Chapter 3.1, Problem 46E

News flash: The following table presents the circulation (in thousands) for the top 25 U.S. daily newspapers in both print and digital editions.

1. Find the mean and median circulation for print editions.
2. Find the mean and median circulation for digital editions.
3. The editor of an Internet news source says that digital circulation is more than half of print circulation. Do the data support this claim?

To determine

To Find: the mean and median circulation for print editions.

Answer to Problem 46E

For print editions, mean circulation = 335.07 and median circulation $=227.7$

Explanation of Solution

Given:

  $\begin{array}{l}125.7,126.3,149.5,159.4,172.0,180.3,184.8,184.8,190.6,192.8,213.8,216.1,227.7,\\ 231.2,241.0,265.8,285.9,300.0,360.5,368.1,431.1,432.9,731.4,1424.4,1480.7\end{array}$

Formula used:

  $\begin{array}{l}\text{Mean}(\overline{x})=\frac{{\displaystyle \sum x_i}}{n}\\ \\ \text{Median}=\left\{\begin{array}{l}\frac{{\left(n+1\right)}^{\text{th }}\text{term}}{2},\text{When}\text{n}\text{is}\text{odd}\\ \frac{{\left(\frac{n}{2}+1\right)}^{\text{th }}\text{term + }{\left(\frac{n}{2}\right)}^{\text{th}}\text{term}}{2},\text{When n is even}\end{array}\right\}\end{array}$

Calculation- Mean and Median circulation for print editions,

  $\begin{array}{l}\overline{x}=\frac{\sum x_i}{n}\\ =\frac{\left[\begin{array}{c}1480.7+300.0+190.6+241.0+125.7+731.4+431.1+265.8+216.1+126.3+1424.4+184.8+231.2\\ +184.8+192.8+432.9+213.8+159.4+227.7+172.0+360.5+368.1+180.3+285.9+149.5\end{array}\right]}{25}\\ =\frac{8876}{25}\\ =355.07\end{array}$

For Median, first arrange data in sequential order, $\begin{array}{l}125.7,126.3,149.5,159.4,172.0,180.3,184.8,184.8,190.6,192.8,213.8,216.1,227.7,\\ 231.2,241.0,265.8,285.9,300.0,360.5,368.1,431.1,432.9,731.4,1424.4,1480.7\end{array}$

  $\begin{array}{c}\text{Median =}{\left(\frac{\left(\text{n+1}\right)}{\text{2}}\right)}^{\text{th}}\text{term}\\ \text{=}{\left(\frac{\text{25+1}}{\text{2}}\right)}^{\text{th}}\text{term}\\ \text{=}{\left(\frac{\text{26}}{\text{2}}\right)}^{\text{th}}\text{term}\\ {\text{=13}}^{\text{th}}\text{term}\\ \text{=227}\text{.7}\end{array}$

Thus, for print editions, mean circulation 335.07 and median circulation $=227.7$ .

To determine

To Find: To Find the mean and median circulation for print editions.

Answer to Problem 46E

For digital editions, mean circulation 163.19 and median circulation $=73.7$

Explanation of Solution

Calculation- Mean and Median circulation for digital editions,

  $\begin{array}{l}\overline{x}=\frac{\sum x_i}{n}\\ =\frac{\left[\begin{array}{l}6.7+7.0+15.5+16.0+17.1+21.6+42.3+46.8+65.9+68.0+69.0+73.5+73.7+77.7\\ +95.5+102.3+112.0+155.7+160.5+177.7+192.8+200.6+249.9+898.1+1133.9\end{array}\right]}{25}\\ =\frac{4079.8}{25}\\ =163.19\end{array}$

For Median, first arrange data in sequential order,

  $\begin{array}{l}6.7,7.0,15.5,16.0,17.1,21.6,42.3,46.8,65.9,68.0,\\ 69.0,73.5,73.7,77.7,95.5,102.3,112.0,155.7,\\ 160.5,177.7,192.8,200.6,249.9,898.1,1133.9\end{array}$

  $\begin{array}{c}\text{Median =}{\left(\frac{\left(\text{n+1}\right)}{\text{2}}\right)}^{\text{th}}\text{term}\\ \text{=}{\left(\frac{\text{25+1}}{\text{2}}\right)}^{\text{th}}\text{term}\\ \text{=}{\left(\frac{\text{26}}{\text{2}}\right)}^{\text{th}}\text{term}\\ {\text{=13}}^{\text{th}}\text{term}\\ \text{=73}\text{.7}\end{array}$

Thus, for digital editions, mean circulation

  $\approx 163.19$ and median circulation $=73.7$ .

To determine

To Check: Whether digital circulation is more than half of print circulation.

Answer to Problem 46E

Digital circulation is not more than half of print circulation.

Explanation of Solution

By looking at mean and median number of print and digital circulation,

For print editions, mean circulation 335.07 and median circulation $=227.7$

For digital editions, mean circulation 163.19 and median circulation $=73.7$

So, it can be said from above data, the claim by an editor that digital circulation is more than half of print circulation is incorrect.