For women aged 18-24, systolic blood pressures (in mm hg) are normally distributed with a mean of 114.8 and a standard deviation of 13.1. if 23 women aged 18-24 are randomly selected, find the probability that their mean systolic blood pressure is between 119 and 122. 7)

Mean, x_bar = 114.8

Standard deviation, S = 13.2

Sample size, n = 23

Range required: 119≤X≤122.7

Using this data;

Z = Sqrt (n) (X-x_bar) / S

Therefore,

Z1 = Sqrt (23) (119-114.8) / 13.2 ≈ 1.53

Z2 = Sqrt (23) (122.7-114.8) / 13.2 ≈ 2.87

From Z-table and Z = 1.53, P (119) = 0.9357

From Z-table and Z = 2.87, P (122.7) = 0.9979

Therefore,

P (119≤X≤122.7) = P (122.7) - P (119) = 0.9979 - 0.9357 = 0.0622 or 6.22%