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MATH 166-506 Fall 2016, EXAM II
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This exam contains 6 pages and 15 problems. If any pages are missing, please inform the instructor immediately.
Directions:
• Enter all requested information on the top of this page, and put your initials on the top of every page, in case
the pages become separated.
• You may not use your books and notes.
• Part I is multiple-choice problems, mark the correct choice on your Scantron using a No. 2 pencil. The Scantron
will not be returned, therefore for you own records, also record your choices on your exam! Each problem is
worth 5 points.
• Part II is work-out problems. Show All your work. Partial credit is possible. A correct answer without any
work will not receive full credit. Please write clear and legibly, and indicate your final answer by boxing it
when possible.
• Unless otherwise noted, your answers must be in fraction form or terminating decimals if possible. Otherwise,
round to 4 decimal place.
• A standard deck of playing cards contains 52 cards (NO jokers). There are 13 cards for each of the four suits:
clubs, spades, diamonds, and hearts. The 13 cards are numbered 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King,
Ace. Clubs and spades are black. Diamonds and hearts are red. A face card is a Jack, Queen, or King and an
Ace is NOT considered as a face card.
THE AGGIE CODE OF HONOR
An Aggie does not lie, cheat or steal, or tolerate those who do.
By signing below you indicate that all your work is your own and that you have neither given nor received help
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DO NOT WRITE BELOW!
Question:
1-11
12
13
14
15
Total
Points:
55
15
8
8
14
100
Score:
Math 166-506
Exam 1 - Page 2 of 6
10/21/16
PART I: Multiple Choice Problems
1. (5 points) Find the number of distinct ways to arrange the letters in ”HAKUNAMATATA”
A. 1,209,600
B. 479,001,600
C. 50,400
12!
D. 1,995,840
5!×2! = 1, 995, 840
E. 47,900,160
2. (5 points) An exam consists of four true/false questions followed by five multiple choice questions each with 3
answers. How many ways can a student answer the exam if they can leave any of the questions blank?
A. 82,944
B. 253,125
(2 + 1)4 × (3 + 1)5 = 34 × 45 = 82944
C. 16,384
D. 3,888
E. None of these
3. (5 points) Pick a marble at a time randomly WITHOUT replacement from a box containing 3 green marbles, 5
red marbles and 2 blue marbles. Assign the random variable A to be the times of picking until a green marble
obtained
A. X = {0, 1, 2, 3, 4, ...}
B. X = {1, 2, 3, 4, 5, 6, 7, 8}
C. X = {0, 1, 2, 3, 4, 5, 6, 7, 8}
D. X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
E. X = {1, 2, 3, 4, 5, ...}
4. (5 points) Redbox selected 200 customers at random and recorded the number of movies which each customer
rented last week. The result is given in the following table. Find the mean value of the number of movies rented.
Number of Movies
Number of Customers
0
20
1
60
2
40
3
20
4
60
A. 40
B. 2
C. 1.183
D. 1.4
E. 2.2
5. (5 points) Classify the random variable for the following two experiments.
a. Pick a marble at a time randomly WITH replacement from a box containing 3 green marbles, 5 red marbles
and 2 blue marbles. Assign the random variable X to be the times of picking until a green marble obtained.
b. Assign the random variable Y to be the height of all the freshman in Texas A& M University this year.
A. X: Finite Discrete, Y: Continuous.
B. X: Infinite Discrete, Y: Continuous .
C. X: Infinite Discrete, Y: Finite Discrete.
D. X: Finite Discrete, Y: Finite Discrete.
E. X: Continuous, Y: Infinite Discrete.
Math 166-506
Exam 1 - Page 3 of 6
10/21/16
6. (5 points) Mary and her daughter go to watch a movie with other 5 friends. In how many ways can these people
sit in a row if Mary want to sit with her daughter?
A. 10080
B. 2520
C. 5040
D. 240
(5 + 1)! × 2! = 6! × 2! = 1440
E. 1440
7. (5 points) A student creates 100 notecards for questions covered on the midterm. For entertainment, he draws
a notecard ramdomly to test his friend (replace the notecard in the pile each time.) If the friend can solve 85 of
the notecards, what is the probability that the friend can solve at least 14 of it after draw 20 times.
A. 0.04537
B. 0.06731
p=85/100=0.85, n=20.
C. 0.02194
D. 0.93269
P (X ≥ 14) = 1 − P (X ≤ 13) = 1- binomcdf( 20, 0.85, 13)=0.97806
E. 0.97806
8. (5 points) Four marbles are selected at random without replacement from a jar containing two white marbles ,
six green marbles and four blue marbles. Find the probability of the event if it contains exactly 2 green marbles
A. 15
B. 0.0303
n(event) = C(6, 2)C(2 + 4, 4 − 2) = C(6, 2)C(6, 2) = 225
C. 0.4545
n(sample space) = C(12, 4) = 495
D. 225
P (event) = 225/495 = 0.4545
E. 0.0189
8
30
10
30
6
30
1
30
−2 −1 0
I
5
30
1
6
30
4
30
−2 −1 0
II
1
2
30
2
6
30
17
30
6
30
6
30
6
30
6
30
−2 −1 0
II
1
2
1
30
2
9. (5 points) What is E(X) for histogram I.
