Chapter 12, Problem 1T

Visits to Social Networking Sites The table shows the number of persons age 2 and older in any U.S. location, in millions, who visited the listed Web sites at least once in June 2013.

Rank	Web Site	Visitors
1	Facebook	144.7
2	LinkedIn	50.6
3	Twitter	39.1
4	Tumblr	35.9
5	Pinterest	27.9
Source: com Score Media Metrix; com Score q Search

Use this information to determine each of the following.

(a) the mean number of visitors per site

(b) the range

(c) the standard deviation

(d) the coefficient of variation

To determine

To calculate: The Mean number of visitors per site using the following information.

Rank	Web Site	Visitor
1	Facebook	144.7
2	LinkedIn	50.6
3	Twitter	39.1
4	Tumblr	35.9
5	Pinterest	27.9

Answer to Problem 1T

Solution:

The required mean is approximately 59.6 million

Explanation of Solution

Formula used:

Mean of n data items  can be found with the formula, $\overline{x}=\frac{{\displaystyle \sum x}}{n}$

Calculation:

Since, mean can be found with the formula, $\overline{x}=\frac{{\displaystyle \sum x}}{n}$

Thus, mean will be:

$\begin{array}{c}\overline{x}=\frac{144.7+50.6+39.1+35.9+27.9}{5}\\ \approx 59.6\end{array}$

To determine

To calculate: The range of the following data.

Rank	Web Site	Visitor
1	Facebook	144.7
2	LinkedIn	50.6
3	Twitter	39.1
4	Tumblr	35.9
5	Pinterest	27.9

Answer to Problem 1T

Solution:

The required range is 116.8 million.

Explanation of Solution

Formula used:

Range is the difference between highest data value and lowest data value.

Calculation:

Since, range is the difference between highest data value and lowest data value

So, range $=144.7-27.9=116.8$

To determine

To calculate: The standard deviation of the following data.

Rank	Web Site	Visitor
1	Facebook	144.7
2	LinkedIn	50.6
3	Twitter	39.1
4	Tumblr	35.9
5	Pinterest	27.9

Answer to Problem 1T

Solution:

The required standard deviation is 48.2 million.

Explanation of Solution

Formula used:

Standard deviation is:

$s=\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}$, where $x_i$ is the data value and $\overline{x}$ is mean

Calculation:

Formula for standard deviation is:

$s=\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}$, where $x_i$ is the data value and $\overline{x}$ is mean

Thus,

$\begin{array}{c}s=\frac{{\left(144.7-59.6\right)}^2+{\left(50.6-59.6\right)}^2+{\left(39.1-59.6\right)}^2+{\left(35.9-59.6\right)}^2+{\left(27.9-59.6\right)}^2}{n-1}\\ =\frac{7242.0+81.0+420.3+561.7+1004.9}{5-1}\\ =\frac{9309.9}{4}\\ =2327.5\end{array}$

Now, take the square root,

$\begin{array}{c}s=\sqrt{2327.5}\\ \approx 48.2\text{ million}\end{array}$

To determine

To calculate: The coefficient of variation of the following data.

Answer to Problem 1T

Solution:

The coefficient of variation is approximately 80.9%.

Explanation of Solution

Formula used:

Formula for coefficient of variation is:

$V=\frac{s}{\overline{x}}\times 100$, where $s$ is standard deviation, $\overline{x}$ is mean.

Calculation:

$\begin{array}{c}V=\frac{48.2}{59.6}\times 100\\ \approx 80.9\%\end{array}$

Hence, the coefficient of variation is approximately 80.9%.