Introduction to the Practice of Statistics - PCC

Introduction to the Practice of Statistics
Sixth Edition
Moore, McCabe
Section 4.2 Homework Answers
4.27 Loaded Die. There are many ways to produce crooked dice. To load a die so that 6 comes up too often
and 1 (whichis opposite 6) comes up too seldom, add a bit of lead to the filling of the spot on the 1 face.
Because the spot is solid plastic, this works even with transparent dice. If a die is loaded so that 6 comes up
with probability 0.2 and the probabilities of the 2, 3, 4, and 5 faces are not affected, what is the assignment of
probabilities to the six faces?
P( six) = 0.2 P(two) = 1/6 P(three) = 1/6, P(four) = 1/6, P(five) = 1/6.
P(one) = 1 – (0.2 + 4/6)
= 2/15
4.28 Race in the census. The 2000 census allowed each person to choose from a long list of races. That is,
in the eyes of the Census Bureau, you belong to whatever race
you say you belong to. "Hispanic/Latino" is a separate category; Hispanics may be of any race. If we choose
a resident of the United States at random, the 2000 census gives these probabilities:
Hispanic Not Hispanic Total
Asian 0.000 0.036 0.036
Black 0.003 0.121 0.124
White 0.060 0.691 0.751
Other 0.062 0.027 0.089
Total 0.125 0.875 1
Let A be the event that a randomly chosen American is Hispanic, and let B be the event that the person
chosen is white.
(a) Verify that the table gives a legitimate assignment of probabilities
All probabilities assigned are between 0 and 1. Also, the sum of all probabilities is 1. The assignment is
legitimate.
(b) What is P(A)? P(A) = 0.125
(c) Describe Bc in words and find P(Bc) by the complement rule.
The event Bc is a person is not white. P(B c ) = 1 – P(B) = 1 – 0.751 = 0.249.
(d) Express "the person chosen is a non-Hispanic white" in terms of events A and B. What is the probability
of this event?
“person is a non-Hispanic White” : A c and B.
P(A c and B) = 0.691.

4.29 Rh blood types. Human blood is typed as O, A, B, or AB and also as Rh-positive or Rh-negative.
ABO type and Rhfactor type are independent because they are governed by different genes. In the
American population, 84% of people are Rh-positive. Use the information about ABO type in Exercise 4.21
to give the probability distribution of blood type (ABO and Rh) for a randomly chosen person.
First I am going to write down the given probabilities using function notation.
P(Rh-positive) = 0.84 thus, P(Rh-negative) = 0.16
Here is the table from problem 4.21.
Blood Type O A B AB
U.S. probability 0.45 0.40 0.11 0.04
Next we go through all possible combinations.
P(Rh-positive AND O) = P(Rh-positve)P(O)
= 0.84(0.45) This calculation is correct because we know we have independence.
We continue in the same manner to get the rest.
P(Rh-positive AND A) = 0.84(0.4)
P(Rh-positive AND B) = 0.84(0.11)
P(Rh-positive AND AB) = 0.84(0.04)
P(Rh-negative AND O) = 0.16(0.45)
P(Rh-negative AND A) = 0.16(0.40)
P(Rh-negative AND B) = 0.16(0.11)
P(Rh-negative AND AB) = 0.16(0.04)
4.30 Are the events independent? Exercise 4.28 assigns probabilities for the ethnic background of a
randomly chosen resident of the United States. Let A be the event that the person chosen is Hispanic, and let
B be the event that he or she is white. Are the events A and B independent? How do you know?
I will use two approaches. The first uses the formula P(A | B) = P(A) this is only true if we have
independence.
P(Hispanic | White) =
P( Hispanic and White)
P(White)
= 0.060
0.751
= 0.0799
But P(Hispanic) = 0.125 which means that the 12.5% of the people in the U.S. in 2000 were Hispanic.
However, if we look at the subset of the population “White” and ask what percentage of whites are
hispanic the value is 7.99%. The percentage of Hipanics drops when we look at the subgroup “white”
thus we do not have independence since the proportion changes. That is what independence deals with
a change in probability. If the probability changes, we do not have independence.

