Chapter 6 Problems

Chapter 6 Problems

Number of Breakdowns
Per Week / Weeks / Probability
0 / 20 / 0.333
1 / 20 / 0.333
2 / 10 / 0.167
3 / 10 / 0.167
Total / 60 / 1.000

Problem 1

Bill Russe, production manager at Ross Manufacturing, maintains detailed records on the number of times each machine breaks down and requires service during the week. Bill’s records show that the Puret grinder has required repair service according to the following distribution. Compute the arithmetic mean and the variance of the number of breakdowns per week.

Number of Breakdowns
Per Week / Probability
x / P(x) / xP(x)
0 / 0.333 / 0.000
1 / 0.333 / 0.333
2 / 0.167 / 0.334
3 / 0.167 / 0.501
Total / 1.168

Solution 1

The arithmetic mean, or expected number of breakdowns per week for the probability distribution, is computed using formula [6-2].

 = [xP(x)] [6-2]

The arithmetic mean number of times the Puret machine breaks down per week is 1.168. The variance of the number of breakdowns is computed using formula [6-4].

2 = [(x-)2P(x)]6-4

Number of
Breakdowns
Per Week / Probability
x / P(x) / (x) / (x )2 / (x )2 P(x)
0 / 0.333 / 0 – 1.168 = 1.168 / 1.364 / (1.364)(0.333) = 0.454212
1 / 0.333 / 1 – 1.168 = 0.168 / 0.028 / (0.028)(0.333) = 0.009324
2 / 0.167 / 2 – 1.168 = +0.832 / 0.692 / (0.692)(0.167) = 0.115564
3 / 0.167 / 3 – 1.168 = +1.832 / 3.356 / (3.356)(0.167) = 0.560452
Total 1.139552

The variance of the number of breakdowns per week is about 1.140. The standard deviation of the number of breakdowns per week is 1.07, found by .

Problem 2

An insurance representative has appointments with four prospective clients tomorrow. From past experience she knows that the probability of making a sale on any appointment is 1 in 5 or 0.20. Use the rule of probability to determine the likelihood that she will sell a policy to 3 of the 4 prospective clients.

Solution 2

First note that the situation described meets the requirements of the binomial probability distribution. The conditions are:

1. There are a fixed number of trialsthe representative visits four customers.

2. There are only two possible outcomes for each trialshe sells a policy or she does not sell a policy.

3. The probability of a success remains constant from trial to trialfor each appointment the probability of selling a policy (a success) is 0.20.

4. The trials are independentif she sells a policy to the second appointment this does not alter the likelihood of selling to the third or the fourth appointment.

If S represents the outcome of a sale and NS the outcome of no sale, one possibility is that no sale is made on the first appointment but sales are made at the last 3.

(NS, S, S, S)

Location
of NS / Order of
Occurrence / Probability of
Occurrence
1 / NS, S, S, S / (0.8)(0.2)(0.2)(0.2) = 0.0064
2 / S, NS, S, S / (0.2)(0.8)(0.2)(0.2) = 0.0064
3 / S, S, NS, S / (0.2)(0.2)(0.8)(0.2) = 0.0064
4 / S, S, S, NS / (0.2)(0.2)(0.2)(0.8) = 0.0064
0.0256

These events are independent; therefore the probability of their joint occurrence is the product of the individual probabilities. Therefore, the likelihood of no sale followed by three sales is (0.8) (0.2) (0.2) (0.2) = 0.0064. However, the requirements of the problem do not stipulate the location of NS. It could be the result of any one of the four appointments. The following summarizes the possible outcomes.

The probability of exactly three sales in the four appointments is the sum of the 4 possibilities. Hence, the probability of selling insurance to 3 out of 4 appointments is 0.0256.

Problem 3

Now let’s use formula [6-7] for the binomial distribution to compute the probability that the sales representative in Problem 2 will sell a policy to exactly 3 out of the 4 prospective clients.

Solution 3

To repeat, formula [6-7] for the binomial probability distribution is:

P(x) = nCxpx(1-p)(n-x)

for x = 0,1,2,3,…,n[6-7]

Where:

denotes a combination of n items selected x at a time

xis the number of successes, 3 in the example.

nis the number of trials, 4 in the example.

pis the probability of a success, 0.20.

