Question

The average mark of candidates in an aptitude test was 128.5 with a standard deviation of 8.2. Three scores extracted from the test are 148, 102, 15zs2. What is the average of the extracted scores that are extreme values (outliers)

Answer 1

Solution:

The average marks of candidates in an aptitude test was 128.5 with a standard deviation of 8.2. Three scores extracted from the test are; 148, 102 and 152.

To find : average of the extracted scores that are the extreme values (outliers)

The average marks of candidates in an aptitude test was 128.5

=> Mean = 128.5

Standard Deviation = 8.2

outlier will be out sides then range Mean ± 3* SD

128.5 ± 8.2 × 3

= 128.5 ± 24.6

Range is ( 103.9 , 153.1)

102 is outside this range

Hence outlier would be 102