Question

The heights of adult men in America are normally distributed, with a mean of 69.5 inches and a standard deviation of 2.65 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.7 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 the 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

Answer 1

a)

µ =    69.5
σ =    2.65
X =6 feet 3 inch = 6*12 + 3 =   75   
                  
Z =(X - µ ) / σ = (   75   -   69.5   ) /    2.65
Z =    2.08

b)

P(X ≤   75   ) = P(Z ≤   2.08   ) =   0.9810 or 98.1%

excel formula for probability from z score is =NORMSDIST(Z)                  

c)

µ =    64.7              
σ =    2.53              
( X = 5 feet 11 inch = 5*12+11 = 71.000   )          
                  
Z =(X - µ ) / σ = (   71   -   64.7   ) /    2.53
Z =    2.49

d)

P(X ≥   71   ) = P(Z ≥   2.490   ) =   P ( Z <   -2.490   ) =    0.0064 or 0.6%

excel formula for probability from z score is =NORMSDIST(Z)