Download Midterm Solution - UC Davis Statistics

UNIVERSITY OF CALIFORNIA DAVIS
Summer Session I. 2014– July 09
Elementary Statistics
STA 013
STUDENT ID:
NAME:
GOOD LUCK!!
Page 1 of ?? – Elementary Statistics (STA 013)
• Show your work for full credit. In problems where you are asked to carry out
a computation, simplify your answers as much as possible and then box your
final answer. Please write clearly.
• Mysterious or unsupported answers will not receive full credit. A correct answer, unsupported by calculations, explanation, or algebraic work will receive no
credit; an incorrect answer supported by substantially correct calculations and
explanations might still receive partial credit.
Question 1
Consider the following pairs of measurements:
x
y
5
4
3
3
-1
0
2
1
7
8
6
5
4
3
(a) [10 points] Find the five number summary for y.
Answer:
min
0
Q1
1
median
3
Q3
5
max
8
Page 2 of ?? – Elementary Statistics (STA 013)
(b) [5 points] Create a boxplot for y. Is there any outlier?
0
2
4
6
8
Answer:
There is no outlier in y.
(c) [12 points] Compute (1) means and (2) variances for both x and y.
Answer:
mean
variance
x
3.71
7.24
y
3.43
6.95
Page 3 of ?? – Elementary Statistics (STA 013)
(d) [5 points] Construct a scatter plot. Without calculating the correlation coefficient, can
you tell what relationship the data have?
Answer:
6
8
●
●
4
y
●
●
2
●
0
●
●
0
2
4
6
x
x and y are positive correlated.
(e) [5 points] Calculate the correlation coefficient of the data.
Answer:
X
Xi Yi = 129
P
Xi Yi − nX̄ Ȳ
Sxy =
= 6.64
n−1
p
p
sd(x) = var(x) = 2.69 and sd(y) = var(y) = 2.64
Sxy
r=
= 0.94
2.69 × 2.64
Page 4 of ?? – Elementary Statistics (STA 013)
Question 2 Suppose that A and B are mutually exclusive events for which P (A) = 0.3 and
P (B) = 0.5. What is the probability that
(a) [5 points] Either A or B occurs.
Answer: Since A and B are mutually exclusive, P (A ∩ B) = 0
P (A ∪ B) = P (A) + P (B) = 0.8
(b) [5 points] Both A and B occur.
Answer: Since A and B are mutually exclusive, P (A ∩ B) = 0
(c) [5 points] P (A ∩ B c )
Answer:
P (A ∩ B c ) = P (A) = 0.3
Page 5 of ?? – Elementary Statistics (STA 013)
Question 3 A cafeteria offers a 3-course meal. One chooses an entree, a side, and a dessert.
The possible choices are give in the following table. For following questions, suppose all simple
events are equally likely.
Course
Entree
Side
Dessert
Choices
Chicken(C) or roast beef(RB)
Pasta(PA) or rice(R) or potatoes(PO)
Ice cream(I) or apple pie(A)
(a) [5 points] How many outcomes are in the sample space.
Answer: There are totally 12 different outcomes by mn rule.
2 × 3 × 2 = 12
(b) [5 points] Let A be the event that ice cream is chosen. Specify the outcomes in A.
Answer:
{chicken, pasta, ice cream}, {chicken, rice, ice cream}, {chicken, potatoes, ice cream}
{roast beef, pasta, ice cream}, {roast beef, rice, ice cream}, {roast beef, potatoes, ice cream}
(c) [5 points] Let B be the event that chicken is chosen. Specify the outcomes in B.
Answer:
{chicken, pasta, ice cream}, {chicken, rice, ice cream}, {chicken, potatoes, ice cream}
{chicken, pasta, apple pie}, {chicken, rice, apple pie}, {chicken, potatoes, apple pie}
Page 6 of ?? – Elementary Statistics (STA 013)
(d) [9 points] Find the probabilities of P (A ∩ B), P (A|B), and P (B|A).
