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İstanbul Kültür University
Faculty of Engineering
MCB1007
Introduction to Probability and Statistics
Second Midterm
Fall 2014-2015
Number:
Name:
Department:
– You have 90 minutes to complete the exam. Please do not leave the examination
room in the first 30 minutes of the exam. There are six questions, of varying credit
(100 points total). Indicate clearly your final answer to each question. You are
allowed to use a calculator. During the exam, please turn off your cell phone(s).
You cannot use the book or your notes. You have one page for “cheat-sheet” notes
at the end of the exam papers.
Good luck!
Question 1.
Question 2.
Question 3.
Emel Yavuz Duman, PhD.
Question 4.
Question 5.
Question 6.
TOTAL
17 points
A statistics professor classifies his students according to their grade point average (GPA)
and their gender. The accompanying table gives the proportion of students falling into
the various categories.
PP
PP
GPA
P
Gender PPPP
Under 2.0
2.0 − 3.0
Over 3.0
0.05
0.10
0.25
0.30
0.10
0.20
Male
Female
One student is selected at random. Let we define two random variables X and Y as
⎧
⎪
⎨0, student has GPA under 2.0
0, student is a male
X=
, Y = 1, student has GPA between 2.0 and 3.0
⎪
1, student is a female
⎩
2, student has GPA over 3.0
Find the conditional variance of Y given that X = 1.
Answer. Joint probability distribution of X and Y is
H
HH
y
H
x
HH
0
1
h(y)
0
1
2
g(x)
0.05 0.25 0.10
0.10 0.30 0.20
0.15 0.55 0.30
0.4
0.6
1
The conditional distribution of Y given X = 1 is w(y|1) =
y = 0, 1, 2. That is, that
w(0|1) =
f (1, y)
f (1, y)
=
for
g(1)
0.60
1
f (1, 1) 3
f (1, 2)
2
f (1, 0)
= , w(1|1) =
= , w(2|1) =
= .
g(1)
6
g(1)
6
g(1)
6
On the other hand
2
E[Y |1] =
2
y 2 w(y|1) = 02 ·
y=0
E[Y |1] =
2
yw(y|1) = 0 ·
y=0
11
1
3
2
+ 12 · + 22 · = ,
6
6
6
6
3
2
7
1
+1· +2· = .
6
6
6
6
Since the conditional variance of Y given that X = 1 is
σY2 |1 = E[Y 2 |1] − (E[Y |1])2
we find that
σY2 |1
MCB1007 - Int. to Prob. and Statistics
11
=
−
6
2
2
17
7
= .
6
36
Second Midterm
17 points
Flip a fair coin 3 times. Let X be the number of heads in the first 2 flips and let Y be
the number of heads on the last 2 flips. Compute Cov(X, Y ).
Answer. With 3 tosses there are 8 outcomes in the sample space S = {HHH, HHT, HT H,
HT T, T HH, T HT, T T H, T T T }. So,
1
1
f (0, 0) = P (T T T ) = , f (0, 1) = P (T T H) = , f (0, 2) = 0,
8
8
2
1
1
f (1, 0) = P (HT T ) = , f (1, 1) = P (T HT ) = P (HT H) = , f (1, 2) = P (T HH) = ,
8
8
8
1
1
f (2, 0) = 0, f (2, 1) = P (HHT ) = , f (2, 2) = P (HHH) = .
8
8
Thus we have the following table
HH
y
H
x HHH
0
1
2
h(y)
0
1/8
1/8
0
2/8
1
2
1/8 0
2/8 1/8
1/8 1/8
4/8 2/8
g(x)
2/8
4/8
2/8
1
On the other hand, since
μX = E(X) =
2
xg(x) = 0 · g(0) + 1 · g(1) + 2 · g(2) = 0 ·
4
2
2
+ 1 + 2 · = 1,
8
8
8
yh(y) = 0 · h(0) + 1 · h(1) + 2 · h(2) = 0 ·
4
2
2
+ 1 + 2 · = 1,
8
8
8
x=0
μY = E(Y ) =
2
y=0
μXY
= E(XY ) =
2 2
xyf (x, y) = 1 · 1 ·
x=0 y=0
1
1
1
10
2
+1·2· +2·1· +2·2· =
8
8
8
8
8
then we have
Cov(X, Y ) = σXY = E(XY ) − E(X)E(Y ) =
MCB1007 - Int. to Prob. and Statistics
3
10
2
−1·1= .
