Download Introduction to the Practice of Statistics

Introduction to the Practice of Statistics
Fifth Edition
Moore, McCabe
Section 4.5 Homework Answers to 98, 99, 100,102, 103,105, 107, 109,110, 111, 112, 113
Working. In the language of government statistics, you are "in the labor force" if you are available for
work and either working or actively seeking work. The unemployment rate is the proportion of the labor
force (not of the entire population) who are unemployed. The table is for people aged 25 years or older.
The table is in thousands of people.
4.98 find the unemployment rate for people with each level of education. How does the unemployment
rate change with education? Explain carefully why your results show that level of education and being
employed are not independent.
Highest Education
Total Population In Labor Force
Employed
Did not finish high school
28,021
12,623
11,552
High School but no college
59,844
38,210
36,249
Less than a bachelors degree
46,777
33,928
32,429
College Graduate
51,568
40,414
39,250
Total
186,210
125,175
119,480
I calculate the unemployment rate as (In Labor Force – Employed)/In Labor Force
Highest Education
In Labor Force
12,623
38,210
33,928
Employed
11,552
36,249
32,429
College Graduate
40,414
Total
125,175
Did not finish high school
High School but no college
Less than a bachelors degree
Unemployment
Rate
0.085
0.051
0.044
Employed Rate
0.915
0.949
0.956
39,250
0.029
0.971
119,480
0.045
0.955
Notice that the unemployment rate changes as we look at different education categories, that is the value
of the unemployment rate depends on which category of education you are considering. Thus, education
level and the unemployment rate are not independent.
Highest Education
Total Population In Labor Force
Employed
Did not finish high school
28,021
12,623
11,552
High School but no college
59,844
38,210
36,249
Less than a bachelors degree
46,777
33,928
32,429
College Graduate
51,568
40,414
39,250
Total
186,210
125,175
119,480
4.99
(a) What is the probability that a randomly chosen person 25 years of age or older is in the labor force?
P(in labor force) =
125,175
≈ 0.6722
186, 210
(b) If you know that the person chosen is a college graduate, what is the conditional probability that he or
she is in the labor force?
P(in labor force | college graduate) =
40414
≈ 0.7837
51568
(c) Are the events "in the labor force" and "college graduate" independent? How do you know?
•t-
No, since P(in labor force | college graduate) ≠ P(in labor force); see page 322.
4.100 You know that a person is employed. What is the conditional probability that he or she is a
college graduate? You know that a second person is a college graduate. What is the conditional
probability that he or she is employed?
P(college graduate | employed) =
39250
= .3285
119480
39250
= 0.9712 This is the answer if you only consider college
40414
graduates that are categorized in being in the labor force.
P(employed | college graduate) =
P(employed | college graduate) =
39250
= 0.7611 This is the answer if you consider the entire
51568
population of college graduates.
4.102 The probability that a randomly chosen student at the University of New Harmony is a woman is
0.6. The probability that the student is studying education is 0.15. The conditional probability that the
student is a woman, given that the student is studying education, is 0.8. What is the conditional
probability that the student is studying education, given that she is a woman?
P(woman) = 0.6, P(study education) = 0.15. P(woman | study education) = 0.8.
P(study education | woman) = ?
Notice that the probability we want is the opposite of the conditional probability given. This would
normally signal for me to create a tree diagram, but, when I tried I realize there is information that is
missing not allowing me to finish the tree. So, next I look to see if a formula will show me the way.
Here is the formula for the conditional probability P(A | B) =
P(study education | woman) =
P(A and B)
P(B)
P(study education AND woman)
P(woman)
=
P(study education)P(woman | study education)
P(woman)
=
(0.15)(0.8)
0.6
this transfers to
4.103 As explained in Exercise 4.60 (page 305), spelling errors in a text can be either nonword errors or
word errors. Nonword errors make up 25% of all errors. A human proofreader will catch 90% of
nonword errors and 70% of word errors. What percent of all errors will the proofreader catch? (Draw a
tree diagram to organize the information given.)
