Question

CNAE, an all-news AM station, finds that the distribution of the lengths of time listeners are tuned to the station follows the normal distribution. The mean of the distribution is 14.6 minutes and the standard deviation is 2.1 minutes. Refer to the table in Appendix B.1. What is the probability that a particular listener will tune in: (Round the intermediate calculations to 2 decimal places and the final answers to 4 decimal places.) a. More than 20 minutes? Probability b. For 20 minutes or less? Probability c. Between 10 and 12 minutes? Probability

Answer 1

Solution :

Given that ,

mean = = 14.6

standard deviation = = 2.1

a) P(x > 20 ) = 1 - p( x< 20)

=1- p P[(x - ) / < (20 - 14.6) / 2.1 ]

=1- P(z < 2.57)

Using z table,

= 1 - 0.9949

= 0.0051

b) P(x < 20 ) = P[(x - ) / < (20 - 14.6) / 2.1 ]

= P(z < 2.57)

Using z table,

= 0 9949

P( 10 < x < 12 ) = P[(10 - 14.6) / 2.1 ) < (x - ) /  < (12 - 14.6) / 2.1 ) ]

= P( - 2.19 < z < - 1.24 )

= P(z < - 1.24) - P(z < - 2.19)

Using z table,

= 0.1075 - 0.0143

= 0.0932