8
8
Second Midterm
10 + 7 points
It is known that 3% of the circuit boards from a production line are defective. If a
random sample of 120 circuit boards is taken from this production line
(a) determine the probability that the sample contains exactly 2 defective boards.
Answer. Substituting x = 2, n = 120, and θ = 0.03 into the formula for the binomial
distribution, we obtain
120!
120
× 0.032 × 0.97118 = 0.1766059636.
b(2; 120, 0.03) =
0.032 (1 − 0.03)118 =
2! · 118!
2
(b) use the Poisson approximation to estimate the probability that the sample contains
2 defective boards and compare this result with (a).
Answer. Substituting x = 2, λ = nθ = 120 × 0.03 =
Poisson distribution, we get
p(2; 3.6) =
18
5
= 3.6 into the formula for
3.62 × e−3.6
= 0.1770577215.
2!
Since n = 120 ≥ 10 and nθ = 3.6 < 10 we see that the approximation obtained in (b)
0.1770577215 is very close the exact probability 0.1766059636.
10 + 5 points
A manufacturer received an order of 250 computer chips. Unfortunately, 12 of the chips
are defective. To test the shipment, the quality-control engineer randomly selects 20
chips from the box of 250 and tests them.
(a) What is the probability of obtaining 3 defective chips? (Leave your answer in terms
of factorials)
Answer. Substituting x = 3, n = 20, N = 250, and M = 12 into the formula for the
hypergeometric distribution, we get
12!
238!
12
238
×
= 3! · 9! 17! · 221! .
h(3; 20, 250, 12) = 3
25017
250!
20
20! · 230!
(b) How many defective chips would you expect to select?
Answer. The mean of the hypergeometric distribution is μ =
number of defective chips is
μ=
MCB1007 - Int. to Prob. and Statistics
nM
.
N
So the expected
20 · 12
24
nM
=
=
= 0.96.
N
250
25
4
Second Midterm
10 + 7 points
Suppose that during practice, a basketball player can make a free throw 80% of the
time. Furthermore, assume that a sequence of free-throw shooting can be thought of
as independent trials. What is the probability that the basketball player makes his (a)
tenth free throw on his twelfth shot, (b) first free throw on his sixth shot?
Answer.
(a) Substituting x = 12, k = 10, and θ = 0.8 into the formula for the negative binomial
distribution, we get
11
11! 10 2
∗
b (12; 10, 0.80) =
0.8 0.2 ≈ 0.236.
0.810 (1 − 0.8)2 =
9!2!
9
(b) Substituting x = 6, and θ = 0.8 into the formula for the geometric distribution, we
get
4
= 2.56 × 10−4.
g(6; 0.8) = 0.8(1 − 0.8)6−1 = 0.8 × 0.25 =
15625
10 + 7 points
(a) Find the moment generating function of the random variable whose probability
density is given by
2e−2x , for x > 0,
f (x) =
0,
elsewhere.
Answer. The moment generating function of the random variable X is
∞
∞
c
tX
tx −2x
x(t−2)
MX (t) = E[e ] =
e 2e dx = 2
e
dx = 2 lim
ex(t−2) dx
c→∞
0
0
0
c
2
2
2
lim ex(t−2) =
lim (ec(t−2) − e0 ) =
(0 − 1)
=
t − 2 c→∞
t − 2 c→∞
t−2
0
2
for t < 2
=
2−t
(b) Use the moment generating function to determine μ1 and μ2 for f (x) given in (a).
dr MX (t) for r = 1, 2 we obtain
Answer. Since μr =
dtr t=0
dMX (t) d
1
−1 = (2(2 − t) ) = (2(2 − t)−2 )t=0 = ,
μ1 =
dt
dt
2
t=0
t=0
d2 MX (t) d
1
−2 −3 μ2 =
(2(2
−
t)
=
)
=
(4(2
−
t)
)
= .
t=0
2
dt
dt
2
t=0
t=0
MCB1007 - Int. to Prob. and Statistics
5
Second Midterm
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