How to calculate proportion with mean and standard deviation in normal distribution

A standard kitchen worktop is designed to cater for people whose heights are between 1.5m. and 2.0m. Given that the heights of adult males are normally distributed with a mean of 1.7m. and standard deviation of 0.10m, and heights of adult females are normally distributed with a mean of 1.65m. and a standard deviation of 0.10m, what proportion of the adult population can use the worktops? (assume equal numbers of males and females)

Hi,

Calculate the percentage included in each range of heights using your TI-83. It has the command normalcdf(min,max,mean,stddev) under DISTR.

For men, normalcdf(1.5,2,1.7,.1) = .9759
For women, normadcdf((1.5,2,1.65,.1) = .9330

(.9759 + .9330)/2 = .9544 or 95.44% of the adult population could use the counter <==ANSWER

I hope that helps!! :-)

Calculate Z-scores for men: 1.5 = 1.7 - 2(0.1) ..and.. 2.0 = 1.7 + 3(0.1)

So, men are 2 standard deviations below and 3 above. That is, 97.5% (= 47.5% below + 50% above) of men can use the worktops.

Calculate Z-scores for women: 1.5 = 1.65 - 1.5 (0.1) .. and .. 2.0 = 1.65 + 3.5 (0.1).

So, women are 1.5 standard deviations below and 3.5 standard deviations above. That is, X.Y% (=A.B% below corresponding to Z-score of -1.5 + 50% above corresponding to Z-score of 3.5)