Question

Refer to Exercise 2.141. Assuming all populations are approximately mound-shaped, for parts a–d, determine whether the values 0, 4, and 12 are outliers.

Short answer

a. Outlier – 12

b. Outlier - 12

c. Outlier - 0

d. No Outliers

Step-by-step solution

Determining whether 0, 4, 12 are outliers when the mean is 2 and the standard deviation is 1 

If the distribution is mound-shaped, most of the population should lie within 3 standard deviations from the mean. Therefore, if the z-score is greater than 3, the value is an outlier.

When the mean is 2 and the standard deviation is 1, the z-scores are -2, 2, and 10 for 0, 4, and 12 respectively. Thus,only 12 is the outlier.

Repeating Step 1 for mean = 4 and s = 2

$\begin{array}{c}\text{Thez}-\text{scorevaluesare}0=-\text{2}\\ \text{4}=0\\ \text{12}=\text{4}\end{array}$

Using the same logic as above, again 12 is an outlier.

Finding the outlier from 0, 4, and 12

$\begin{array}{c}\text{Thez}-\text{scorevaluesare}0=-\text{4}\\ \text{4}=-2\\ \text{12}=2\end{array}$

Following the above procedure, 0 is an outlier.

Identifying the outlier with µ = 8 and σ = 8

$\begin{array}{c}\text{Thez}-\text{scorevaluesare}0=-\text{1}\\ \text{4}=-0.5\\ \text{12}=0.5\end{array}$

None of the values are outliers because they lie between (-3,3).