Using the Standard Normal Table 4 find the z-score that has 79.39% of the distribution's area to the left.

79.39 = 0.7939 = 0.79

= 0.7852 + 0.5000 = 1.2852

The area to the left of 79.39% is 1.29

2. Find the z-score for which 60% of the distribution's area lies between -z and z

60% = 0.60

= 0.7257 + 0.5000 = 1.2257

-0.60 = 0.2743 + 0.5000 = 0.7743

1.2257 - 0.7743 = 0.4514

The z-score of the area between -z and z is 0.4514

2.Because of heavy online Christmas shopping, FEDEX is dealing with many late deliveries. If the mean number of packages delivered each day is Florence and the surrounding area is 250,000 packages with a standard deviation of 800, how should FEDEX word it's agreement with the shipping companies so that 90% of deliveries arrive on time?

mean = 250,000 standard deviation = 80090% = 0.90

90 - 250,000/800 = -312.3875

0.90 + 0.50 = 1.4

I think that FEDEX should world their agreement by telling the public that they are dealing with a lot of packages and that they are sorry for the inconvenience if some of their packages are late.