The heights of adult men in america are normally distributed, with a mean of 69.1 inches and standard deviation of 2.65 inches. what percentage of adult males in america are under 5 feet 7 inches tall? round your answer as a percentage to the nearest whole number.

Answer is 21%

Step-by-step explanation:

Let X be the heights of adult men in America

X is Normal with mean =69.1 and std dev = 2.65 inches

Percentage of adult males who are under 5 ft 7inches

= height less than 67 inches (since 1 ft = 12 inches)

P(X<67)

=P(Z<

=P(Z<-0.79)

=0.5-0.2852

=0.2148

=21.48%

Hence 21% nearly of males in America are shorter than 5'7"

given hc ∥ gd.¯¯¯¯¯¯which statement is true

step-by-step explanation:

∠cbe and ∠deb are same-side interior angles

what are you trying to solve here

step-by-step explanation: