Chapter 14.2, Problem 29E

(a)

To determine

To find: The mean of the number of goals scored by a team.

Answer to Problem 29E

The approximately mean is $215.2$ .

Explanation of Solution

Given:

The given table is as:

Goals	Number of Teams
$\begin{array}{l}\text{160-180}\\ \text{180-200}\\ \text{200-220}\\ \text{220-240}\\ \text{240-260}\\ \text{260-280}\end{array}$	$\begin{array}{l}1\\ 6\\ 10\\ 6\\ 3\\ 1\end{array}$

Calculation:

The mean is calculated as:

  $\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x_i{f}_i}}{{\displaystyle \sum f_i}}\\ =\frac{1}{2}\left(1\cdot 170+6\cdot 190+10\cdot 210+6\cdot 230+3\cdot 250+1\cdot 270\right)\\ \approx 215.2\end{array}$

The approximately mean is $215.2$ .

(b)

To determine

To find: The median class of the frequency distribution.

Answer to Problem 29E

The median class of the frequency distribution is $200\text{-220}$ .

Explanation of Solution

First make a table of cumulative frequency teams.

Goals	Number of Teams	Cumulative frequency
$\begin{array}{l}\text{160-180}\\ \text{180-200}\\ \text{200-220}\\ \text{220-240}\\ \text{240-260}\\ \text{260-280}\end{array}$	$\begin{array}{l}1\\ 6\\ 10\\ 6\\ 3\\ 1\end{array}$	$\begin{array}{l}1\\ 7\\ 17\\ 23\\ 26\\ 27\end{array}$

Notice that $\frac{27}{2}=13.5$ has been gathered in the $200\text{-220}$ class. So this is the median class.

(c)

To determine

To find: The median of the number of goals scored by a team.

Answer to Problem 29E

The median is $213$ .

Explanation of Solution

The median class is $200\text{-220}$ .

The median is calculated as”

  $\begin{array}{c}\text{Median}=l+\left(\frac{\frac{n}{2}-cf}{f}\right)h\\ =200+\left(\frac{13.5-7}{10}\right)20\\ =213\end{array}$

The median is $213$ .

(d)

To determine

To find: The mean and median of actual data.

Answer to Problem 29E

The mean and median are given below.

Explanation of Solution

The actual data is as:

  $\begin{array}{l}268,248,245,2472,239,236,231,230,217,215,214,211,210,\\ 210,207,205,202,200,196,194,192,190,189,184,179\end{array}$

The mean and median is calculated as:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_i}}{n}\\ =\frac{\begin{array}{l}268+248+245+2472+239+236+231+230+217+215+214+211+210+\\ 210+207+205+202+200+196+194+192+190+189+184+179\end{array}}{25}\\ \approx 215.9\end{array}$

After arranging the data in ascending order we will find the median as:

  $\begin{array}{c}{M}_d=\left(\frac{n+1}{2}\right)\text{th term}\\ =13\text{th term}\\ =211\end{array}$

The mean and median are given above.

(e)

To determine

To find: The comparison between the measure of central tendency found by using the frequency chart with the measures of central tendency found by using the actual data.

Answer to Problem 29E

The explanation is given below.

Explanation of Solution

The mean calculated using the frequency distribution is very close to the one calculated with the actual data. The median calculated with the actual data is less than the one calculated with the frequency distribution.

The explanation is given above.