60. A toll bridge charges $1.00 for passenger cars and $2.50 for other vehicles. Suppose that during daytime hours, 60%
of all vehicles are passenger cars. If 25 vehicles cross the bridge during a particular daytime period, what is the resulting expected toll revenue? [Hint:Let
of passenger cars; then the toll revenue h(X) is a linear function of X.]
61. A student who is trying to write a paper for a course has a choice of two topics, A and B. If topic A is chosen, the student will order two books through interlibrary loan, whereas if topic B is chosen, the student will order four books. The student believes that a good paper necessitates receiving and using at least half the books ordered for either topic chosen. If the probability that a book ordered through interlibrary loan actually arrives in time is .9 and books arrive independently of one another, which topic should the student choose to maximize the probability of writing a good paper? What if the arrival probability is only .5 instead of .9?
62. a.For fixed n,are there values of for which
? Explain why this is so.
b. For what value of pis V(X) maximized? [Hint: Either graph V(X) as a function of por else take a derivative.]
63. a.Show that .
b. Show that .
[Hint:At most x S’s is equivalent to at least F’s.]
c. What do parts (a) and (b) imply about the necessity of including values of pgreater than .5 in Appendix Table A.1?
64. Show that when X is a binomial random variable. [Hint:First express E(X) as a sum with lower limit . Then factor out np,let so that the sum is from y50to y5n21, and show that the sum equals 1.]
y5x21 x51
E(X)5np
(n2x) B(x; n, 12p)512B(n2x21; n, p) b(x; n, 12p)5b(n2x; n, p)
V(X)50
p(0#p#1)
X5the number
65. Customers at a gas station pay with a credit card (A), debit card (B), or cash (C). Assume that successive customers make independent choices, with , , and
.
a. Among the next 100 customers, what are the mean and variance of the number who pay with a debit card?
Explain your reasoning.
b. Answer part (a) for the number among the 100 who don’t pay with cash.
66. An airport limousine can accommodate up to four passengers on any one trip. The company will accept a maximum of six reservations for a trip, and a passenger must have a reserva- tion. From previous records, 20% of all those making reservations do not appear for the trip. Answer the following questions, assuming independence wherever appropriate.
a. If six reservations are made, what is the probability that at least one individual with a reservation cannot be accommodated on the trip?
b. If six reservations are made, what is the expected num- ber of available places when the limousine departs?
c. Suppose the probability distribution of the number of reservations made is given in the accompanying table.
P(C)5.3
P(B)5.2 P(A)5.5
The hypergeometric and negative binomial distributions are both related to the binomial distribution. The binomial distribution is the approximate probability model for sampling without replacement from a finite dichotomous (S–F) popula- tion provided the sample size n is small relative to the population size N; the hypergeometric distribution is the exact probability model for the number of S’s in the sample. The binomial rv Xis the number of S’s when the number nof trials is fixed, whereas the negative binomial distribution arises from fixing the number of S’s desired and letting the number of trials be random.
Example 3.35
2. Each individual can be characterized as a success (S) or a failure (F), and there are Msuccesses in the population.
3. A sample of nindividuals is selected without replacement in such a way that each subset of size nis equally likely to be chosen.
The random variable of interest is of S’s in the sample. The probability distribution of Xdepends on the parameters n, M,and N,so we wish to obtain .
During a particular period a university’s information technology office received 20 service orders for problems with printers, of which 8 were laser printers and 12 were inkjet models. A sample of 5 of these service orders is to be selected for inclusion in a customer satisfaction survey. Suppose that the 5 are selected in a completely random fashion, so that any particular subset of size 5 has the same chance of being selected as does any other subset. What then is the probability that exactly
of the selected service orders were for inkjet printers?
Here, the population size is , the sample size is , and the number of S’s and F’s in the population are and , respectively. Consider the value . Because all outcomes (each consisting of 5 particular orders) are equally likely,
The number of possible outcomes in the experiment is the number of ways of selecting 5 from the 20 objects without regard to order—that is, . To count the number of outcomes having , note that there are ways of selecting 2 of the inkjet orders, and for each such way there are ways of selecting the 3 laser orders to fill out the sample. The product rule from Chapter 2 then gives as the number of outcomes with , so
■
In general, if the sample size nis smaller than the number of successes in the pop- ulation (M), then the largest possible Xvalue is n.However, if (e.g., a sample size of 25 and only 15 successes in the population), then Xcan be at most M.Similarly, whenever the number of population failures exceeds the sample size, the smallest possible Xvalue is 0 (since all sampled individuals might then be failures).
