contestada

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.

Respuesta :

shinmin
The solution for this problem is:It's given that the heights are normally distributed. 
5 feet 7 inches = (5*12 +7) inches = (60+7) inches = 67 inches. 
The z-score is (67 - 69.1)/(2.65) = -1.136. The probability is 0.127978 or 12.8%, making probability of a height over 67 inches

Answer:

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<[tex]\frac{67-69.1}{2.65} =-0.79[/tex]

=P(Z<-0.79)

=0.5-0.2852

=0.2148

=21.48%

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

Otras preguntas