Solutions to Problem Set #8
Section 9.1
8. A club serves dinner to members only. They are seated at 12-seat tables. The manager
observes over a long period of time that 95 percent of the time there are between six and nine
full tables of members, and the remainder of the time the numbers are equally likely to fall
above or below this range. Assume that each member decides to come with a given probability
p, and that the decisions are independent. How many members are there? What is p?
The problem states that 95 percent of the time there are between six and nine full tables,
which translates to there being somewhere between 72 and 108 people at the club. We know
from class that a 95% confidence interval corresponds to considering the values that are
between two standard deviations to the left of µ and two standard deviations to the right of
µ. Since 108 is two standard deviations to the right of µ and 72 is two standard deviations to
the left, then µ (being in the center) must be 90. For a Bernoulli trials process, it is always
the case that µ = pn, thus p = 90
.
n
The first question that we want to answer is “which value of n yields a 95% confidence
interval of (72, 108)?” Since a 95% confidence
� pq interval corresponds to being 2 standard
deviations from µ then the expression p − 2 n will give us the lower bound of the “95%
confidence interval divided by n,” which is 72
. We can therefore solve the equation
n
�
90 n−90
· n
90
72
−2 n
=
n
n
n
to obtain n = 900. Thus, p =
90
900
= .1.
14. A restaurant feeds 400 customers per day. On the average 20 percent of the customers
order apple pie.
(a) Give a range (called a 95 percent confidence interval) for the number of pieces of
apple pie ordered on a given day such that you can be 95 percent sure that the actual number
will fall in this range.
� Since n = 400 and p = .2 then µ = (400)(.2) = 80. The standard deviation will be
(.2)(.8)
= .02. Thus, since a 95% confidence interval corresponds to being 2 standard
400
�
deviations from µ then we can use the formula p − 2SD = .2 − 2 (.2)(.8)
to obtain a lower
400
bound of .16 for the “95% confidence interval divided by n.” In other words, the lower
bound of the 95% confidence interval should be (.16)(400) = 64. A similar argument (using
the formula p + 2SD) shows that the upper bound should be 96. Thus, the range for the
95% confidence interval is: (64, 96).
(b) How many customers must the restaurant have, on the average, to be at least 95
percent sure that the number of customers ordering pie on that day falls in the 19 to 21
percent range?
We just need to “solve for SD” in the equation that gives the lower bound for the 95%
confidence interval and then use the formula for SD to recuperate n. Since p = .2, we have
1
Solutions to Problem Set #8
�
.2 − 2SD = .19, so SD = .005. Using the formula SD = .005 = (.2)(.8)
, we obtain n = 6400.
n
Note: using the formula for the upper bound of the 95% confidence interval would have
yielded the same result.
Section 9.2
2. A random walker starts at 0 on the x-axis and at each time unit moves 1 step to the
right or 1 step to the left with probability 1/2. Estimate the probability that, after 100 steps,
the walker is more than 10 steps from the starting position.
Let Xi be 1 if the walker moves right in the ith step and −1 if he moves left. Then
Xi has parameters µ = 0 and σ 2 = E(Xi2 ) − (E(Xi ))2 = [ 12 (1)2 − 12 (−1)2 ] − 02 = 1. Now,
Sn = X1 + X2 + · · · + Xn represents the walker’s position after n steps. When n = 100,
we have E(S100 ) = 100µ = 0 since µ = 0 and V (Sn ) = 100σ 2 = 100. Hence, the standard
deviation is 10. Since the random walk is determined by a discrete independent trials process
(i.e. you could use a coin flip to determine whether the walker moves left or right), we can
use the Central Limit Theorem to approximate the probability. Thus, we have
P (| S100 |≥ 10) = 1 − P (| S100 < 10) = 1 − N A(−1, 1) ≈ 1 − .6826 = .3174.
Section 11.1
4. For Example 11.6, find the probability that the grandson of a man from Harvard went
to Harvard.
(2)
Since the grandson is the “son of the son,” then we’re looking for pHH . We note that
(2)
pHH = pHH · pHH + pHY · pY H + pHD · pDH = (.8)(.8) + (.2)(.3) + (0)(.2) = .7.
2
3
4
n
7. Find the
� matrices
� P , P , P and P for the Markov chain determined
�
�by the transition
1 0
0 1
matrix P =
. Do the same for the transition matrix P =
. Interpret what
0 1
1 0
happens in each of these processes.
�
�
1 0
All powers of P =
are just
0 1
�
�
1 0
.
0 1
�
�
0 1
For P =
, even powers are
1 0
�
�
0 1
1 0
and odd powers are
�
1 0
0 1
2
�
.
Solutions to Problem Set #8
8. A certain calculating machine uses only the digits 0 and 1. It is supposed to transmit
only one of these digits through several stages. However, at every stage, there is a probability
p that the digit that enters this stage will be changed when it leaves and a probability q = 1−p
that it won’t. Form a Markov chain to represent the process of transmission by taking as
states the digits 0 and 1. What is the matrix of transition probabilities?
This problem is identical to the “telephone” question
� from
� class. Our states in this case
q p
are the digits 0 and 1. Our transition matrix is P =
.
p q
3