Use Excel to answer each of the following questions. In a murder trial in Los An

Question

Use Excel to answer each of the following questions. In a murder trial in Los Angeles, a shoe expert stated that the range of heights of men with a size 12 shoe is 71 inches to 76 inches. Suppose the heights of all men wearing size 12 shoes are normally distributed with a mean of 73.5 inches and a standard deviation of 1 inch. Assume each of the questions relates to a randomly selected man who wears a size 12 shoe. What is the z-score associated with a man who is 75" tall? a) 1.50 b) 1.99 c) 2.05 d) 5.57 What is the probability that a randomly selected man with a size 12 shoe is less than 75" tall? a) .3392 b) 0.9332 c) 1.05 d) 1.50 What is the probability that a randomly selected man with a size 12 shoe is greater than 75" tall? a) 0.8660 b) 0.9960 c) 0.0668 d) 1.6680 What is the z-score associated with a man who is 71" tall? a) +2.50 b) -2.50 c) +1.50 d) -1.50 What is the probability that a randomly selected man with a size 12 shoe is less than 71" tall? a) 0.0062 b) 0.6002 c) 0.2206 d) 0.8820 What is the probability that a randomly selected man with a size 12 shoe is between 71" and 75" tall? a) 0.7290 b) .07750 c) 0.5000 d) 0.9270 What is the z-score associated with a man who is 76" tall? a) +2.50 b) -2.50 c) +1.50 d) -1.50 What is the probability that a randomly selected man with a size 12 shoe is outside of the range of between 71 and 76 inches? a) 0.1240 b) 0.0124 c) 2.50 d) -1.50 A machine is used to cut a metal automobile part to its desired length. The machine can be set so that the mean length of the part will be any value that is desired. The standard deviation of the lengths always runs at 0.02 inches. What is the z-score associated with only 0.4% of the parts being cut shorter than some value? a) +2.65 b) -2.50 c) -2.65 d) -1.50 Where should the mean be set if we want only 0.4% of the parts cut by the machine to be shorter than 15 inches long? a) 13.00 b) 14.00 c) 15.05 d) 18.05

Explanation / Answer

Q1

Z= (73.5-75)/1 = 1.5

Option a)

Q2. 0.9332 oprion b)

P(Z<75) = 0.5+P(0<Z<1.5) = 0.5+0.4332 = 0.9332

Q3

0.0668 option c)

P(Z>75)= 0.5-P(0<z<1.5) = 0.5 - 0.4332 = 0.0668

Q4

Z = {71-73.5}/1 = -2.5

option b

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