Answer:
A ∩ B ={(chicken, pasta, ice cream), ((chicken, rice, ice cream)),
(chicken, potatoes, ice cream)}
3
1
=
12
4
P (A ∩ B)
1
P (A|B) =
=
P (B)
2
1
P (A ∩ B)
=
P (B|A) =
P (A)
2
P (A ∩ B) =
(e) [6 points] Are event A and event B independent with each other? Explain.
Answer: Yes, they are independent. Since P (A ∩ B) = P (A)P (B).
Page 7 of ?? – Elementary Statistics (STA 013)
Question 4 A worker has asked her supervisor for a letter of recommendation for a new job.
She estimates that there is an 80 percent chance that she will get the job if she receives a strong
recommendation, a 40 percent chance if she receives a moderately good recommendation, and
a 10 percent chance if she receives a weak recommendation. She further estimates that the
probabilities that the recommendation will be strong, moderate, or weak are 0.7, 0.2, and 0.1
respectively.
From the problem, we can get following information:
P (job |strong recommendation) = 0.8
P (job |moderately good recommendation) = 0.4
P (job |weak recommendation) = 0.1
P (strong recommendation) = 0.7, P (moderately good recommendation) = 0.2,
P (weak recommendation) = 0.1
(a) [6 points] What is the probability that she will receive the new job offer?
Answer:
P (job) =P (job |strong recommendation) × P (strong recommendation)+
P (job |moderately good recommendation) × P (moderately good recommendation)+
P (job |weak recommendation) × P (weak recommendation) = 0.65
(b) [6 points] Given that she does receive the offer, how likely should she feel that she received
a strong recommendation?
Answer:
P (strong recommendation |job) =
P (strong recommendation ∩ job)
= 0.86
P (job)
(c) [6 points] Given that she does not receive the job offer, how likely should she feel that she
received a weak recommendation?
Answer:
P (no job) = 1 − P (job) = 0.35
P (no job |weak recommendation) = 1 − P (job |weak recommendation) = 0.9
P (weak recommendation ∩ no job)
= 0.26
P (weak recommendation |no job) =
P (no job)
Page 8 of ?? – Elementary Statistics (STA 013)
Extra Credit Genes relating to albinism are denoted by A and a. Only those people who
receive the a gene from both parents will be albino. Persons having the gene pair A, a are
normal in appearance and, because they can pass on the trait to their offspring, are called
carriers. Suppose that a normal couple has two children, exactly one of whom is an albino.
Suppose that the nonalbino child mates with a person who is known to be a carrier for albinism.
Consider the children they will have:
(a) [7 points] What is the probability that their first offspring is an albino?
Answer: 1/6
Since the normal couple has one albino child, they are both carrier, having gene pair (A,
a). Hence, the nonalbino child has 1/3 chance to have gene (A, A) and 2/3 chance to
have gene (A, a). The (A, A) child has probability 0 to have an albino child and the (A,
a) child will have 1/4 chance to get an albino child when he mates with a carrier.
Therefore the first probability that the first offspring is an albino is
2 1
1
× =
3 4
6
(b) [8 points] What is the conditional probability that their second offspring is an albino given
that their firstborn is not?
Answer: 3/20
The probability of the first offspring is nonalbino is 5/6.
P (f irst nonalbino) = 1 − P (f irst albino) = 1 −
1
5
=
6
6
The probability of the first one is nonalbino and the second one is albino is 1/8. Since
we already know that they have one albino child, the gene of both parents are (A, a).
Therefore, the probability of having a nonalbino child is 1/2 and the probability of having
an albino child is 1/4.
P (first nonalbino and second albino) = P (f irst nonalbino)P (second albino) =
P (second albino |f irst nonalbino) =
1
1 1
× =
2 4
8
P (first nonalbino and second albino)
3
=
P (f irst nonalbino)
20
——— End of Examination ———
Page 9 of ?? – Elementary Statistics (STA 013)