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 = 114.8standard deviation, s = 13.2sample size, n = 23range required: 119≤x≤122.7using this data; z = sqrt (n)(x-x_bar)/stherefore,z1 = sqrt (23)(119-114.8)/13.2  ≈ 1.53z2 = sqrt (23)(122.7-114.8)/13.2  ≈ 2.87from z-table and z = 1.53, p(119) = 0.9357from z-table and z = 2.87, p(122.7) = 0.9979therefore,p(119≤x≤122.7) = p(122.7)-p(119) = 0.9979 - 0.9357 = 0.0622 or 6.22%

What am i doing here. add?