Question

A sample of 20 glass bottles of a particular type was selected, and the internal pressure strength of each bottle was determined. Consider the following partial sample information.

• median = 201.7
• lower fourth = 198.0
• upper fourth = 216.4

Three smallest observations	125.3	187.5	193.7
Three largest observations	221.4	230.9	250.5

(a)

Are there any outliers in the sample?

Yes No    

Any extreme outliers?

Yes No    

Answer 1

We have to find fouth spread fs.

fs = Upper forth - Lower fouth

fs = 216.4 - 198 = 18.4

we have that,

1.5fs = 1.5*18.4 = 27.6

3fs = 3*18.4 = 55.2

using the above infromation, we can calculate the mild and extreme outliers.

x1 = upper fourth; x2 = lower fourth

x1+1.5fs = 216.4+27.6 = 244 (mild outliers)

x2-1.5fs = 198-27.6 = 170.4 (mild outliers)

x1+3fs = 216.4+55.2 = 271.6 (extreme outliers)

x2+3fs = 198-55.2 = 142.8 (extreme outliers)

The mild outliers all are values between 142.8 and 170.4 as well as between 244 and 271.6

The extreme outliers are all values below 142.8 and bigger than 271.6

There is an extreme outliers 125.3 in the listed observation, because it is smaller than 142.8

There is a mild outliers 250.5

because 244 < 250.5 < 271.6