Download Introduction to the Practice of Statistics

Introduction to the Practice of Statistics
Sixth Edition
Moore, McCabe
Section 4.3 Homework Answers
4.50 Texas hold 'em. The game of Texas hold 'em starts with each player receiving two cards. Here is
the probability distribution for the number of aces in two-card hands:
Number of aces
Probability
0
0.559
1
0.382
2
0.059
(a) Verify that this assignment of probabilities satisfies the requirement that the sum of the
probabilities for a discrete distribution must be 1.
0.559 + 0.382 + 0.059 = 1
(b) Make a probability histogram for this distribution.
c) What is the probability that a hand contains at least one ace? Show two different ways to calculate
this probability.
P(at least one ace) = P(1 ace or 2 aces)
= P(1 ace) + P(2 aces)
= 0.382 + 0.059
= 0.441
P(at least one ace) = 1 – P(no ace)
= 1 – 0.559
= 0.441
4.52 Length of human pregnancies. The length of human pregnancies from conception to birth varies
according to a distribution that is approximately Normal with mean 266 days and standard deviation 16
days. Call the length of a randomly chosen pregnancy Y.
(a) Make a sketch of the density curve for this random variable.
µ−3σ
µ−2σ
218
234
µ−σ
250
µ
µ+σ
266
282
µ+2σ
298
µ+3σ
314
µ+
Y
(b) What is P(F > 300)?
300 - 266 

P(F > 300) = P  Z >

16


218
= P(Z > 2.125)
234
250
266
282
298
314
300
= 0.0168
4.57 Non-Standarde Dice
Table contains Sum
of die
die 2
die 1
1
2
3
4
5
6
0
1
2
3
4
5
6
0
1
2
3
4
5
6
0
1
2
3
4
5
6
6
7
8
9
10
11
12
6
7
8
9
10
11
12
6
7
8
9
10
11
12
{Y | 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} There are 36 possilbe outcomes, but only 12 unique
outcomes; every outcome has three chances of appearing. For example P(Y = 1) = 3/36 = 1/12.
X
1
2
3
4
5
6
7
8
9
10
11
12
P(X) 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12
Y
4.58 Dungeons & Dragons. Role-playing games like Dungeons & Dragons use many different types of
dice, usually having either 4, 6, 8, 10, 12, or 20 sides. Roll a balanced 8-sided die and a balanced 6sided die and add the spots on the up-faces. Call the sum X. What is the probability distribution of the
random variable Y?
First lets organize the situation so we can easily see what are all the possible outcomes, which will
lead us to the sample space of the random variable Y.
1
2
3
4
5
6
1
(1,1)
(2,1)
(3,1)
(4,1)
(5,1)
(6,1)
2
(1,2)
(2,2)
(3,2)
(4,2)
(5,2)
(6,2)
3
(1,3)
(2,3)
(3,3)
(4,3)
(5,3)
(6,3)
4
(1,4)
(2,4)
(3,4)
(4,4)
(5,4)
(6,4)
5
(1,5)
(2,5)
(3,5)
(4,5)
(5,5)
(6,5)
6
(1,6)
(2,6)
(3,6)
(4,6)
(5,6)
(6,6)
7
(1,7)
(2,7)
(3,7)
(4,7)
(5,7)
(6,7)
8
(1,8)
(2,8)
(3,8)
(4,8)
(5,8)
(6,8)
Each of the outcomes in the cells is equally likely to occur since the two die are perfectly
balanced. There are 48 cells, thus, each cell has a 1/48 chance of appearing.
X
2
3
4
5
6
7
8
9
10
11
12
13
14
P(X) 1/48 2/48 3/48 4/48 5/48 6/48 6/48 6/48 5/48 4/48 3/48 2/48 1/48
4.60 Select the members of a student advisory board. Weary of low turnout in student elections, a
college administrator decides to choose a SRS of three students to form an advisory board that
represents student opinion. Suppose that 40% of all students oppose the use of student fees to fund
student interest groups, and that the opinion of the three students on the board are independent. Then
the probability is 0.4 that each opposes the funding of interest groups.
a) Call the three students A, B, and C. What is the probability that A and B support funding and C
opposes it?
P(A supports AND B supports AND C oppose) = (0.6)(0.6)(0.4)
b) List all possible combinations of opinions that can be held by students A, B, and C. Then give
the probabilities of each of these outcomes. Note that they are not all equally likely.
Define the ordered pair (SSO) to be A supports, B Supports and C opposes.
The sample space is then {SSS, SSO, SOS, OSS, SOO, OSO, OOS, OOO}
P(SSS) = (0.6)3,
P(SSO) = (0.6)2(0.4),
P(SOS) = (0.6)2(0.4),
P(OSS) = (0.6)2(0.4), P(SOO) = (0.6)(.4)2,
P(OSO) = (0.6)(.4)2,
P(OOS) = (0.6)(.4)2, P(OOO) = (0.4)3.
c) Let the random variable X be the number of student representatives who oppose the funding of
interest groups. Give the probability distribution of X.
X
P(X)
0
(0.6)3
1
3(0.6)2(0.4)
2
3(0.6)(.4)2
3
(0.4)3
d) Express the event “a majority of the advisory board opposes funding” in terms of X and find its
probability.
P(X ≥ 2) = P(X = 2 OR X = 3)
= P(X = 2) + P(X = 3)
= 3(0.6)(.4)2 + (0.4)3
4.64 The sum of two uniform random numbers. Generate two random numbers between 0 and 1 and
take Y to be their sum. Then Y is a continuous random variable that can take any value between 0 and
2. The density curve of Y is the triangle shown in Figure 4.12.
(a) Verify by geometry that the area under this curve 1.
Area of a triangle = 0.5(base)(height)
Area of triangle in this problem = 0.5(2)(1)
=1
(b) What is the probability that Y is less than 1? (Sketch the density curve, shade the area that represents
the probability, then find that area. Do this for (c) also.)
P(Y < 1) = 0.5(1)(1)
= 0.5
(c) What is the probability that Y is less than 1.5?
Here is another good example of how the complement concept can work.
I want to calculate P(Y < 1.5), but I recognize that it would be simpler if we calculated instead
P(Y > 1.5) the complement of what I really want.
Thus P(Y < 1.5) = 1 – P(Y > 1.5)
= 1 – (0.5)(0.5)(0.5)
= 1 – 0.125
= 0.875
4.66 Normal approximation for a sample proportions. A sample survey contacted an SRS of
663 registered voters in Oregon shortly after an election and asked respondents whether they had
voted. Voter records show that 56% of registered voters had actually voted. We will see in the
next chapter that in this situation the proportion p̂ of the sample who voted has approximately the
normal distribution with mean µ = 0.56 and standard deviation
σ = 0.019.
a) If the respondents answer truthfully, what is P(0.52 ≤ p̂ ≤ 0.60)? This is the probability that
the statistic p̂ estimates the parameter 0.56 within plus or minus 0.04.
0.60-0.56 
 0.52-0.56
P(0.52 ≤ pˆ ≤ 0.60) = P 
≤ Z≤

0.019 
 0.019
= P(-2.105 ≤ Z ≤ 2.105)
= 0.9647
b) In fact, 72% of the respondents said they had voted ( p̂ = 0.72). If respondents answer
truthfully, what is P( p̂ ≥ 0.72)? This probability is so small that it is good evidence that some
people who did not vote claimed that they did vote.
0.72-0.56 

P(pˆ ≥ 0.72) = P  Z ≥

0.019 

= P(Z ≥ 8.42)
≈ 0