Question

In a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of 68.8 inches and a standard deviation of 3.0 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 is (Round to four decimal places as needed.) (b) Find the probability that a study participant has a height that is between 65 and 71 inches The probability that the study participant selected at random is between 65 and 71 inches tall is (Round to four decimal places as needed.) (c) Find the probability that a study participant has a height that is more than 71 inches. The probability that the study participant selected at random is more than 71 inches tallis (Round to four decimal places as needed.) (d) Identify any unusual events. Explain your reasoning. Choose the correct answer below O A. The events in parts (a) and (c) are unusual because its probabilities are less than 0.05. OB. The event in part (a) is unusual because its probability is less than 0.05. OC. There are no unusual events because all the probabilities are greater than 0.05. OD. The events in parts (a), (b), and (c) are unusual because all of their probabilities are less than 0.05

Answer 1

Part a)
X ~ N ( µ = 68.8 , σ = 3 )
P ( X < 65 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 65 - 68.8 ) / 3
Z = -1.2667
P ( ( X - µ ) / σ ) < ( 65 - 68.8 ) / 3 )
P ( X < 65 ) = P ( Z < -1.2667 )
P ( X < 65 ) = 0.1026

Part b)
X ~ N ( µ = 68.8 , σ = 3 )
P ( 65 < X < 71 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 65 - 68.8 ) / 3
Z = -1.2667
Z = ( 71 - 68.8 ) / 3
Z = 0.7333
P ( -1.27 < Z < 0.73 )
P ( 65 < X < 71 ) = P ( Z < 0.73 ) - P ( Z < -1.27 )
P ( 65 < X < 71 ) = 0.7683 - 0.1026
P ( 65 < X < 71 ) = 0.6657

Part c)
X ~ N ( µ = 68.8 , σ = 3 )
P ( X > 71 ) = 1 - P ( X < 71 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 71 - 68.8 ) / 3
Z = 0.7333
P ( ( X - µ ) / σ ) > ( 71 - 68.8 ) / 3 )
P ( Z > 0.7333 )
P ( X > 71 ) = 1 - P ( Z < 0.7333 )
P ( X > 71 ) = 1 - 0.7683
P ( X > 71 ) = 0.2317

Part d)

C. There are no unusual events because all the probabilities are greater than 0.05.

Answer 2

71-67.2 divide by 2.0