130k views
4 votes
The amount of time a bank teller spends with each customer has a population mean  = 3.10 minutes and standard deviation  = 0.40 minute. if a random sample of 16 customers is selected without replacement from a population of 500 customers,

a. what is the probability that the average time spent per customer will be at least 3 minutes?
b. there is an 85% chance that the sample mean will be below how many minutes?

User Peska
by
7.8k points

1 Answer

4 votes

a. First we solve the finite population correction factor σx from the formula:

σx = [s / sqrt(n)] * sqrt [(N – n) / (N – 1)]

σx = [0.40 / sqrt(16)] * sqrt[(500 – 16) / (500 – 1)]

σx = 0.0985

Then we compute for z score when x ≥ 3 minutes:

z = (x – m) / σx

z = (3 – 3.10) / 0.0985

z = -1.015

From the tables, the probability (p value) using right tailed test is:

P = 0.845 = 84.5%


b. At P = 0.85, the z score is z = 1.04

1.04 = (x – 3.10) / 0.0985

x = 3.20

Hence there is a 85% chance it will be below 3.20 minutes

User Schubie
by
8.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories