Question #25132

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

Expert's answer

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

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

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