MAT 320 Exam #1
NAME:
Directions: Read each problem carefully and answers all the problems. Round decimal
answers to 3 places where applicable. Show all work you want credit for. Justify steps in
proofs. Use backs’ of pages if needed.
1.
57
73
Below are grades from a test. Make a box-plot for the data. Be sure to label the boxplot, including the median, Q1, Q3 and any outliers. Justify why they are outliers.
80
82
61
88
76
74
95
89
51
60
73
70
93
63
83
24
96
57
98
84
65
80
Q1 = 62
Median = 75 Q3 = 86
IQR = 24
LIF = 26
UIF = 122
24 is a potential outlier because 24 < 26
Note left whisker goes to 51, not 24 and right whisker goes to 98 not 122
2.
For the data in problem 1, make a histogram using classes of size 10, with the first
class starting at 20 and including 20. Label the x and y axes and make a frequency
distribution.
Class
Freq
[20, 30)
1
[30, 40)
0
[40, 50)
0
[50, 60)
3
[60, 70)
4
[70, 80)
5
[80, 90)
7
[90, 100)
4
3.
Prove that for any sets A, B, and C where P(A ∩ B) ≠ 0:
P(A ∩ B ∩ C) = P(A) P(B | A) P(C | A ∩ B)
Proof
P(A ∩ B ∩ C) = P(C | A ∩ B) P(A ∩ B) by Conditional Probability
P(A ∩ B ∩ C) = P(C | A ∩ B) P(B | A) P(A) by Conditional Probability
4.
Prove that for any sets A and B where P(B) ≠ 0:
P( A | B) 
P( A)  P( B)  1
P( B)
Proof
P(A | B) = P(A ∩ B) / P(B) by Conditional Probability
P(A | B) = [P(A) + P(B) – P(A U B)]/P(B) by Additive Rule
Since P(A U B) ≤ 1 then
P(A | B) ≥ [P(A) + P(B) – 1]/P(B)
Note: you cannot assume it is true and show it is valid. This is a much weaker statement!
5.
A polygraph is a lie detector machine. Given that the suspect is lying, the polygraph
is accurate 90% of the time (polygraph reports a lie). Given the suspect is telling the
truth the polygraph is accurate 98% of the time (polygraph reports truth). In a
random sample of 100 suspects only 2 are lying. What is the probability that a
suspect is telling the truth, if the polygraph reports the suspect is lying? What is the
probability that a suspect is lying, if the polygraph reports the suspect is telling the
truth?
Let T = {suspect told truth} S = {polygraph recorded truth}
Given:
P(S´|T´) = .90 P(S | T) = .98
P(T´) = .02
P(T) = .98
so P(S |T´) = .10
P(S´|T) = .02
a.
P(T | S´) = P(S´|T) P(T) / [P(S´|T) P(T) + P(S´|T´) P(T´)]
P(T | S´) = .02(.98) / [.02(.98) + .90(.02)]
P(T | S´) = .0196 / .0376 = .5212
b.
P(T´ | S) = P(S|T´) P(T´) / [P(S|T´) P(T´) + P(S|T) P(T)]
P(T´ | S) = .10(.02) / [.10(.02) + .98(.98)]
P(T´ | S) = .002 / .9624 = .0020
6.
On a lot there are 24 used cars, 16 are domestic and 8 are foreign. Five cars are
selected at random. Find the following probabilities:
a. All five selected are domestic.
b. Exactly three cars selected are domestic.
c. At least three cars selected are domestic.
a.
b.
c.
P(5 Domestic) = 16C5 * 8C0 / 24C5 = 4368/42504 = .1028
P(3 Domestic) = 16C3 * 8C2 / 24C5 = 15680/42504 = .3689
P(at least 3 Domestic) = P(3) + P(4) + P(5)
P(4 Domestic) = 16C4 * 8C1 / 24C5 = 14560/42504
P(at least 3 Domestic) = 34608/42504 = .8142
7.
You ask a friend to water your plant while you are away on vacation. Without water
your plant will die with probability 0.8; with water it will die with probability 0.15.
You are 90% sure that your friend will remember to water the plant.
a. Find the probability that the plant lives.
b. If the plant dies, find the probability that your friend forgot to water it.
c. Find the probability that the plant lives or your friend waters the plant.
D = {plant dies}
W = {plant is watered}
P(D | W´) = .8
P(D | W) = .15
P(W) = .9
P(D´ | W) = .2
P(D´ | W) = .85
P(W´) = .1
a.
P(D´) = P(D´ ∩ W) + P(D´ ∩ W´)
P(D´) = P(D´|W)P(W) + P(D´| W´)P(W´)
P(D´) = .85(.9) + .2(.1) = .785
b.
P(W´|D) = P(W´ ∩ D) / P(D) = P(D|W´)P(W´)/P(D)
P(W´|D) = .8 * .1 / .215 = .372
c.
P(D´ U W) = P(D´) + P(W) – P(D´ ∩ W)
P(D´ U W) = .785 + .9 - .765 = .92
8.
A closet contains 8 different pairs of shoes. If 4 shoes are randomly selected, what
is the probability that there will be:
a. Two complete pairs of shoes?
b. At least 1 complete pair of shoes?
c. Exactly 1 complete pair of shoes?
a.
P(2 pairs) = 8C2 / 16C4 = 28 / 1820 = .0154
b.
P(At least 1 pair) = P(1) + P(2) = (672 + 28) / 1820 = 700 / 1820 = .3846
or
P(At least 1 pair) = 1 – P(0) = 1 – [(16 * 14 * 12 * 10)/4!/16C4]
c.
P(Exactly 1 pair) = (8 * 7C2 * 4) / 16C4 = 672 / 1820 = .3692
or
P(Exactly 1 pair) = (8 * (14C2 – 7))/16C4
Note that P(0 pairs) = P(exactly 1 pair) + P(exactly 2 pairs) = 1
1120/1820 + 672/1820 + 28/1820 = 1