Question

The time it takes to process phone orders in a small florist/gift shop is normally distributed with a mean of 6 minutes and a standard deviation of 1.24 minutes.

a. What cutoff value would separate the 2.5% of orders that take the most time to process?

b. What cutoff value would separate the 16% of orders that take the least time to process?

c. What cutoff values would separate the 95% of orders that are in the middle of the distribution with respect to processing time?

Answer 1

Solution:

We are given

Mean = 6

SD = 1.24

Part a

The critical z value for upper 2.5% area is 1.96 (by using z-table)

X = mean + z*SD

X = 6 + 1.96*1.24

X = 8.4304

Answer: 8.43

Part b

The critical z value for lower 16% area is -0.99446 (by using z-table).

X = mean + z*SD

X = 6 + (-0.99446)*1.24

X = 4.76687

Answer: 4.77

Part c

The critical z values for middle 95% area are given as -1.96 and 1.96.

First find Lower X

X = mean + z*SD

X = 6 - 1.96*1.24

X = 3.5696

Now find upper X

X = mean + z*SD

X = 6 + 1.96*1.24

X = 8.4304

Answer: 3.57 and 8.43