Assume the zscores of the bone mineral density test values a

Assume the z-scores of the bone mineral density test values are normally distributed with a mean of 0 and a standard deviation of 1. Consider that bone density test reading between -1.00 and 2.50 reveals that the subject has osteopenia (or some bone loss). Find the probability that a randomly selected subject has a reading between -1.00 and -2.50.

Solution

Mean = 0
SD = 1

z = (x - u) / SD
z = (-1 - 0) / 1
z = -1
P(z < -1) = 0.1587

z = (x - u) / SD
z = (-2.5 - 0) / 1
z = -2.5
P(z < -2.5) = 0.0062

P(-2.5 < z < -1) = 0.1587 - 0.0062

0.1525