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)?

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

Step-by-step 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!!!

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .

1 is c

2 is a

3 is a

step-by-step explanation: