Question

According to survey​ data, the distribution of arm spans for males is approximately Normal with a mean of

71.4 inches and a standard deviation of 3.5 inches.

1. What percentage of men have arm spans between 66 and 76 ​inches?

The percentage of men with arms spans between 66 and 76 inches is_________ %.

(Round to one decimal place as​ needed.)

2. A particular professional basketball player has an arm span of almost 89 inches. Find the​ z-score for this​ person's arm span. ___________

What percentage of males have an arm span at least as long as this​ person? ___________

Answer 1

Ans:

Normal distribution with mean=71.4 and standard deviation=3.5

a)

z(66)=(66-71.4)/3.5=-1.543

z(76)=(76-71.4)/3.5=1.314

P(-1.543<z<1.314)=P(z<1.314)-P(z<-1.543)

=0.9056-0.0614=0.8442

b)

z score=(89-71.4)/3.5

z score=5.03

P(z>=5.03)=0.00000 or 0%