Chapter 6 Problems
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.
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)]
Number of
Breakdowns
Per Week
x
0
1
2
3
Number of
Breakdowns
Per Week
0
1
2
3
Total
Weeks
Probability
20
20
10
10
60
0.333
0.333
0.167
0.167
1.000
Number of
Breakdowns
Per Week
Probability
x
0
1
2
3
P(x)
0.333
0.333
0.167
0.167
Total
xP(x)
0.000
0.333
0.334
0.501
1.168
6-4
Probability
(x  )
P(x)
0.333
0.333
0.167
0.167
0 – 1.168 =
1 – 1.168 =
2 – 1.168 =
3 – 1.168 =
1.168
0.168
+0.832
+1.832
(x   )2
(x   )2 P(x)
1.364
0.028
0.692
3.356
(1.364)(0.333) = 0.454212
(0.028)(0.333) = 0.009324
(0.692)(0.167) = 0.115564
(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
1139552
.
 10674
.
 107
. .
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)
These events are independent;
Location
Order of
Probability of
therefore the probability of their joint
of NS
Occurrence
Occurrence
occurrence is the product of the
1
NS, S, S, S
(0.8)(0.2)(0.2)(0.2) = 0.0064
individual probabilities. Therefore, the
2
S, NS, S, S
(0.2)(0.8)(0.2)(0.2) = 0.0064
likelihood of no sale followed by three
3
S, S, NS, S
(0.2)(0.2)(0.8)(0.2) = 0.0064
sales is (0.8) (0.2) (0.2) (0.2) = 0.0064.
4
S, S, S, NS
(0.2)(0.2)(0.2)(0.8) = 0.0064
However, the requirements of the
0.0256
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:
n
Cx
x
n
p
(1p)
denotes a combination of n items selected x at a time
is the number of successes, 3 in the example.
is the number of trials, 4 in the example.
is the probability of a success, 0.20.
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.
P( x) n C x ( p) x (1  p) n  x 

n!
( p) x (1  p) n  x
x!(n  x)!
4!
(.20) 3 (0.80) 43  0.0256
3!(4  3)!
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
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
Binomial Probability
Distribution
n=4
p = 0.20
Number of
Successes (x)
Probability
0
1
2
3
4
0.410
0.410
0.154
0.026
0.002
*1.000
Refer to Appendix A, the binomial table. Find the table where n, the
*Slight discrepancy due to
number of trials, is 4. Next, find the row where x = 0, and move
rounding
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.
bg bg
P( more than 2)  P 3  P 4
 0.026  0.002
 0.028
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
0.64  0.80 .
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]:
P( x) 
 x e 
[6 - 10]
x!
Where:
x is 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
e is a mathematical constant equal to 2.71828.
The probability that exactly two claims are filed is 0.0164, found by
P  2 
 x e 
x!
 0.2   2.718 
2

2!
0.2

 0.04  0.81874   0.0164
 2 1
Probability
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.
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
1
2
3
4
Poisson Probability
Distribution
 = 0.2
Number of
Probability
Claims
0
1
2
3
4
0.8187
0.1637
0.0164
0.0011
0.0001
1.0000
5
Number of Claims
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:
P  x 
 S Cx  N S Cn x 
 N Cn 
[5  6]