(1p) is the probability of a failure, 0.80 found by ( 1  0.20).

The formula is applied to find the probability of selling an insurance policy to exactly 3 out of 4 potential customers.

Thus the probability is 0.0256 that the representative will be able to sell policies to exactly 3 out of the 4 clients visited. This is the same probability as computed earlier. Clearly, formula [6-7] leads more directly to a solution and better accommodates the situation where the number of trials is large.

Problem 4

Binomial Probability Distribution
n = 4 / p = 0.20
Number of
Successes (x) / Probability
0 / 0.410
1 / 0.410
2 / 0.154
3 / 0.026
4 / 0.002
*1.000
*Slight discrepancy due to rounding

In Problems 2 and 3 the probability of 3 sales resulting from 4 appointments was computed using both the rules of addition and multiplication and the binomial formula. A more convenient way of arriving at the probabilities for 0, 1, 2, 3, or 4 sales out of 4 appointments is to refer to a binomial table. We use the binomial table to determine the probabilities for all possible outcomes.

Solution 4

Refer to Appendix A, the binomial table. Find the table where n, the number of trials, is 4. Next, find the row where x = 0, and move horizontally to the column headed p = 0.20. The probability of 0 sales is 0.410. The list for all possible outcome number of successes is shown at the right.

Problem 5

Use the information regarding the insurance representative, where n = 4 and p = 0.20, to compute the probability that the representative sells more than two policies. Also determine the mean and variance of the number of policyholders.

Solution 5

The binomial table (Appendix A) can be used to determine the probability. First, note that the solution must include the probability that exactly 3 policies are sold and exactly 4 policies are sold, but not 2. From Appendix A, P(3) = 0.026 and P(4) = 0.002. The rule of addition is then used to combine these mutually exclusive events.

Thus the probability that a representative sells more than 2 policies is 0.028. Suppose the question asked is: “What is the probability of selling three or more policies in four trials?” Since there are no outcomes between “greater than 2” and “less than 3”, the answer is exactly the same (0.028).

To determine the mean of a binomial we use text formula [6-8].

 = np = 4(0.20) = 0.80

Thus if the sales representative has several days with 4 appointments, typically he/she will sell 0.80 policies.

To determine the variance of a binomial we use text formula [6-9].

2 = np(1-p) = 4(0.20)(0.80) = 0.64

So the standard deviation is .

Problem 6

Alden & Associates write weekend trip insurance at a very nominal charge. Records show that the probability a motorist will have an accident during the weekend and will file a claim is quite small (0.0005). Suppose Alden wrote 400 policies for the forthcoming weekend. Compute the probability that exactly two claims will be filed. Depict this distribution in the form of a chart.

Solution 6

The Poisson distribution is appropriate for this problem because the probability of filing a claim is small (p= 0.0005), and the number of trials n is large (400).

The Poisson distribution is described by formula [6-10]:

Where:

xis the number of successes (claims filed). In this example x = 2

is the expected or mean number of claims to be filed  = np= (400) (0.0005) = 0.2

eis a mathematical constant equal to 2.71828.

The probability that exactly two claims are filed is 0.0164, found by

Poisson Probability Distribution
 = 0.2
Number of Claims / Probability
0 / 0.8187
1 / 0.1637
2 / 0.0164
3 / 0.0011
4 / 0.0001
1.0000

This indicates that the probability is somewhat small (about 0.0164) that exactly 2 claims will be filed. The complete Poisson distribution is shown at the right and a graph for the case where  = 0.20 is shown below. Note the shape of the graph. It is positively skewed, and as the number of claims increases, the probability of a claim decreases.

Problem 7

The government of an underdeveloped country has 8 loans payable, 5 of which are overdue. If a representative of the International Monetary Fund randomly selects 3 loans, what is the probability that exactly 2 are overdue?

Solution 7

Note in this problem that successive observations are not independent. That is, the outcome of one sampled item influences the next sampled item. Because the observations are not independent, the binomial distribution is not appropriate and the hypergeometric distribution is used. To repeat the formula for the hypergeometric distribution as shown in the CD-ROM’s Additional Content section Ch 6B: