Chapter 4, Problem 12CAP

(a)

To determine

Find the range of data in the sample.

Answer to Problem 12CAP

The range of data in the sample is 21.

Explanation of Solution

Calculation:

The data measures the number of minutes (per day) that a small hypothetical population of college students spends online. The sample consists of persons A, B, C, F, G, H, I, and J.

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 77. The range is,

$\begin{array}{c}\text{Range}=L-S\\ =98-77\\ =21\end{array}$

Hence, the value of range is 21.

(b)

To determine

Find the IQR of data in this sample.

Answer to Problem 12CAP

The IQR of data in this sample is 11.

Explanation of Solution

Calculation:

Arrange the data in the numeric order 77, 82, 88, 88, 92, 94, 98, 98. That is $n=8$.

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,

$\begin{array}{c}\frac{n+1}{2}=\frac{8+1}{2}\\ =\frac{9}{2}\\ =4.5\end{array}$

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 77, 82, 88, 88, now $n=4$

The position of first quartile for data is,

$\begin{array}{c}\frac{n+1}{2}=\frac{4+1}{2}\\ =\frac{5}{2}\\ =2.5\end{array}$

The value that is in position 2.5 is 82 and 88. The value of lower quartile $\left(Q_1\right)$ is the average of the 82 and 88.

The lower quartile $\left(Q_1\right)$ is,

$\begin{array}{c}{Q}_1=\frac{82+88}{2}\\ =\frac{170}{2}\\ =85\end{array}$

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, 92, 94, 98, 98, now $n=4$

The position of upper quartile for data is,

$\begin{array}{c}{Q}_3=\frac{n+1}{2}\\ =\frac{4+1}{2}\\ =\frac{5}{2}\\ =2.5\end{array}$

The value that is in position o is 94 and 98. The value of upper quartile $\left(Q_3\right)$ is the average of the 94 and 98.

The lower quartile $\left(Q_3\right)$ is,

$\begin{array}{c}{Q}_3=\frac{94+98}{2}\\ =\frac{192}{2}\\ =96\end{array}$

The interquartile range is,

$\begin{array}{c}IQR=Q_3-Q_1\\ =96-85\\ =11\end{array}$

Hence, the IQR of data in this sample is 11.

(c)

To determine

Find the SIQR of data in this sample.

Answer to Problem 12CAP

The SIQR of data in this sample is 5.5.

Explanation of Solution

Calculation:

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

From part 2, the IQR value is 11.

The semi-interquartile range is,

$\begin{array}{c}SIQR=\frac{IQR}{2}\\ =\frac{11}{2}\\ =5.5\end{array}$

Hence, the SIQR of data in this sample is 5.5.

(d)

To determine

Find the sample variance.

Answer to Problem 12CAP

The sample variance is 55.41.

Explanation of Solution

Calculation:

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

The sample variance is, $s^2=\frac{SS}{n-1}$.

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

The population meanis,

$\begin{array}{c}M=\frac{{\displaystyle \sum x}}{n}\\ =\frac{98+77+88+92+94+98+88+82}{8}\\ =\frac{717}{8}\\ =89.625\end{array}$

The squared deviation for score 98 is,

$\begin{array}{c}\text{Squared deviation}={\left(98-89.625\right)}^2\\ ={\left(8.375\right)}^2\\ =70.14\end{array}$

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

Minutes	Squared deviation
98	${\left(98-89.625\right)}^2=70.14$
77	${\left(77-89.625\right)}^2=159.39$
88	${\left(88-89.625\right)}^2=2.64$
92	${\left(92-89.625\right)}^2=5.64$
94	${\left(94-89.625\right)}^2=19.14$
98	${\left(98-89.625\right)}^2=70.14$
88	${\left(88-89.625\right)}^2=2.64$
82	${\left(82-89.625\right)}^2=58.14$

Table 1

The sum of squared deviation is,

$\begin{array}{c}SS={\displaystyle \sum {\left(X-M\right)}^2}\\ =70.14+159.39+2.64+5.64+19.14+70.14+2.64+58.14\\ =387.87\end{array}$

The sample varianceis,

$\begin{array}{c}{s}^2=\frac{SS}{n-1}\\ =\frac{387.87}{8-1}\\ =\frac{387.87}{7}\\ =55.41\end{array}$

Hence, the sample varianceis 55.41.

(e)

To determine

Find the sample standard deviation.

Answer to Problem 12CAP

The sample standard deviationis 7.44.

Explanation of Solution

Calculation:

The samplevariance is 55.41.

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

The standard deviation is,

$\begin{array}{c}s=\sqrt{55.41}\\ =7.44\end{array}$

Hence, the sample standard deviation is 7.44.