Question

I can understand questions a through d but cant seem to figure out question e. Please help me solve this and explain how its done please
The heights of adult men in a certain country are normally distributed with a mean of 70.1 inches and a standard deviation of 2.9 inches. Complete parts (a) through (e) below. a. What are the standard score and percentile of a height of 72 inches? The standard score is z= .66. (Round to two decimal places as needed.) The percentile is the 74.54 percentile. (Round to two decimal places as needed.) b. What are the standard score and percentile of a height of 66 inches? The standard score is z= -1.41. (Round to two decimal places as needed.) The percentile is the 7.93 percentile. (Round to two decimal places as needed.) c. What percentage of the adult male population of this country is taller than 6'1"'?

Answer 1

Total population - 92 million men so, Now, to find at least as tall as baskept ball Player who is 6 g" tall, Height = 81 inches ( 2 feet = 12 inches) X=81 cell = 70. 2 S.D. (0) - 2.9, Z = X-4 so, meam ů 6 2 = 89-70.1 2.9 Z=.3.7586 z = 3.76

From Normal Distribution Table, For z = 376, Percentage of men tallen tham SO 9" tall is (0.008% So, From population of 22 million men (120,00,000) -0.008.% of popula ation men are at least tall as 6'gl tall. So, Total men 120,00, 000 X 0.008 100 Total men 960
BY THIS WAY, ANSWER OF GIVEN PROBLEM.