c) 56 and 88
95% is 2 standard deviations above and below the mean.
72 ± 2(8)
= 72 ± 16
= 56 and 88
C) 56 and 88
Option is is the correct answer.
The score of Mr.Myles is between 56 and 88.
Given : When Mrs. Myles gave a test, the scores were normally distributed with a mean of 72 and a standard deviation of 8.
To find : The class interval at 95% of her students scored between which two scores?
We have given,
The class interval at 95% is given by
Substitute the values in the formula,
Therefore, The score of Mr.Myles is between 56 and 88.
68% of data is within 1 std deviation of the mean
95% of data is within 2 std deviation of the mean
99.7% of data is within 3 std deviation of the mean
In this case 95% of the cases would be within two std deviations of the mean
mean - 8 and mean + 8
72 - 8 = 64 and 72 + 8 = 80
then 95% of the scores are between 64% and 80% on the test.
the answer is c
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when mrs. myles gave a test, the scores were normally distribute...