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# Answer to Question #25132 in Statistics and Probability for joe smith

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
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

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