Chapter 3, Problem 12RE

Advertising costs: The amounts spent (in billions) on media advertising in the United States for a sample of categories are presented in the following table.

1. Find the mean amount spent on advertising.
2. Find the median amount spent on advertising.
3. Find the sample variance of the advertising amounts.
4. Find the sample standard deviation of the advertising amounts.
5. Find the first quartile of the advertising amounts.
6. Find the third quartile of the advertising amounts.
7. Find the 40th percentile of the advertising amounts.
8. Find the 65th percentile of the advertising amounts.

To determine

To find mean.

Answer to Problem 12RE

Mean is 8.75.

Explanation of Solution

Given table:

Amount spent
16.35
14.84
8.98
8.66
7.89
6.84
6.57
6.34
6.19
4.86

Formula used:

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

Here n = 10.

Putting all values in formula,

  $\overline{x}=\frac{87.52}{10}=8.75$

Therefore, mean is 8.75.

To determine

The median of given of amount spent.

Answer to Problem 12RE

Median is7.365.

Explanation of Solution

Given data sorted in ascending order:

Amount spent
4.86
6.19
6.34
6.57
6.84
7.89
8.66
8.98
14.84
16.35

Calculation:

Here n = 10 which is an even number.

Formula of median for even number is,

  $\begin{array}{l}Median={\left(\frac{n+1}{2}\right)}^{th}item\\ Median={\left(\frac{10+1}{2}\right)}^{th}item\\ Median={\left(5.5\right)}^{th}item\\ Median=\frac{5^{th}item+6^{th}item}{2}\\ Median=\frac{6.84+7.89}{2}\\ Median=7.365\end{array}$

To determine

To find sample variance.

Answer to Problem 12RE

Sample variance is 14.62.

Explanation of Solution

Creating table for sample standard deviation:

$x$	$x-\overline{x}$	${(x-\overline{x})}^2$
16.35	1.03	1.0609
14.84	-0.48	0.2304
8.98	-6.34	40.1956
8.66	-6.66	44.3556
7.89	-7.43	55.2049
6.84	-8.48	71.9104
6.57	-8.75	76.5625
6.34	-8.98	80.6404
6.19	-9.13	83.3569
4.86	-10.46	109.4116
Sum		131.5

Formula is,

  $s^2=\frac{{\displaystyle \sum {(x_i-\overline{x})}^2}}{n-1}$

Calculation:

From the table,

  ${\displaystyle \sum {(x_i-\overline{x})}^2}=131.5$

Put in a formula,

  $s^2=\frac{131.5}{10}=14.62$

Sample variance = 14.62

To determine

The sample standard deviation.

Answer to Problem 12RE

Sample standard deviation is 3.82.

Explanation of Solution

Sample variance is 14.62

Formula for sample standard deviation is,

  $\begin{array}{l}s=\sqrt{s^2}\\ s=\sqrt{14.62}=3.82\end{array}$

To determine

To find first quartile.

Answer to Problem 12RE

First quartile is 6.30.

Explanation of Solution

Given data sorted in ascending order:

Amount spent
4.86
6.19
6.34
6.57
6.84
7.89
8.66
8.98
14.84
16.35

Formula:

  $\begin{array}{l}{Q}_1=\frac{1}{4}{(}^{n}value\\ Q_1=\frac{1}{4}{(}^{10}value\\ Q_1=\frac{1}{4}{(}^{11}value\\ Q_1={(}^{2.75}value\\ Q_1=2^{nd}value+\frac{3}{4}\times (3^{rd}value-2^{nd}value)\\ Q_1=6.19+\frac{3}{4}\times (6.34-6.19)\\ Q_1=6.30\end{array}$

The first quartile is 6.30.

To determine

The third quartile.

Answer to Problem 12RE

Third quartile is 10.445.

Explanation of Solution

Formula and calculation:

  $\begin{array}{l}{Q}_3=\frac{3}{4}{(}^{n}value\\ Q_3=\frac{3}{4}{(}^{10}value\\ Q_3=\frac{3}{4}{(}^{11}value\\ Q_3={(}^{8.25}value\\ Q_3=8.98+(8.25-8)x(14.84-8.98)\\ Q_3=10.445\end{array}$

To determine

To find 40^th percentile is 6.678.

Answer to Problem 12RE

40^th percentile.

Explanation of Solution

Given data sorted in ascending order:

Amount spent
4.86
6.19
6.34
6.57
6.84
7.89
8.66
8.98
14.84
16.35

Formula:

  $\begin{array}{l}PercentilePosition=\frac{(n+1)P}{100}\\ PercentilePosition=\frac{(10+1)\times 40}{100}\\ PercentilePosition=4.4\end{array}$

Since the position found is not an integer, the method of interpolation needs to be used. The 40% percentile is located between the values in positions 4 and 5. Those values, based on the data organized in ascending order, are 6.57 and 6.84.

The value of 4.4 - 4 = 0.4 corresponds to the proportion of the distance between 6.57 and 6.84 where the percentile we are looking for is located at.

Therefore,

  $P_{40}=6.57+0.4\times (6.84-6.57)=6.678$

To determine

The 65^th percentile.

Answer to Problem 12RE

65^th percentile is 8.708.

Explanation of Solution

  $\begin{array}{l}PercentilePosition=\frac{(n+1)P}{100}\\ PercentilePosition=\frac{(10+1)\times 65}{100}\\ PercentilePosition=7.15\end{array}$

Since the position found is not integer, the method of interpolation needs to be used. The 65% percentile is located between the values in the positions 7 and 8. Those values, based on the data organized in ascending order, are 8.66 and 8.98.

The value of 7.15 - 7 = 0.15 corresponds to the proportion of the distance between 8.66 and 8.98 where the percentile we are looking for is located at.

  $P_{65}=8.66+0.15\times (8.98-8.66)=8.708$