Question

In a survey of a group of​ men, the heights in the​ 20-29 age group were normally​ distributed, with a mean of 67.1 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 68 inches. The probability that the study participant selected at random is less than 68 inches tall is ___ (round to 4 decimal points)

(b) Find the probability that a study participant has a height that is between 68-72 inches. The probability that the study participant selected at random is between 68-72 inches is __ (round to 4 decimal points)

(c) Find the probability that a study participant has a height that is more than 72 inches. The probability that the study participant selected at random is more than 72 inches is __ (round to 4 decimal points)

(d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.

A. The event in part (a) is unusual because its probability is less than 0.05.

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

C. The events in parts (a), (b), and (c) are unusual because all of the probabilities are less than 0.05

D. The events in parts (a) and (c) are unusual because its probabilities are less than 0.05

Answer 1

Solution :

Given that ,

mean = = 67.1

standard deviation = = 3.0

P(x < 68 ) = P[(x - ) / < ( 68 - 67.1) / 3.0 ]

= P(z < 0.3 )

Using z table,

= 0.6179

Probability = 0.6179

( b )

P( 68 < x < 72 )

= P[( 68 - 67.1) / 3.0 ) < (x - ) /  < ( 72 - 67.1) / 3.0 ) ]

= P( 0.3 < z < 1.63 )

= P(z < 1.63) - P(z < 0.3)

Using z table,

= 0.9484 - 0.6179

= 0.3305

Probability = 0.3305

( c )

P(x > 72 ) = 1 - P( x < 72 )

= 1- P[(x - ) / < ( 72 - 67.1 ) / 3.0 ]

= 1- P(z < 1.63)

Using z table,

= 1 - 0.9484

= 0.0516

Probability = 0.0516

( d )

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