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 15.0 minutes and the standard deviation is 3.5 minutes. Refer to the table in Appendix A.1. What is the probability that a particular listener will tune in: a. More than 20 minutes? (Round intermediate calculations to 2 decimal places. Round final answer to 4 decimal places.) Probability b. For 20 minutes or less? (Round intermediate calculations to 2 decimal places. Round final answer to 4 decimal places.) Probability c. Between 10 and 12 minutes? (Round intermediate calculations to 2 decimal places. Round final answer to 4 decimal places.)
Explanation / Answer
Let X be the length of time a listener is tuned to the station. We must find: a) Prob(X>20) Using the procedure I explained before, we'll try to leaave the equation in terms of a standard normally distributed random variable. Prob( X > 20 ) =Prob( X-15 > 20-15 ) =Prob( (X-15)/3.5 > (20-15)/3.5 ) So that's it, the left-hand side of the inequality is a standard normally distributed random variable. We'll call it Z. So we must find: Prob ( Z > (20-15)/3.5 ) =Prob ( Z > 1.43 ) Since the table I provided in the link gives the probabilites of Z being smaller than a given number (instead of greater, as in this equation) and greater than zero, we transform the equation in order to be able to use this table: Prob ( Z > 1.43 ) =1 - Prob(Z < 1.43 ) =1 - [ Prob(Z