Chapter 3.1, Problem 43E

Commercial bank: Following arc amounts spend (in millions of dollars)on media advertising in the United States by a sample of 10 companies in the first quarter of 2013.

1. Find the mean amount spent on advertising.
2. Find the median amount spent on advertising.
3. 1f the amount of 722.3 was incorrectly listed as 7225. How would this affect the mean? How would it affect the median?

To determine

Find the mean amount spent on advertising.

Answer to Problem 43E

The mean amount spend on advertising is $\$373.49$

Explanation of Solution

Given information:

$295.9$

$722.5$

$280.3$

$286.9$

$362.9$

$331.2$

$394.6$

$257.0$

$463.5$

$340.1$

Formula used:

$\overline{x}=\frac{{\displaystyle \sum x}}{n}$

Calculation:

Find the mean amount spend on advertising by using the formula is given below.

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{295.9+722.5+280.3+286.9+362.9+311.2+331.2+394.6+257+463.5+340.1}{10}\\ =\frac{3,734.9}{10}\\ =\$373.49\end{array}$

Thus the mean amount spend on advertising is $\$373.49$.

To determine

Find the media aount spent on advertising.

Answer to Problem 43E

The median amount spend on advertising is $\$335.65$

Explanation of Solution

Given information:

$295.9$

$722.5$

$280.3$

$286.9$

$362.9$

$331.2$

$394.6$

$257.0$

$463.5$

$340.1$

Formula used:

The median is the middle value and lies at the ${\left(\frac{n+1}{2}\right)}^{th}$ position, when the data are in ascending order.

Calculation:

Find the median amount spend on advertising.

Here, there are $10$ observations in the data. That is, the sample size, $n\left(=10\right)$ Is even.

Thus, the median is the middle value and lies at the ${\left(\frac{n+1}{2}\right)}^{th}$ position, when the data are in ascending order.

First, arrange the data in ascending order $257,280.3,286.9,2959,3312,340.1,362.9,394.6,463.5,722.5$

Hence, the median lies at the ${\left(\frac{10+1}{2}\right)}^{th}$ position. Therefore, the median is the mean of 5^{t h} and 6^{t h} position. That is.

$\begin{array}{c}\text{Median}=\frac{331.2+340.1}{2}\\ =\frac{671.3}{2}\\ =335.65\end{array}$

Thus, the median amount spend on advertising is $335.65$.

To determine

If the amount of $722.5$ was incorrectly listed as $7225$, how would this affect the mean? How would it affect the median?

Answer to Problem 43E

If tile amount of $722.5$ was incorrectly listed as $7225$ which is the extreme value in the given data, the value of mean would be affected and the value of median would not be affected. That is, there is no change in median but the value of mean is increased to $\$1,023.74$.

Explanation of Solution

Given information:

The amount of $722.5$ was incorrectly listed as $7225$.

Formula used:

$\overline{x}=\frac{{\displaystyle \sum x}}{n}$

Calculation:

Find the mean amount spend on advertising by using the formula is given below:

The amount of $722.5$ was incorrectly listed as $7225$. Therefore, the value of mean is,

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{295.9+7225+280.3+286.9+362.9+331.2+394.6+257+463.5+340.1}{0}\\ =\frac{10,237.4}{10}\\ =\$1,023.74\end{array}$

Thus, the mean amount spend on advertising is $\$1,023.74$

Find the median amount spend on advertising.

The amount of $722.5$ was incorrectly listed as $7225$.

Here, there are $10$ observations in the data. That is, the sample size $n\left(=10\right)$ is even.

Thus, the median is the middle value and lies at the ${\left(\frac{n+1}{2}\right)}^{th}$ position, when the data are in ascending order. First, arrange the data in ascending order.

$257,280.3,286.9,295.9,331.2,340.1,362.9,394.6,463.5,7225$

Hence, the median lies at the ${\left(\frac{10+1}{2}\right)}^{th}$ position Therefore, the median is the mean of 5^{t h} and 6^{t h} position That is,

$\begin{array}{c}\text{Median}=\frac{331.2+340.1}{2}\\ =\frac{671.3}{2}\\ =335.65\end{array}$

Hence, the median amount spend on advertising is $\$335.65$.

If the amount of $722.5$ was incorrectly listed as $7225$ which is the extreme value in the given data, the value of mean would be affected and the value of median would not be affected. That is, there is no change in median but the value of mean is increased to $\$1,023.74$.