Question

A normal population has a mean of 81 and a standard deviation of 6. You select a sample of 36. Use Appendix B.1 for the z-values.

Compute the probability that the sample mean is: (Round the z-values to 2 decimal places and the final answers to 4 decimal places.)

a. Less than 79.

Probability            

b. Between 79 and 83.

Probability            

c. Between 83 and 84.

Probability            

d. Greater than 84.

Probability            

Answer 1

µ = 81

sd = 6

n = 36

a)

                            

                             = P(Z < -2)

                             = 0.0228

b)

                                        

                                         = P(-2 < Z < 2)

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

                                         = 0.9772 - 0.0228

                                         = 0.9544

c)

                                        

                                         = P(2 < Z < 3)

                                         = P(Z < 3) - P(Z < 2)

                                         = 0.9987 - 0.9772

                                         = 0.0215

d)

                             

                              = P(Z > 3)

                              = 1 - P(Z < 3)

                              = 1 - 0.9987

                              = 0.0013