Problem 1: A retail outlet wanted to know whether its weekly advertisement in th
ID: 3219750 • Letter: P
Question
Problem 1: A retail outlet wanted to know whether its weekly advertisement in the daily newspaper works. To acquire this critical information, the store manager surveyed the people who enter the store and determined whether each individual saw the ad and whether a purchase was made. From the information developed, the manager produced the following table of joint probabilities.
Ads
Purchase Status
Total
Purchase
No Purchase
See the Ads
0.22
0.33
0.55
Do not see the Ads
0.09
0.36
0.45
Total
0.31
0.69
1.00
1. What is the probability that the customer purchase?
What is probability that the customer purchase if he sees the ads?
Are the events of “See the Ads” and “Purchase” independent? Show why or why not.
Are the ads effective? Explain
Problem 2: It has been reported that 8% of the work force will retire this year. Consider a random sample of 150 workers.
a. What is the probability that at least five of them will retire this year?
b. What are the mean and the standard deviation of the number of people who will retire this year?
Problem 3: A construction company has submitted bids on two separate state contracts, A and B. The company feels that it has a 70% chance of winning contract A, and a 40% chance of winning contract B. Furthermore, the company believes that it has 60% chance of winning contract B given that it wins contract A.
a. What is the probability that the company will win both of the contracts?
b. Is winning contract B independent of winning contract A? Explain.
Problem 4: Let X represents the number of children in a household. The probability distribution of X is as follows:
x
1
2
3
4
5
6
p(x)
0.22
0.28
0.19
0.14
0.11
0.06
a. What is the probability that a randomly selected household will have at most 2 children?
b. What is the expected number of children in the household?
Ads
Purchase Status
Total
Purchase
No Purchase
See the Ads
0.22
0.33
0.55
Do not see the Ads
0.09
0.36
0.45
Total
0.31
0.69
1.00
Explanation / Answer
Solution:-
Problem 2:
p = 0.08, n = 150, x = 5
a) The probability that at least five of them will retire this year is 0.994.
By applying binomial distribution:-
P(x, n, p) = nCx*p x *(1 - p)(n - x)
P(x > 5) = 0.994
b) The mean and the standard deviation of the number of people who will retire this year is 12 and 3.323.
Mean = n × p
Mean = 0.08 × 150
Mean = 12
= Sqrt(n × p × (1 - p))
= 3.323