The mean score of a competency test is 65, with a standard deviation of 4. use the empirical rule to find the percentage of scores between 53 and 77. (assume the data set has a bell-shaped distribution.)

To solve this problem, we use the t statistic. The z score is calculated as:

z = (x – u) / s

where x is the sample value = 53 and 77, u is the mean score = 65 while s is the standard deviation = 4

when x = 53

z = (53 – 65) / 4 = -3

From the standard distribution tables, the p value at z = -3 is:

P (z=-3) = 0.0013

when x = 77

z = (77 – 65) / 4 = 3

From the standard distribution tables, the p value at z = 3 is:

P (z=3) = 0.9987

The percentage P between the two would be the difference:

P (53 < x < 77) = 0.9987 – 0.0013 = 0.9974

0.9974 or 99.74%

i got 10.28 but that rounds up to 10.3