Chapter 3, Problem 62SE

The average number of times Americans dine out in a week fell from 4.0 in 2008 to 3.8 in 2012 (Zagat.com, April 1, 2012). The number of times a sample of 20 families dined out last week provides the following data.
a.Compute the mean and median.
b.Compute the first and third quartiles.
c.Compute the range and interquartile range.
d.Compute the variance and standard deviation.
e.The skewness measure for these data is 0.34. Comment on the shape of this distribution. Is it the shape you would expect? Why or why not?
f.Do the data contain outliers?

To determine

(a)

To find the mean and median of the given data.

Answer to Problem 62SE

Mean = 3

Median = 3

Explanation of Solution

Given:

The data:

6	1	5	3	7	3	0	3	1	3
4	1	2	4	1	0	5	6	3	1

Formula used:

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

Calculation:

  $\begin{array}{c}\text{Mean = }\frac{59}{20}\\ \Rightarrow 2.95\end{array}$

Now, let's calculate median,

In order to calculate we need to first arrange the data in ascending order

0	0	1	1	1	1	1	2	3	3
3	3	3	4	4	5	5	6	6	7

  $\begin{array}{c}\text{Median = }\frac{{\left(\frac{20}{2}\right)}^{\text{th}\text{term}}+{\left(\frac{20}{2}+1\right)}^{\text{th}\text{term}}}{2}\\ \Rightarrow \frac{{10}^{\text{th}\text{term}}+{11}^{\text{th term}}}{2}\\ \Rightarrow \frac{3+3}{2}\\ \Rightarrow 3\end{array}$

Conclusion:

The mean and median for the given data is 3.

To determine

(b)

To find the first and third quartile of the given data.

Answer to Problem 62SE

  $\begin{array}{l}{Q}_1=1\\ Q_3=4.75\end{array}$

Explanation of Solution

Given:

The data:

6	1	5	3	7	3	0	3	1	3
4	1	2	4	1	0	5	6	3	1

Formula used:

  $\begin{array}{l}{\text{Q}}_1={\left[\frac{1}{4}\left(n+1\right)\right]}^{\text{th term}}\\ {\text{Q}}_3={\left[\frac{3}{4}\left(n+1\right)\right]}^{\text{th term}}\end{array}$

Calculation:

We need to first arrange the data in ascending order;

0	0	1	1	1	1	1	2	3	3
3	3	3	4	4	5	5	6	6	7

  $\begin{array}{c}{\text{Q}}_1={\left[\frac{1}{4}\left(n+1\right)\right]}^{\text{th term}}\\ \Rightarrow {\left[\frac{1}{4}\left(20+1\right)\right]}^{\text{th term}}\\ \Rightarrow {5.25}^{\text{th}\text{term}}\\ \Rightarrow 5^{\text{th term}}+\left(6^{\text{th term}}-5^{\text{th term}}\right)\times \frac{1}{4}\\ \Rightarrow 1+\left(1-1\right)\times \frac{1}{4}\\ \Rightarrow 1\end{array}$

  $\begin{array}{c}{\text{Q}}_1={\left[\frac{3}{4}\left(n+1\right)\right]}^{\text{th term}}\\ \Rightarrow {\left[\frac{3}{4}\left(20+1\right)\right]}^{\text{th term}}\\ \Rightarrow {15.75}^{\text{th}\text{term}}\\ \Rightarrow {15}^{\text{th term}}+\left({16}^{\text{th term}}-{15}^{\text{th term}}\right)\times \frac{3}{4}\\ \Rightarrow 4+\left(5-4\right)\times \frac{3}{4}\\ \Rightarrow 4.75\end{array}$

Conclusion:

The value of first quartile is 1 and third quartile is 4.75.

To determine

(c)

To find the value of Range and Interquartile Range.

