Chapter 4, Problem 11CAP

(a)

To determine

Find the range of data in the population.

Answer to Problem 11CAP

The range of data in the populationis 74.

Explanation of Solution

Calculation:

The data measures the number of minutes (per day)that a small hypothetical population of college students spendsonline.

Range:

The difference between the largest value in the data set and smallest value in the data set is termed as range.

The formula for range is, $\text{Range}=L-S$,L denotes the largest value and S denotes the smallest value.

The largest value is 98 and the smallest value is 24. The range is,

$\begin{array}{c}\text{Range}=L-S\\ =98-24\\ =74\end{array}$

Hence, the value of range is 74.

(b)

To determine

Find the IQR of data in this population.

Answer to Problem 11CAP

The IQR of data in this population is 17.

Explanation of Solution

Calculation:

Arrange the data in the numeric order 24, 65, 77, 82, 88, 88, 92, 94, 98, 98. That is $n=10$.

The formula to find the interquartile range is, $IQR=Q_3-Q_1$

Follow the below steps to calculate IQR for given data:

The position of median for data is,

$\begin{array}{c}\frac{n+1}{2}=\frac{10+1}{2}\\ =\frac{11}{2}\\ =5.5\end{array}$

First quartile:

The median of the lower half of the values from $Q_2$ is termed as first quartile.

Compute position of $Q_1=\frac{n+1}{2}$ for the values below position of $Q_2$. The values are 24, 65, 77, 82, 88, now $n=5$

The position of first quartile for data is,

$\begin{array}{c}\frac{n+1}{2}=\frac{5+1}{2}\\ =\frac{6}{2}\\ =3\end{array}$

The value that is in position 3 is 77. The value of lower quartile $\left(Q_1\right)$ is 77.

Third quartile:

The median of the upper half of the values from $Q_2$ is termed as third quartile.

Compute position of $Q_3=\frac{n+1}{2}$ for the values above position of $Q_2$. The values are, 88, 92, 94, 98, 98, now $n=5$

The position of third quartile for data is,

$\begin{array}{c}\frac{n+1}{2}=\frac{5+1}{2}\\ =\frac{6}{2}\\ =3\end{array}$

The value that is in position 3 is 94. The value of upper quartile $\left(Q_3\right)$ is 94.

The interquartile range is,

$\begin{array}{c}IQR=Q_3-Q_1\\ =94-77\\ =17\end{array}$

Hence, the IQR of data in this population is 17.

(c)

To determine

Find the SIQR of data in this population.

Answer to Problem 11CAP

The SIQR of data in this population is 8.5.

Explanation of Solution

Calculation:

The formula to find the semi-interquartile range is, $SIQR=\frac{IQR}{2}$

From part 2, the IQR value is 17.

The semi-interquartile range is,

$\begin{array}{c}SIQR=\frac{IQR}{2}\\ =\frac{17}{2}\\ =8.5\end{array}$

Hence, the SIQR of data in this population is 8.5.

(d)

To determine

Find the population variance.

Answer to Problem 11CAP

The population variance is 448.64.

Explanation of Solution

Calculation:

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

Assume that, the random variable X denotes the scores of students.

The formula for population variance is, ${\sigma }^2=\frac{SS}{N}$.

Follow the below steps to find the sum of the squared deviation for given data.

The population meanis,

$\begin{array}{c}\mu =\frac{{\displaystyle \sum x}}{N}\\ =\frac{98+77+88+65+24+92+94+98+88+82}{10}\\ =\frac{806}{10}\\ =80.6\end{array}$

The squared deviation for score 98 is,

$\begin{array}{c}\text{Squared deviation}={\left(98-80.6\right)}^2\\ ={\left(17.4\right)}^2\\ =302.76\end{array}$

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

Scores	Squared deviation
98	${\left(98-80.6\right)}^2=302.76$
77	${\left(77-80.6\right)}^2=12.96$
88	${\left(88-80.6\right)}^2=54.76$
65	${\left(65-80.6\right)}^2=243.36$
24	${\left(24-80.6\right)}^2=3,203.56$
92	${\left(92-80.6\right)}^2=129.96$
94	${\left(94-80.6\right)}^2=179.56$
98	${\left(98-80.6\right)}^2=302.76$
88	${\left(88-80.6\right)}^2=54.76$
82	${\left(82-80.6\right)}^2=1.96$

Table 1

The sum of squared deviation is,

$\begin{array}{c}SS={\displaystyle \sum {\left(X-\mu \right)}^2}\\ =302.76+12.96+54.76+243.36+3,203.56+129.96+179.56+302.76+54.76+1.96\\ =4,486.4\end{array}$

The population varianceis,

$\begin{array}{c}{\sigma }^2=\frac{SS}{N}\\ =\frac{4,486.4}{10}\\ =448.64\end{array}$

Hence, the population varianceis 448.64.

(e)

To determine

Find the population standard deviation.

Answer to Problem 11CAP

The population standard deviationis 21.18.

Explanation of Solution

Calculation:

The population variance is 448.64.

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

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{448.64}\\ =21.18\end{array}$

Hence, the population standard deviation is 21.18.