Another option to find if we have independence uses the formula P(A and B) = P(A)P(B) which only
works if we have independence.
P(Hispanic AND White) = 0.06 from the table.
Now we can get the two other values from the table as well: P(Hispanic) = 0.125 P(White) = 0.751
P(Hispanic)P(White) = 0.751(0.125)
= 0.0939
Since P(Hispanic AND White) ≠ P(Hispanic)P(White) we do not have independence.
4.34 Universal blood donors. People with type O-negative blood are universal donors. That is, any patient
can receive a transfusion of O-negative blood. Only 7% of the American population have O-negative
blood. If 10 people appear at random to give blood, what is the probability that at least 1 of them is a
universal donor?
What is the sample space? What are we observing? How many people have O-negative blood out of
ten people; S:{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} What do we want? P(1 or 2 or 3 or 4 or… or 10) we want
the probability of observing one person or two people or three … with O-negative blood. It is easier to
answer the complement question: What is the complement of the event we want to witness {1, 2, 3, 4,
5, …, 9, 10}? Well the complement is to observe 0 people having this type of blood.
P(0) = P(1 st person no AND … AND 10 th person no) recall we have independence so…
= (1 – 0.07) 10
= .4840
Thus, P(1 or 2 or 3 or 4 or… or 10) = 1 – 0.484
= 0.516
4.35 Disappearing Internet sites. Internet sites often vanish or move, so that references to them can't be
followed. In fact, 13% of Internet sites referenced in major scientific journals are lost within two years after
publication.9 If a paper contains seven Internet references, what is the probability that all seven are still good
two years later? What specific assumptions did you make in order to calculate this probability?
P(link lost, within 2 yrs) = 0.13
P(link not lost, within 2 yrs) = 0.87
P(link not lost AND link not lost AND … AND link not lost) = 0.87 7 ≈ 0.3773
In order to calculate this I had to assume I had independence, which means the following. I am assuming
that if a link is lost within two years, the probability of another link not being lost has not changed from the
stated 13%. On a practical matter, suppose you knew that all seven links resided on the same server, then
the independence assumption would not apply. Why? Because one can reason that if the links all reside on
the same server, the manager of the server might decide to sever all links, or if the server goes down, then all
the links go down. The fact that they all reside on the same server, would put in question the independence
of their survival.

4.36 Random digit dialing. Most sample surveys use random digit dialing equipment to call residential
telephone numbers at random. The telephone polling firm Zogby International reports that the
probability that a call reaches a live person is 0.2. Calls are independent.
(a) A polling firm places 5 calls. What is the probability that none of them reaches a person?
What is the probability that a person is not reached, P(no reach) = 1 – 0.2 = 0.8
The question asks for none of the five calls to reach a person.
P(no reach AND no reach AND no reach AND no reach AND no reach) = 0.8 5 ≈ 0.3277
(b) When calls are made to New York City, the probability of reaching a person is only 0.08. What is the
probability that none of 5 calls made to New York City reaches a person?
P(no reach AND no reach AND no reach AND no reach AND no reach) = 0.92 5 ≈ 0.6591
4.39 Random walks and stock prices. The "random walk" theory of securities prices holds that price
movements in disjoint time periods are independent of each other. Suppose that we record only whether the
price is up or down each year, and that the probability that our portfolio rises in price in any one year is 0.65.
(This probability is approximately correct for a portfolio containing equal dollar amounts of all common .
stocks listed on the New York Stock Exchange.)
First I am going to organize the given information.
P(price up) = 0.65 P(price down) = 0.35 (we will have to assume that if the price remains flat that
we can place it in one of the two mentioned categories).
Notice we have independence.
(a) What is the probability that our portfolio goes up for 3 consecutive years?
P(price up AND price up AND price up) = 0.65 3 ≈ 0.2746
(b) If you know that the portfolio has risen in price 2 years in a row, what probability do you assign
to the event that if will go down next year?
P(price down) = 0.35 since I have independence (it does not matter what occurred the previous
year); that is, P(price down | price down) = 0.35.
(c) What is the probability that the portfolio's value moves in. the same direction in both of the next
2 years?
P(down And down OR up And up) = P(down And down) + P(up And up)
= 0.35 2 + 0.65 2
= 0.545