Answer:
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
