Question

At the beginning of the month, the amount of time a customer waits in line at Starbucks normally distributed with a mean of ten minutes and a standard deviation of three minutes.

What is the 80th percentile of waiting times (in minutes)?
a. 0.8416 b. 7.48 c. 12.52 d. 80

The median waiting time (in minutes) for one customer is:
a. 50 b. 10 c. 5 d. 3

Find the probability that the average wait time for five customers is at most 5.5 minutes.
a. 0.0004 b. 0.0668 c. 0.9332 d. 0.9996

Answer 1

(b) in normal distribution mean = median

= median = 10 option (b)

(c) z value for 5.5 = x-mean/(sigma/root(n))

= =    = 0.37

hence probility for z less then 0.37(as avarage wait mean time should be less then 5.5)   

from the normal destribution table = -3.354

hence from the table p = 0.0004

option (a)