The average mark of candidates in an aptitude test was 138.5 with a standard deviation of 10.6. Three scores extracted from the test are; 178, 122, 100. What is the average of the extracted scores that are extreme values (outliers)? Correct your answer to the nearest whole number.
Answers
Answer:
Step-by-step explanation:
Given : The average mark of candidates in an aptitude test was 138.5 with a standard deviation of 10.6. Three scores extracted from the test are; 178, 122, 100
To find : . What is the average of the extracted scores that are extreme values (outliers)
Solution:
Mean = 138.5
SD = 10.6
Data values outside Z score ± 3 are outliers
Z score = ( value - mean)/Standard Deviation
-3 = ( Value - 138.5)/10.6
=> - 31.8 = Value - 138.5
=> Value = 106.7
3 = ( Value - 138.5)/10.6
=> 31.8 = Value - 138.5
=> Value = 170.3
Data range is 106.7 , 170.3
Hence 100 & 178 are outliers
Average of outliers = ( 100 + 178)/2 = 278/2 = 139
139 is the average of outliers
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