Chapter 3, Problem 3.1LB

The following set of data is from a sample of n = 5:

7 4 9 8 2

1. Calculate the mean, median, and mode.
2. Calculate the range, variance. standard deviation, and coefficient of variation.
3. Calculate the Z scores. Are there any outliers?
4. Describe the shape of the data set.

To determine

To compute: The mean, median, and mode for the given data set.

Answer to Problem 3.1LB

Mean is 6, median is 7 for the given dataset.

Explanation of Solution

Given:

Given data set is:

7,4,9,8, and 2.

Formula:

  $\text{Mean}\left(\mu \right)=\frac{{\displaystyle \sum x}}{n}$

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

Calculation:

Mean could be calculated as for the given data set:

  $\begin{array}{c}\text{Mean}\left(\mu \right)=\frac{7+4+9+8+2}{5}\\ =6\end{array}$

Median could be calculated as for the given data set:

Arrange the data in ascending ordered:

2,4,7,8, and 9.

  $\begin{array}{c}\text{Median}={\left(\frac{n+1}{2}\right)}^{th}term\\ =3^{rd}term\\ =7\end{array}$

Thus, Median is 7.

Mode:

There is no mode.

To determine

To conclude:

The range, variance, standard deviation, and coefficient of variation.

Answer to Problem 3.1LB

The is 7, variance is 11.33, standard deviation is 3.366 for the given data set.

Explanation of Solution

Given:

From the above given data set,

Calculation:

The range could be calculated as:

  $\begin{array}{c}R=\max -\min \\ =9-2\\ =7\end{array}$

Formula:

  ${\sigma }^2=\frac{1}{n-1}{\displaystyle \sum {\left(x-\mu \right)}^2}$

The standard deviation $\left({\sigma }^2\right)$ could be calculated as:

  $\begin{array}{c}{\sigma }^2=\frac{1}{n-1}{\displaystyle \sum {\left(x-\mu \right)}^2}\\ =\frac{{\left(7-6\right)}^2+{\left(4-6\right)}^2+{\left(9-6\right)}^2+{\left(8-6\right)}^2+{\left(2-6\right)}^2}{4-1}\\ =11.33\end{array}$

The standard deviation $\left(\sigma \right)$ could be calculated as:

  $\begin{array}{c}\left(\sigma \right)=\sqrt{{\sigma }^2}\\ =\sqrt{11.33}\\ =3.3660\end{array}$

To determine

To compute:

The Z-score, there is any outliers.

Answer to Problem 3.1LB

z scores are -1.18, -0.29, 0.89, -0.59, 0.59.

Explanation of Solution

Given:

from the above part,

Formula:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

Z score can be calculated:

x =2,7,9,4,8

  $z=\frac{x-\mu }{\sigma }$

Thus,

  $z=\frac{x-6}{3.366}$

Thus,

Z =-1.18, -0.29,0.89,-0.59,0.59

There is one value outlier 2 because 2 does not lie in the interval $\left(\mu -\sigma ,\mu +\sigma \right)$ .

To determine

To explain:

The shape of the data set.

Explanation of Solution

Given:

from the above given data set:

Graph:

The graph could be construct for the given data set as:

  

Interpretation:

Shape:

It seems that there would be two nods.