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Instructor : Donglei Du
ADM 2623 Midterm I (Nov 4-2015)
Student Name:_______________________
Student ID________
Duration: 80 minutes
________________
1. Problem 1. Short Explanations (8 marks).
a. (2 points) What is a random variable?
ANS: a function from the sample space to numbers
b. (2 points) What is the difference between the general and special addition rule?
ANS: general rule can be applied to any events and special rule is only good for mutually exclusive
evetns
c.
(2 points) What is the difference between independence and mutually
exclusiveness between two events?
ANS: Independence means the joint probability of two events is equal to the product of the two
event probabilities (or one event has no impact on the other in terms, or the conditional
probability is equal to the unconditional probability), while mutually exclusiveness means two
events are disjoint.
d. (2 points) What is the difference between Chebyshev Theorem and the empirical
rule?
ANS: Chebyshev Theorem is applicable to any data set while the empirical rule only works for
normally distributed data (bell-shaped symmetric data)
2. Problem 2. (4 marks). Thirty AA batteries were tested to determine how long they would last. The
results, to the nearest minute, were recorded as follows:
423, 369, 387, 411, 393, 394, 371, 377, 389, 409,
392, 408, 431, 401, 363, 391, 405, 382, 400, 381,
399, 415, 428, 422, 396, 372, 410, 419, 386, 390
Generate a frequency table for this raw data
i. (1 point) Calculate the number of classes:
ANS: k=5
ii. (1 point) Calculate the class width
ANS: w=14
iii. (1 point) Calculate the lower limit of the first class
ANS: l=362
iv. (1 point) draw a frequency distribution table below:
class
freq
[362,376)
4
[376,390)
6
[390,404)
9
[404,418)
6
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[418,432]
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3. Problem 3. (6 marks). Given the following grouped data:
Class Limits
Frequency
Class
fx
(f)
midpoint
(x)
[40, 50)
1
45
45
[50, 60)
2
55
110
[60, 70)
3
195
65
[70 80)
4
300
75
[80, 90]
6
510
85
[90,100]
4
380
95
fx^2
2025
6050
12675
22500
43350
36100
20
1540
122700
a. (2 points) Calculate the mean for this grouped data
ANS: mean = 1540/20 = 77
b. (2 points) Calculate the median for this grouped dat.
ANS: =70+10*(20/2-6)/4=80
c. (2 points) Calculate the standard deviation for this grouped data assuming it is a
sample.
ANS: Var = (122700-1540^2/20)/(20-1) 216.8421053
and SD=14.72555959
Problem 4. (4 marks) Most people who are used to receiving e-mail containing a word "replica"
know that this message is likely to be spam, more precisely a proposal to sell counterfeit copies
of well-known brands of watches. The spam detection software, however, does not "know" such
facts; all it can do is compute probabilities. Recent statistics show that the probability of any
message being spam is 80%; the probability that a spam message contains the word "replica" is
90%, while the probability that a non-spam message contains the word "replica" is 5%. Suppose
that now you receive a message with a “replica” in it, what is the chance that this message is a
spam? (hint: Bayes Theorem)?
ANS: Let S be the event representing all spams, and R be the event represent all messages
containing the word “Replica”. Then
P(S)=0.8, P(R|S)=0.9, P(R|~S)=0.05 (here ~ represents complement)
Based on Bayes theorem
𝑃(𝑆|𝑅) =
𝑃(𝑅|𝑆) ∗ 𝑃(𝑆)
0.9 ∗ 0.8
0.72
=
=
≈ 0.9863014
𝑃(𝑅|𝑆) ∗ 𝑃(𝑆) + 𝑃(𝑅|~𝑆) ∗ 𝑃(~𝑆) 0.9 ∗ 0.8 + 0.05 ∗ 0.2 0.73
Problem 5. (8 marks) Short calculations
a. (2 points). Given a random variable with probability distribution as follows
X
P(X)
2
0.4
-2
0.1
3
0.5
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Find the probability distribution of the new random variable Y=2X-3
ANS:
Y
1
-7
3
b.
P(Y)
0.4
0.1
0.5
(2 points) Given a joint probability distribution as follows, please explain if X
and Y are independent or not?
(X,Y)
-1
-2
-3
1
0.01
0.02
0.15
0.18
2
0.19
0.18
0.03
0.40
3
0.20
0.20
0.02
0.42
0.4
0.4
0.2
ANS: They are not independent because some joint probability is not equal to the
product of the two corresponding marginal probabilities, such as
P(X=-1, Y=1)=0.01 is not equal to P(X=-1)*P(Y=1)=0.4(0.18
c. (2 points)? In a group of 101 students, 49 are juniors, 50 are female, and 22 are
female juniors. Find the probability that a student picked from this group at
random is either a junior or a female.
ANS: Note that P(junior) = 49/101 and P(female) = 50/101, and P(junior and female) =
22/101. Thus
P(junior or female) = 49101 + 50/101 – 22/101 = 77/101
d. (2 points) Let's try a new probability experiment. This time, consider a bag of
marbles, containing 10 red, 20 blue, and 15 green marbles. Suppose that two marbles
are drawn without replacement. (The first marble is not put back in the bag before
drawing the second.) What is the probability that both marbles drawn are red
ANS: Let's define a couple events:
E = first marble is red
F = second marble is red
We want P(E and F). Using the General Multiplication Rule, we see
P(E and F) = P(E) • P(F|E) = (10/45) • (9/44) ≈ 0.0455