A. 0.1
B. 1.5667
C. -0.1
D. 0.5667
E. 0.2
10. (5 points) From the three histograms above, which has the largest variance
A. I
B. II
C. III
D. I and III
E. All are equal
11. (5 points) The amount of chips in an 8-ounce bag is normally distributed with a mean of 8 ounces and a standard
deviation of 0.3 ounce. What is the probability that a bag has at most 8.1 ounces?
A. −3.7403 × 10−15
µ = 8, σ = 0.3
B. 0.6306
C. 0.5000
P (X ≤ 8.1) = normcdf( -1E99, 8.1, 8, 0.3) = 0.6306
D. -0.5398
E. 0.8352
Math 166-506
Exam 1 - Page 4 of 6
10/21/16
PART II: Work Out Problems
12. (15 points) An unfair coin has the property that the probability of flipping a head is 20%. We flip the coin 200
times.
probability of flipping a head (p) = 0.2
(a) What is the probability that more than 150 or fewer than 140 of these record are Tail ?
Solution 1:
event E: more than 150 tails ⇒ at most 51 heads, P (E) = P (X ≤ 51)=binomcdf(200, 0.2, 51)=0.9506
event F: fewer than 140 tails ⇒ at least 61 heads,
P (F ) = P (X ≥ 61) = 1 − P (X ≤ 60)= 1 - binomcdf(200, 0.2, 60) = 2.7791 × 10−4
Since E ∩ F = ∅, P (E ∪ F ) = P (E) + P (F ) = 0.9509
Solution 2:
It’s TAIL, the probability of shown is 80%.
event E: more than 150 tails, P (E) = P (Y > 150) = 1 − P (Y ≤ 150) = 1 - binomcdf(200, 0.8, 150)=0.9506
event F: fewer than 140 tails, P (F ) = P (Y < 140) = P (Y ≤ 139)= binomcdf(200, 0.8, 139) = 2.7791 × 10−4
Since E ∩ F = ∅, P (E ∪ F ) = P (E) + P (F ) = 0.9509
(b) What is the probability that more than 35 of these record are Head ?
P (X > 35) = 1 − P (X ≤ 35) = 1 - binomcdf(200, 0.2, 35) = 0.7849
(c) What is the probability that at least 85 but at most 155 are Tail ?
Solution 1:
At least 85 but at most 155 are tails ⇒ at least 45 but at most 115 are heads
P (45 ≤ X ≤ 115) = P (X ≤ 115) − P (X ≤ 44)
= binomcdf(200, 0.2, 115) - binomcdf(200, 0.2, 44) = 0.2113
Solution 2:
It’s TAIL, the probability of shown is 80%.
P (85 ≤ Y ≤ 155) = P (Y ≤ 155) − P (Y ≤ 84)
= binomcdf(200, 0.8, 155) - binomcdf(200, 0.8, 84) = 0.2113
Math 166-506
Exam 1 - Page 5 of 6
10/21/16
13. (8 points) A lottery has a grand prize of $200,000, four runner-up prizes of $40,000 each, seven third-place prizes
of $4000 each, and eighteen consolation prizes of $400 each. If 1,000,000 tickets are sold for $1 each and the
probability of any one ticket winning is the same as that of any other ticket winning, find the expected return
on a $1 ticket. (Round your answer to four decimal places.)
People should pay $1 to get a ticket.
1
chance to win $200,000,
There’s 1000000
chance to win $400
Therefore the expect money to win is
4
1000000
1
1000000
chance to win $40,000,
7
1000000
chance to win $4000, and
18
1000000
4
7
18
× 200000 + 1000000
× 40000 + 1000000
× 4000 + 1000000
× 400 = 0.3952
Te expected return = 0.3948 - 1 = - 0.6048
14. (8 points) A teacher wishes to ”curve” a test whose grades were normally distributed with a mean of 69 and
standard deviation of 10. The top 10% of the class will get an A, the next 30% of the class will get a B. Find
the cutoff for A and B. (Round your answers to two decimal places.)
µ = 69, σ = 10, Let the cutoff for A is SA , and the cutoff for B is SB
P (X ≤ SA ) = 1 − 10% = 0.9 ⇒ SA = invNorm(0.9, 69, 10) = 81.82
P (SB ≤ X ≤ SA ) = 30% = P (X ≤ SA ) − P (X ≤ SB ) = 0.9 − P (X ≤ SB )
⇒ P (X ≤ SB ) = 0.6 ⇒ SB = invNorm(0.6, 69, 10) = 71.53
ANS: The cutoff for A is 81.82, and the cutoff for B is 71.53
Math 166-506
Exam 1 - Page 6 of 6
10/21/16
15. (14 points) A box contains only the red cards from a standard deck of playing card. Draw two cards from the
box. Let the random variable X represent the number of Aces drawn.
(a) Find the probability distribution of X.
X
P(X)
0
276
325
P (X = 0) =
P (X = 1) =
P (X = 2) =
1
48
325
2
1
325
C(24,2)C(2,0)
C(26,2)
C(24,1)C(2,1)
C(26,2)
C(24,0)C(2,2)
C(26,2)
=
=
=
276
325 ,
48
325 ,
1
325
(b) Find mean, median, mode, and Var(X)
Mean: 0.1538, Median: 0, Mode: 0, σ(X) = 0.3692 ⇒ Var(X)=σ(X)2 = 0.1363