Question

Test scores of students in a statistics class have a mean of 78 and a standard deviation equal to 6. What percent of the scores are either less than 64 or greater than 90?       

[A]       approx. 5%      [B]       at least 75%     [C]       at most 25%    [D]       approx. 95%

Answer 1

[A]       approx. 5%  

Solution:

We are given

Mean = 78

SD = 6

We have to find P(X<64) + P(X>90)

For X = 64

Z = (X – mean) / SD

Z = (64 -78)/6

Z = -2.33333 ≈ -2

For X = 90

Z = (90 - 78)/6

Z = 2

We know that, according to empirical rule, area lies between two standard deviations from mean is approximately 95%.

So, area lies within 64 and 90 is approximately 95%.

So, area below 64 or above 90 is given as 1 – 0.95 = 0.05 or approximately 5%.