P(nonword) = 0.25, P(catch | nonword) = 0.9 P(catch | word) = 0.7
nonword
0.25
Question: P(catch)
P(catch) = 0.25(0.9) + 0.75(0.7) = 0.75
0.75
word
0.9
0.1
catch
not catch
0.7
catch
0.3
not catch
4.104 The voters in a large city are 40% white, 40% black, and 20% Hispanic. (Hispanics may be of any
race in official statistics, but in this case we are speaking of political blocks.) A black mayoral candidate
anticipates attracting 30% of the white vote, 90% of the black vote, and 50% of the Hispanic vote. Draw
a tree diagram with probabilities for the race (white, black, or Hispanic) and vote (for or against the
candidate) of a randomly chosen voter. What percent of the overall vote does the candidate expect to get?
P(white ) = 0.4, P(black) = 0.4 , P(Hispanic) = 0.2
P(get vote | white) = 0.3, P(get vote | black) = 0.9, P(get vote | Hispanic) = 0.5
The question is P(get vote) = ?
P(get vote) = 0.4(0.3) + 0.4(0.9) + 0.2(0.5)
= 0.58
4.105 At a self-service gas station, 40% of the customers pump regular gas, 35% pump midgrade, and
25% pump premium gas. Of those who pump regular, 30% pay at least $20. Of those who pump
midgrade, 50% pay at least $20. And of those who pump premium, 60% pay at least '$20. What is the
probability that the next customer pays at least $20?
P(regular) = 0.4, P(midgrade) = 0.35, P(premium) = 0.25.
P(pay ≥ $20 | regular) = 0.3 P(pay ≥ $20 | midgrade) = 0.5
P(pay ≥ $20 | premium) = 0.6
The question is P(pay ≥ $20) = ?
0.3
regular
pay < $20
0.4
0.35
0.25
0.5
mid
prem
pay ≥ $20
pay ≥ $20
pay < $20
0.6
pay ≥ $20
P(pay ≥ $20) = 0.4(0.3) + 0.35(0.5) + 0.25(0.6) = 0.445
4.107 In the setting of Exercise 4.105, what percent of customers, who pay at least $20, pump premium?
(Write this as a conditional probability.)
P(pump premium | pay ≥ $20) =
0.25(0.6)
0.15
=
≈ 0.3371
0.4(0.3) + 0.35(0.5) + 0.25(0.6)
0.445
4.109 Albinism. People with albinism have little pigment in their skin, hair, and eyes. The gene that
governs albinism has two forms (called alleles), which we denote by a and A. Each person has a pair of
these genes, one inherited from each parent. A child inherits one of each parent's two alleles,
independently with probability 0.5. Albinism is a recessive trait, so a person is albino only if the inherited
pair is aa
(a) Beth's parents are not albino but she has an albino brother. This implies that both of Beth's
parents have type Aa. Why?
Because if Beth’s parents are not albino, but she has an albino brother (aa), then each parent must
have the allele a.
(b) Which of the types aa, Aa, AA could a child of Beth's
parents have? What is the probability of each type?
Since each type is equally likely, and each is inherited
independently of each other then
A
a
A
AA
Aa
a
Aa
aa
P(AA) = 0.5(0.5) = ¼ , P(Aa) = (0.5)(0.5) + (0.5)(0.5) = ½ , and P(aa) = ¼.
(c) Beth is not albino. What are the conditional probabilities for Beth's
possible genetic types, given this fact? (Use the definition of
conditional probability.)
Notice that we can rule out aa, since Beth is not an Albino. So having
ruled out that possibility the new probabilities are:
A
a
A
AA
Aa
a
Aa
aa
P(AA | not aa) = 1/3. P(Aa | not aa) = 2/3.
4.110 Albinism, continued. Beth knows the probabilities for her genetic types from part (c) of the
previous exercise. She marries Bob, who is albino. Bob's genetic type must be aa.
(a) What is the conditional probability that a child of Beth and Bob is non-albino if Beth has type
Aa? What is the conditional probability of a non-albino child if Beth has type AA? '
BOB
BOB
Beth
a
a
A
Aa
Aa
a
aa
aa
P(no aa child | Beth Aa) = 1/2
Beth
a
a
A
Aa
Aa
A
Aa
Aa
P(no aa child | Beth AA) = 1
(b) Beth and Bob's first child is non-albino. What is the conditional probability that Beth is a carrier, type
Aa?
