Question

Given that Kansas City men's height follows a normal distribution with its mean of 69 inches and standard deviation of 2.5 inches, are the following statements true or false?

Omaha men's height is known to follow a normal distribution with its mean of 68.5 inches and standard deviation of 2.7 inches. A 71 inch tall man would place higher in percentile in Kansas City than in Omaha

If the median is at 67.5 inches, the distribution is positively skewed

Answer 1

A 71 inch tall man would place higher in percentile in Kansas City than in Omaha :_

STATEMENT IS FALSE

z score for Kansas city

z = (x-mean)/standard deviation...................where x = 71, mean = 69 and standard deviation = 2.5

= (71-69)/2.5

= 0.8

z score for Omaha men's height

z = (x-mean)/standard deviation...................where x = 71, mean = 68.5 and standard deviation = 2.7

= (71-68.5)/2.7

= 0.96

So,, z score is higher for Omaha men's height as compared to Kansas city. Therefore, Omaha men's will be placed higher in percentile

If the median is at 67.5 inches, the distribution is positively skewed

When mean is greater than median, we have a positive skewed distribution, So this statement is TRUE