In a survey of a group of men, the heights in the 20-29 age group were normaly d
ID: 3319617 • Letter: I
Question
In a survey of a group of men, the heights in the 20-29 age group were normaly distributed, with a mean of 69.7 inches and a standard deviation of 4.0 inches. A study participant is randomly selected.Complete parts (a) through (d) below. (a) Find the probability that a study participant has a height that is less than 65 inches The probability that the study participant selected at random is less than 65 inches tall isRound to four decimal places as needed.) (b) Find the probability that a study participant has a height that is between 65 and 70 inches The probability that the study participant selected at random is between 65 and 70 inches tall is [ (c) Find the probability that a study participant has a height that is more than 70 inches. Round to four decimal places as needed) The probability that the study parti ant selected at random is more than 70 inches tall is Round to four dec mal places as needed. (d) Identify any unusual events. Explain your reasoning. Choose the correct answer below 0 A. B. ° C. 0 D. The event in part (a) is unusual because its probability is less than 0.05. There are no unusual events because all the probablities are greater than 0.05. The events in parts (a), (b), and (c) are unusual because all of their probabilities are less than 0.05. The events in parts (a) and (c) are unusual because its probabilities are less than 0.05.Explanation / Answer
Mean= 69.7 inches
Standard deviation = 4.0 inches
P(X < A) = P(Z < (A - mean)/standard deviation)
a) P(X < 65) = P(Z < (65 - 69.7)/4)
= P(Z < -1.175)
= 0.1120
b) P(65 < X < 70) = P(X < 70) - P(X < 65)
= P(Z < (70 - 69.7)/4) - 0.1120
= P(Z < 0.075) - 0.1120
= 0.5299 - 0.1120
= 0.4179
c) P(X > 70) = 1 - P(X < 70)
= 1 - 0.5299
= 0.4701
d) B. There are no unusual events because all the probabilities are greater than 0.05
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