However, if , the smallest possible Xvalue is . Thus, the pos-
sible values of Xsatisfy the restriction . An
argument parallel to that of the previous example gives the pmf of X.
max (0, n2 (N2M))#x# min (n, M) n2(N2 M)
N2M ,n
(N2M)
M ,n h(2; 5, 12, 20)5
Q12 2RQ8
3R Q20
5R 5 77
3235.238 X5 2
A122BA83B A83B A122B
X52
A205B P(X5 2)5 h(2; 5, 12, 20)5 number of outcomes having X5 2
number of possible outcomes x5 2
N2M 58 M5 12
(inkjet5 S)
n5 5 N5 20
x (x50, 1, 2, 3, 4, or 5) P(X5x)5h(x; n, M, N)
X5 the number
PROPOSITION If Xis the number of S’s in a completely random sample of size ndrawn from a population consisting of M S’s and F’s, then the probability distri- bution of X,called the hypergeometric distribution,is given by
(3.15)
for x,an integer, satisfying max (0, n2N1 M)#x# min (n, M). P(X5x)5h(x; n, M, N)5
QM
xRQN2 M n2xR QN
nR (N2M)
PROPOSITION Example 3.36
In Example 3.35, , , and , so h(x; 5, 12, 20) for
can be obtained by substituting these numbers into Equation (3.15).
Five individuals from an animal population thought to be near extinction in a cer- tain region have been caught, tagged, and released to mix into the population.
After they have had an opportunity to mix, a random sample of 10 of these animals is selected. Let of tagged animals in the second sample. If there are actually 25 animals of this type in the region, what is the probability that
(a) ? (b) ?
The parameter values are , (5 tagged animals in the population),
and , so
For part (a),
For part (b),
■ Various statistical software packages will easily generate hypergeometric probabilities (tabulation is cumbersome because of the three parameters).
As in the binomial case, there are simple expressions for E(X) and V(X) for hypergeometric rv’s.
5.057 1.2571 .3855 .699 P(X#2)5 P(X5 0, 1, or 2)5
g
2x50
h(x; 10, 5, 25) P(X52) 5h(2; 10, 5, 25)5
Q5 2RQ20
8R Q25
10R 5 .385 h(x; 10, 5, 25)5
Q5 xRQ 20
102 xR Q25
10R x50, 1, 2, 3, 4, 5 N525
M 55 n5 10 X#2
X52
X5the number x50, 1, 2, 3, 4, 5
N520 M5 12
n55
The mean and variance of the hypergeometric rv Xhaving pmf h(x; n, M, N) are E(X)5 n
#
MN V(X)5 aN2 n N2 1b
#
n
#
MN
#
a12 M Nb
The ratio M/Nis the proportion of S’s in the population. If we replace M/Nby pin E(X) and V(X), we get
(3.16)
Expression (3.16) shows that the means of the binomial and hypergeometric rv’s are equal, whereas the variances of the two rv’s differ by the factor , often called the finite population correction factor.This factor is less than 1, so the (N2 n)/(N2 1) V(X)5 aN2 n
N2 1b
#
np(12 p) E(X)5 np
Example 3.37
(Example 3.36 continued)
hypergeometric variable has smaller variance than does the binomial rv. The correction factor can be written , which is approximately 1 when nis small relative to N.
In the animal-tagging example, , , and , so and
If the sampling was carried out with replacement, .
Suppose the population size Nis not actually known, so the value xis observed and we wish to estimate N.It is reasonable to equate the observed sample proportion of S’s, x/n,with the population proportion, M/N,giving the estimate
If , , and , then . ■
Our general rule of thumb in Section 3.4 stated that if sampling was without replacement but n/Nwas at most .05, then the binomial distribution could be used to compute approximate probabilities involving the number of S’s in the sample. A more precise statement is as follows: Let the population size, N,and number of pop- ulation S’s, M, get large with the ratio M/N approaching p. Then h(x; n, M, N) approaches b(x; n, p); so for n/Nsmall, the two are approximately equal provided that pis not too near either 0 or 1. This is the rationale for the rule.