Question

suppose we measured the height of 3000 women and found that the data were normally distributed with a mean of 64 inches and a standard deviation of 4 inches. answer the following
what proportion of women can be expected to have heights less than 57, 61, 67.4, 72 inches

Answer 1

To find proportion of women can be expected to have heights less than some value we have to integrate the probability density function from -infinity to that value, that is integral

int (1/(sigma*sqrt(2*Pi)) * exp(-(x-mu)^2/(2sigma^2)) dx) from -infinity to some value

where sigma = 4 (standard deviation) and mu = 64 (mean)
doing this numerically we obtain:

less then 57 inches
int (1/(sigma*sqrt(2*Pi)) * exp(-(x-mu)^2/(2sigma^2)) dx) from -infinity to 57 is 0.0401
that is 120 women, 4% of all

less then 61 - 0.2266
int (1/(sigma*sqrt(2*Pi)) * exp(-(x-mu)^2/(2sigma^2)) dx) from -infinity to 61 is 0.2266
that is 680 women, 22.3% of all

less then 67.4 - 0.8023
int (1/(sigma*sqrt(2*Pi)) * exp(-(x-mu)^2/(2sigma^2)) dx) from -infinity to 67.4 is 0.8023
that is 2407 women, 80.2% of all

less then 72 - 0.9772
int (1/(sigma*sqrt(2*Pi)) * exp(-(x-mu)^2/(2sigma^2)) dx) from -infinity to 72 is 0.9772
that is 2932 women, 97.7% of all