Question

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 5.1 minutes and the standard deviation was 0.40 minutes. Use Appendix B.3.

1. What is the probability that calls last between 5.1 and 5.8 minutes? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

2. What is the probability that calls last more than 5.8 minutes? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

3. What is the probability that calls last between 5.8 and 6.5 minutes? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

4. What is the probability that calls last between 4.5 and 6.5 minutes? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

5. As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 3 percent of the calls. What is this time?

Answer 1

a) P(5.1 < X < 5.8)

= P((5.1 - )/ < (X - )/ < (5.8 - )/)

= P((5.1 - 5.1)/0.4 < Z < (5.8 - 5.1)/0.4)

= P(0 < Z < 1.75)

= P(Z < 1.75) - P(Z < 0)

= 0.9599 - 0.5

= 0.4599

b) P(X > 5.8)

= P((X - )/ > (5.8 - )/)

= P(Z > (5.8 - 5.1)/0.4)

= P(Z > 1.75)

= 1 - P(Z < 1.75)

= 1 - 0.9599

= 0.0401

c) P(5.8 < X < 6.5)

= P((5.8 - )/ < (X - )/ < (6.5 - )/)

= P((5.8 - 5.1)/0.4 < Z < (6.5 - 5.1)/0.4)

= P(1.75 < Z < 3.5)

= P(Z < 3.5) - P(Z < 1.75)

= 1 - 0.9599

= 0.0401

d) P(4.5 < X < 6.5)

= P((4.5 - )/ < (X - )/ < (6.5 - )/)

= P((4.5 - 5.1)/0.4 < Z < (6.5 - 5.1)/0.4)

= P(-1.5 < Z < 3.5)

= P(Z < 3.5) - P(Z < -1.5)

= 1 - 0.0668

= 0.9332

e) P(X > x) = 0.03

or, P((X - )/ > (x - )/) = 0.03

or, P(Z > (x - 5.1)/0.4) = 0.03

or, P(Z < (x - 5.1)/0.4) = 0.97

or, (x - 5.1)/0.4 = 1.88

or, x = 1.88 * 0.4 + 5.1

or, x = 5.852