Question

The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 32 million with a standard deviation of 6 million.

What is the probability next week's show will:

Have between 38 and 44 million viewers? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.)

Have at least 23 million viewers? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.)

Exceed 47 million viewers? (Round your z-score computation to 2 decimal places and final answer to 4 decimal places.)

Answer 1

µ = 32

sd = 6

a)

                                        

                                         = P(1 < Z < 2)

                                         = P(Z < 2) - P(Z < 1)

                                         = 0.9772 - 0.8413

                                         = 0.1359

b)

                           

                            = P(Z > -1.5)

                            = 1 - P(Z < -1.5)

                            = 1 - 0.0668

                            = 0.9332

c)

                            

                             = P(Z > 2.5)

                             = 1 - P(Z < 2.5)

                             = 1 - 0.9938

                             = 0.0062