Answer to Problem 62SE

  $\begin{array}{l}\text{Range = 7}\\ \text{Interquartile Range = 3}\text{.75}\end{array}$

Explanation of Solution

Given:

The data:

6	1	5	3	7	3	0	3	1	3
4	1	2	4	1	0	5	6	3	1

Formula used:

  $\begin{array}{l}{\text{Q}}_1={\left[\frac{1}{4}\left(n+1\right)\right]}^{\text{th term}}\\ {\text{Q}}_3={\left[\frac{3}{4}\left(n+1\right)\right]}^{\text{th term}}\end{array}$

Calculation:

We need to first arrange the data in ascending order;

0	0	1	1	1	1	1	2	3	3
3	3	3	4	4	5	5	6	6	7

  $\begin{array}{c}\text{Range = Highest Value - Lowest Value}\\ \Rightarrow \text{7 - 0}\\ \Rightarrow 7\end{array}$

  $\begin{array}{c}{\text{Interquartile Range = Q}}_3-{\text{Q}}_1\\ \Rightarrow 4.75-1\\ \Rightarrow 3.75\end{array}$

Conclusion:

The value of range is 7 and interquartile range = 3.75.

To determine

(d)

To find the value of variance and Standard deviation.

Answer to Problem 62SE

  $\begin{array}{l}\text{Standard}\text{Deviation = 2}\text{.09}\\ \text{Variance = 4}\text{.366}\end{array}$

Explanation of Solution

Given:

The data:

6	1	5	3	7	3	0	3	1	3
4	1	2	4	1	0	5	6	3	1

Formula used:

  $\begin{array}{l}\text{Standard}\text{Deviation}\text{=}\sqrt{\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n-1}}\\ \text{Variance = }\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n-1}\end{array}$

Calculation:

	$x$	$\left(x_i-\overline{x}\right)$	${\left(x_i-\overline{x}\right)}^2$
	0	-2.95	8.7025
	0	-2.95	8.7025
	1	-1.95	3.8025
	1	-1.95	3.8025
	1	-1.95	3.8025
	1	-1.95	3.8025
	1	-1.95	3.8025
	2	-0.95	0.9025
	3	0.05	0.0025
	3	0.05	0.0025
	3	0.05	0.0025
	3	0.05	0.0025
	3	0.05	0.0025
	4	1.05	1.1025
	4	1.05	1.1025
	5	2.05	4.2025
	5	2.05	4.2025
	6	3.05	9.3025
	6	3.05	9.3025
	7	4.05	16.4025
Sum	59	0.00	82.95
Mean	2.95

  $\begin{array}{c}\text{Standard Deviation = }\sqrt{\frac{82.95}{19}}\\ \Rightarrow 2.09\end{array}$

  $\begin{array}{c}\text{Variance = }{\left(\text{Standard}\text{Deviation}\right)}^2\\ \Rightarrow {2.09}^2\\ \Rightarrow 4.366\end{array}$

Conclusion:

The value of Standard Deviation is 2.09 and Variance is 4.366.

To determine

(e)

To identify the shape of the distribution when skewness is equal to 0.34.

Answer to Problem 62SE

Left Skewed or Negatively Skewed

Explanation of Solution

Given:

The data:

6	1	5	3	7	3	0	3	1	3
4	1	2	4	1	0	5	6	3	1

It has been given that the value of skewness is 0.34 and as we have calculated the value of mean and median where the value of median is 3 and mean is 2.95, When the median is greater than mean then the shape of the distribution is negatively skewed or left skewed.

Conclusion:

The shape of the given data is negatively skewed or left skewed.

To determine

(f)

To identify whether there is any outlier in the given data or not.

Answer to Problem 62SE

No outliers exist

Explanation of Solution

Given:

The data:

6	1	5	3	7	3	0	3	1	3
4	1	2	4	1	0	5	6	3	1

Formula used:

  $\begin{array}{l}Outlier<{\text{Q}}_{\text{1}}-1.5\left(\text{IQR}\right)\\ Outlier>{\text{Q}}_3+1.5\left(\text{IQR}\right)\end{array}$

  $\begin{array}{l}{\text{Q}}_{\text{1}}-1.5\left(\text{IQR}\right)\\ \Rightarrow 1-1.5\left(3.75\right)\\ \Rightarrow -4.625\end{array}$

  $\begin{array}{l}{\text{Q}}_3+1.5\left(\text{IQR}\right)\\ \Rightarrow 4.75+1.5\left(3.75\right)\\ \Rightarrow 10.375\end{array}$

Since in the given data there is no value which is less than -4.625 and greater than 10.375. Therefore, there is no outliers exists in the given data set.

Conclusion:

The data contains no outliers.