Chapter 3.1, Problem 50E

Take in a show: The following table presents the weekly attendance, in thousands, at Broadway shows during a recent season.

1. Find the mean and median weekly attendance.
2. Based on the mean and median, would you expect that a histogram would be skewed to the left, skewed to the right, or approximately symmetric? Explain.
3. Construct a frequency histogram. Do the results agree with your expectation?

To determine

To find mean weekly attendance

Answer to Problem 50E

Mean weekly attendance is 222.49

Explanation of Solution

Given:

$x$
149.4	196.4	231.6	244.1
162.5	196.6	232.1	245.9
180.9	197	234.9	246.5
181.6	205.3	235.2	251
183	208.5	235.8	252.1
183.1	213.6	236.8	254.9
184.2	214	238.5	256
185.5	214.9	239.3	257.5
187.7	218.2	240.2	260.2
189.9	219.7	240.4	265.6
190.4	220.6	240.9	269.2
191.3	222.4	240.9	269.4
192.2	228.3	241	292.4

Formula:

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

Calculation:

Here n = 52.

Values of Xi are weekly attendance.

Putting all values in formula,

  $\overline{x}=\frac{11569.6}{52}=222.49$

Therefore, mean state income is 222.49

To determine

To find median weekly attendance

Answer to Problem 50E

Median weekly attendance is 229.5

Explanation of Solution

Formula:

  $Median={\left(\frac{n+1}{2}\right)}^{th}item$

Calculation:

Sorted data in ascending order:

$x$
149.4	196.4	231.6	244.1
162.5	196.6	232.1	245.9
180.9	197	234.9	246.5
181.6	205.3	235.2	251
183	208.5	235.8	252.1
183.1	213.6	236.8	254.9
184.2	214	238.5	256
185.5	214.9	239.3	257.5
187.7	218.2	240.2	260.2
189.9	219.7	240.4	265.6
190.4	220.6	240.9	269.2
191.3	222.4	240.9	269.4
192.2	228.3	241	292.4

Here n = 52

Putting all values in formula,

  $\begin{array}{l}Median={\left(\frac{n+1}{2}\right)}^{th}item\\ Median={\left(\frac{52+1}{2}\right)}^{th}item\\ Median={26.5}^{th}item\\ Median=\frac{228.3+231.6}{2}\\ Median=229.95\end{array}$

Therefore, median weekly attendance is 229.95

To determine

To find skewness of the data

Answer to Problem 50E

The data is left skewed

Explanation of Solution

Here,

Mean = 222.49

Median = 229.95

Therefore, the mean is smaller than the median, the data are skewed left

To determine

To construct a frequency histogram

Answer to Problem 50E

The data is left skewed. Hence expectation meet

Explanation of Solution

Frequency histogram from given data:

  

From frequency histogram, the tail of the data seems to be highly on left side. Therefore, shape of the distribution of data is left skewed. Hence expectation meet from frequency histogram