtain the population mean for all scores.
• Take the square of the difference for each score from the population mean to obtain the squared deviation.
• Sum of squared deviations to obtain the sum of squared deviations.
• Divide sum of squared deviation by N to obtain the population variance.

The formula to obtain the sum of squared deviation is, $SS={\displaystyle \sum {\left(X-\mu \right)}^2}$.

The formula to obtain the sample variance is, ${\sigma }^2=\frac{SS}{N}$.

The formula to obtain the sample standard deviation is, $\sigma =\sqrt{{\sigma }^2}$.

The sample mean is,

$\begin{array}{c}\mu =\frac{{\displaystyle \sum x}}{N}\\ =\frac{23+32+44+\cdots +50+40}{16}\\ =\frac{540}{16}\\ =33.75\end{array}$

The squared deviation is,

$\begin{array}{c}\text{Squared deviation}={\left(X-\mu \right)}^2\\ ={\left(23-33.75\right)}^2\\ ={\left(-10.75\right)}^2\\ =115.5625\end{array}$

Similarly, the squared deviation for the remaining scores is obtained as shown in table (1).

Ages	Squared deviation
23	${\left(23-33.75\right)}^2=115.5625$
32	${\left(32-33.75\right)}^2=3.0625$
44	${\left(44-33.75\right)}^2=105.0625$
20	${\left(20-33.75\right)}^2=189.0625$
25	${\left(25-33.75\right)}^2=76.5625$
14	${\left(14-33.75\right)}^2=390.0625$
29	${\left(29-33.75\right)}^2=22.5625$
41	${\left(41-33.75\right)}^2=52.5625$
43	${\left(43-33.75\right)}^2=85.5625$
21	${\left(21-33.75\right)}^2=162.5625$
39	${\left(39-33.75\right)}^2=27.5625$
33	${\left(33-33.75\right)}^2=0.5625$
48	${\left(48-33.75\right)}^2=203.0625$
38	${\left(38-33.75\right)}^2=18.0625$
50	${\left(50-33.75\right)}^2=264.0625$
40	${\left(40-33.75\right)}^2=39.0625$

Table 1

The sum of squared deviation is,

$\begin{array}{c}SS={\displaystyle \sum {\left(X-\mu \right)}^2}\\ =115.5625+3.0625+105.0625+\cdots +264.0625+39.0625\\ =1,755\end{array}$

Thus, the value of SS by assuming 16 students constitute a population is 1,755.

The variance is,

$\begin{array}{c}{\sigma }^2=\frac{1,755}{16}\\ =109.69\end{array}$

Thus, the variance by assuming 16 students constitute a population is 109.69.

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{109.69}\\ =10.47\end{array}$

Thus, the standard deviation by assuming 16 students constitute a population is 10.47.

To determine

Find the value of SS by assuming 16 students constitute a sample.

Find the variance by assuming 16 students constitute a sample.

Find the standard deviation by assuming 16 students constitute a sample.

Answer to Problem 29CAP

The value of SS by assuming 16 students constitute a sample is 1,755.

The variance by assuming 16 students constitute a sample is 117.

The standard deviation by assuming 16 students constitute a sample is 10.82.

Explanation of Solution

The data is 16 students records the time to complete the exam is considered.

The data is 23, 32, 44, 20, 25, 14, 29, 41, 43, 21, 39, 33, 48, 38, 50, and 40.

Calculation:

Follow the below steps to find the sample variance for given data.

• Obtain the sample mean for all scores.
• Take the square of the difference for each score from the sample mean to obtain the squared deviation.
• Sum of squared deviations to obtain the sum of squared deviations.
• Divide sum of squared deviation by $\left(n-1\right)$ to obtain the sample variance.

The formula to obtain the sum of squared deviation is, $SS={\displaystyle \sum {\left(X-M\right)}^2}$.

The formula to obtain the sample variance is, $s^2=\frac{SS}{\left(n-1\right)}$.

The formula to obtain the sample standard deviation is, $SD=\sqrt{s^2}$.

The sample mean is,

$\begin{array}{c}M=\frac{{\displaystyle \sum x}}{n}\\ =\frac{23+32+44+\cdots +50+40}{16}\\ =\frac{540}{16}\\ =33.75\end{array}$

The squared deviation is,

$\begin{array}{c}\text{Squared deviation}={\left(X-M\right)}^2\\ ={\left(23-33.75\right)}^2\\ ={\left(-10.75\right)}^2\\ =115.5625\end{array}$

Similarly, the squared deviation for the remaining scores is obtained as shown in table (1).

Ages	Squared deviation
23	${\left(23-33.75\right)}^2=115.5625$
32	${\left(32-33.75\right)}^2=3.0625$
44	${\left(44-33.75\right)}^2=105.0625$
20	${\left(20-33.75\right)}^2=189.0625$
25	${\left(25-33.75\right)}^2=76.5625$
14	${\left(14-33.75\right)}^2=390.0625$
29	${\left(29-33.75\right)}^2=22.5625$
41	${\left(41-33.75\right)}^2=52.5625$
43	${\left(43-33.75\right)}^2=85.5625$
21	${\left(21-33.75\right)}^2=162.5625$
39	${\left(39-33.75\right)}^2=27.5625$
33	${\left(33-33.75\right)}^2=0.5625$
48	${\left(48-33.75\right)}^2=203.0625$
38	${\left(38-33.75\right)}^2=18.0625$
50	${\left(50-33.75\right)}^2=264.0625$
40	${\left(40-33.75\right)}^2=39.0625$

Table 1

The sum of squared deviation is,

$\begin{array}{c}SS={\displaystyle \sum {\left(X-M\right)}^2}\\ =115.5625+3.0625+105.0625+\cdots +264.0625+39.0625\\ =1,755\end{array}$

Thus, the value of SS by assuming 16 students constitute a sample is 1,755.

The variance is,

$\begin{array}{c}{s}^2=\frac{1,755}{16-1}\\ =\frac{1,755}{15}\\ =177\end{array}$

Thus, the variance by assuming 16 students constitute a sample is 177.

The standard deviation is,

$\begin{array}{c}SD=\sqrt{177}\\ =10.82\end{array}$

Thus, the standard deviation by assuming 16 students constitute a sample is 10.82.