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3 votes
The average mark of c

andidates in an aptitude test was 128.5 with a standard deviation of

8.2. Three scores extracted from the test are; 148, 102, 152. What is the average of the

extracted scores that are extreme values (outliers)?​

User Nadh
by
8.1k points

1 Answer

6 votes

Answer:

The average of the extracted scores that are extreme values (outliers) = 102

Explanation:

With the logical assumption that the population size is large enough, for a normal distribution,

68% of the data lies within 1 standard deviation of the mean.

95% of the data lies between 2 standard deviations of the mean.

99.7% of the data lies within 3 standard deviations of the mean.

So, the outliers for a normal distribution are usually beyond 3 standard deviations of the mean.

The mean = 128.5

Standard deviation = 8.2

The range of scores within 3 standard deviations of the mean is obtainable thus

(Mean ± 3standard deviations)

3 × standard deviations = 3 × 8.2 = 24.6

(128.5 ± 24.6) = (103.9, 153.1)

The 3 extracted scores are 148, 102 and 152. The only extreme value of these 3 extracted scores is 102. The two other scores are within the range of 3 standard deviations of the mean.

Hence, the average of the extracted scores that are extreme values (outliers) = 102

Hope this Helps!!!

User Blklight
by
7.7k points
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