Statistics homework help needed

Question

The heights of adult men in America are normally distributed, with a mean of 69.7 inches and a standard deviation of 2.61 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.3 inches and a standard deviation of 2.53 inches.

a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?

z =

b) What percentage of men are SHORTER than 6 feet 3 inches? Round to nearest tenth of a percent.

%

c) If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?

z =

d) What percentage of women are TALLER than 5 feet 11 inches? Round to nearest tenth of a percent.

%

e) Who is relatively taller: a 6'3" American man or a 5'11" American woman? Defend your choice in a meaningful sentence.

Answer 1

First Step, gather the parameters. (you should be familiar with what each symbol stands for)

(all in inches) µ_men =69.7 σ_men = 2.61 µ_women = 64.3 σ_women = 2.53

Second step, get necessary formulae

other formulae. Z = (x - µ) / σ where x is measure of interest; Lefttail CDF = NORM.DIST(x,µ,σ,1)

Third step, use proper formulae , solve

a) x = 75 inches. Z = (75-69.7)/2.61 Z = 2.03

b) x = 75 using CDF. (in excel NORM.DIST(75,69.7,2.61,1) P(x≤ 75) = .979 or 97.9%

(also using Z values NORM.DIST(2.03,0,1,1) )

c) x = 71. Z = (71-64.3)/2.53 Z = 2.64822 or rounded = 2.65

d) x = 71 right tail cdf = 1- NORM.DIST(2.65,0,1,1) or 0.004045 (rounded to 0.4%)

e) which of the two examples is relatively taller (larger chance of being less than measured values.

Remember, for man it was 97.9% , for woman it was 99.6%. (1 - .4%). SO: what do you think?

e)