Question

determine the 2.5th percentile of the normal distribution with a mean of 72.5 and a standard deviation of 15 as a decimal rounded to the nearest tenth.

Answer 1

43.1

Explanation:

2.5 % = .025 look for this value on the z-score table to find this corresponds to z = - 1.96 Standard Deviations

72.5 - 1.96 * 15 = 43.1