I used the information from problem 109c to help me answer this question. What I had to realize
determine was the “starting point”, which turned out to be Beth’s possible genetic makeup. How did I
now this? Because part (a), of this problem, dealt with the chances of Bob and Beth having an albino
child, depending (given) on the assumed genetic make up of Beth. I realized that the problem a, material
was allowing me to create the second branch of the tree.
The question then is P(Beth Aa | no aa child), the graph is below, and then I used it to answer the
question by reading the tree.
aa
1/2
Aa
2/3
No aa
1/2
1/3
AA
0
aa
1
No aa
 2  1 
  
 3  2 
P(Beth Aa | no aa child) =
= 0.5
 2  1   1 
   +  (1)
 3  2   3 
4.111 Cystic fibrosis. Cystic fibrosis is a lung disorder that often results in death. It is inherited but can
be inherited only if both parents are carriers of an abnormal gene. In 1989, the CF gene that is abnormal
in carriers of cystic fibrosis was identified. The probability that a randomly chosen person of European
ancestry carries an abnormal CF gene is 1/25. (The probability is less in other ethnic groups) The
CF20m test detects most but not all harmful mutations of the CF gene. The test is positive for 90% of
people who are carriers. It is (ignoring human error) never positive for people who are not carriers.
Jason tests positive. What is the probability that he is a carrier?
P(carries gene) = 1/25 Note that I am assuming from this point on that we are only considering
someone from European ancestry.
pos
0.9
1/25
P(positive test | carries gene) = 0.9
P(positive test | NOT carries gene) = 0
24/25
/3
P(carries gene | positive test) = ?
carry gene
0.1
Not cg
0
1
neg
pos
neg
You can logically see that the answer is 1, since the test is never positive if you are a carrier. Creating a
tree diagram will allow you to see this as well.
 1 
  ( 0.9 )
 25 
P(carries gene | positive test) =
=1
 1 
 24 
  ( 0.9 ) +   (0)
 25 
 25 
4. 112 Cystic fibrosis, continued. Jason knows that he is a carrier of cystic fibrosis. His wife, Julianne,
has a brother with cystic fibrosis, which means the probability is 2/3 that she is a carrier. If Julianne is a
carrier, each child she has with Jason has probability ¼ of having cystic fibrosis. Is she is not a carrier,
her children can not have the disease. Jason and Julianne have one child, who does not have cystic
fibrosis. This information reduces the probability that Julianne is a carrier; given that she and Jason have
one child who does not have cystic fibrosis.
This problem involves Julianne which may carry the gene for Cystic Fibrosis. The known facts are
that her husband Jason is a carrier, her brother is also a carrier, and that they have one child who
does not have the cystic fibrosis.
P(carry gene) = 2/3 P(cystic F. | carry gene) = ¼,
P(cystic F. | not carry gene) = 0
The question is given that their child does not have cystic
fibrosis, what is she is still a carrier?
 2  3 
  
 3  4 
P(carry gene | not cystic F.) =
= 0.6
 2  3   1 
    +   (1)
 3  4   3 
4.113 Muscular dystrophy. Muscular dystrophy is an incurable muscle-wasting disease. The most
common and serious type, called DMD, is caused by a sex-linked recessive mutation. Specifically:
women can be carriers but do not get the disease; a son of a carrier has probability 0.5 of having DMD; a
daughter has probability 0.5 of being a carrier. As many as 1/3 of DMD cases, however, are due to
spontaneous mutations in sons of mothers who are not carriers. Toni has one son, who has DMD
In the absence of other information, the probability is 1/3 that the son is the victim of a spontaneous
mutation and 2/3 that Toni is a carrier. There is a screening test called the CK test that is positive with
probability 0.7 if a woman is a carrier and with probability 0.1 if she is not. Toni’s CK test is positive.
What is the probability that she is a carrier?
P(son has DMD | mom carrier) = 0.5
P(daughter a carrier of DMD | mom carrier) =
0.5
P(son has DMD | mom not a carrier) = 0.3333
Toni has one son who has DMD
P(CK test positive | mom is a carrier) = 0.7
P(CK test is positive | mom not a carrier) = 0.1
P(Toni carrier | test positive) = ?
After I wrote all this down, and I began constructing the tree, I realized that I did not need much of the
information that was given.
2
( 0.7 )
3
P(Toni carrier | test positive) =
2
1
(0.7) + ( 0.1)
3
3
= 0.9333