Question

Directions:

Show your work AND explain your reasoning using complete English sentences. Explanations must directly follow the solution for each part of the problem.

1.The mean height of men is 69.7 inches. Assume that the heights of men are normally distributed, with a standard deviation of 2.9 in.

a. 10% of men are taller than which height? Round your answer to 1 decimal place.

b. Find the percentage of men that are 6’0” or taller. Round your final answer to 2 decimal places.

c. How does the approximate percentage for the height of men 6’0” or taller (from the normal distribution) compare with the actual percentage of baseball players 6’0” or taller (from Problem 1)?

d. Based on your normal distribution calculation from part (b), approximately how many Major League Baseball players would you expect to be 6’0” or taller? Round your answer to the nearest whole number.

Answer 1

Let X be the random variable, heights of men in inches. Then X~ N(69.7, 2.92

Mean = 69.7

Stdev = 2.9 a. 10% of mean are taller than lets say height h

So , P(x>h) = .10 or P(x<=h) = .90

So, from the Z tables we know that for a cumulative probability of .90 we have a Z of 1.282

Hence, (x-Mu)/Sigma = 1.282

(x-69.7)/2.9 = 1.282

x = 69.7 + 2.9*1.282 = 73.42

So, 10% of men are taller than 73.4 inches

b. 1 foot = 12 inches, so 6 ft = 72 inches

So, %age of people greater than 6 ft = P(x>72) = P(z > (72-69.7)/2.9) = P(z>.793) = .2139 or 21.39% of population has height more than 6 ft

c. and d. ) these parts are not solvable , as we don't know the normal distribution of baseball players.