s t a n b u l K ü l t ü r U n i v e r s i t y Faculty of Engineering
IE 3101 - IE310 - MAT003 - MBT5001
SECOND MIDTERM
Fall 2012-2013
SOLUTIONS
Directions You have 90 minutes to complete the exam. Please do not leave the examination room in the rst 30 minutes of the exam. There are six questions, of varying credit (100 points total). Indicate clearly your nal answer to each question. You are allowed to use a calculator. During the exam, please turn o 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. The answer key to this exam will be posted on Department of Mathematics and Computer Science board after the exam.
Good luck! Emel Yavuz Duman, PhD.
Question 1. Question 4.
Question 2. Question 5.
Question 3. Question 6.
TOTAL
stanbul Kültür University
Que ion 1. 20points
The joint probability distribution of X and Y is given by f (x, y) = x + y
21 , x = 1, 2, 3, and y = 1, 2.
Find the conditional variance of X given Y = 1.
Answer.
HHX HHHH
Y 1 2 g(x)
1 2/21 3/21 5/21
2 3/21 4/21 7/21
3 4/21 5/21 9/21
h(y) 9/21 12/21 1
f (x|1) = f (x,1)h(1) = f (x,1)9/21 = 219f (x, 1)⇒ f(1|1) = 29, f (2|1) = 39, f (3|1) = 49 E[X|1] = µX|1 =∑3
x=1xf (x|1) = 1·f(1|1)+2·f(2|1)+3·f(3|1) = 1·29+2·39+3·49 = 209 E[X2|1] =∑3
x=1x2f (x|1) = 12·f(1|1)+22·f(2|1)+32·f(3|1) = 12·92+22·39+32·49 = 509 σX|1 = E[X2|1] − (E[X|1])2 = 509 −(20
9
)2
= 5081.
Que ion 2. 20points
If X is the number of heads and Y is the number of heads minus the number of tails obtained in three ips of a balanced coin nd the covariance of X and Y .
Answer. There are 8 possible outcomes:
HHH: X = 3, Y = 3; HHT: X = 2, Y = 1; HT H: X = 2, Y = 1;
HT T: X = 1, Y =−1; T T T : X = 0, Y =−3; T T H: X = 1, Y = −1;
T HT: X = 1, Y =−1; T HH: X = 2, Y = +1.
The the joint probability distribution is
HHX HHHH
Y −3 −1 1 3 g(x)
0 1/8 0 0 0 1/8
1 0 3/8 0 0 3/8
2 0 0 3/8 0 3/8
3 0 0 0 1/8 1/8
h(y) 1/8 3/8 3/8 1/8 1
Since
E(XY ) = µ′1,1 =∑
x
∑
yxyf (x, y) = 0(−3)18 + 1(−1)38 + 2(1)38 + 3(3)18 = 128, E(X) = µx =∑
xxg(x) = 0· 18 + 1· 38 + 2· 38 + 3· 18 = 128, E(Y ) = µy =∑
yyh(y) = (−3) ·18 + (−1) ·38 + 1·38 + 3· 18 = 0, thus
σXY = E(XY )− E(X)E(Y ) = µ′1,1− µxµy = 12 8 − 12
8 · 0 = 12 8 = 3
2.
IE 3101 - IE310 - MAT003 - MBT5001 2
stanbul Kültür University
Que ion 3. 15points
The number of marriage licenses issued in a certain city during the month of June may be looked upon as a random variable with µ = 124 and σ = 7.5. According to Chebyshev's theorem, with what part probability can we assert that between 64 and 184 marriage licenses will be issued there during the month of June?
Answer. Since 64 < X < 184, then
64− 124 < X − 124 < 184 − 124 ⇒ −60 < X − 124 < 60 ⇒ |X − 124| < 60.
Therefore,
P (|X − 124| < 60) = 1 − P (|X − 124|{z}
µ
| ≥ 60|{z}
ε
)≥ 1 −σ2
ε2 = 1− 7.52
602 = 1− 1 64 = 63
64. So, the probability of between 64 and 184 marriage licenses will be issued there during the month of June is at least 6364.
Que ion 4. 15points
The probability density function of a random variable X is given by
f (x) =
1/3, 0 < x ≤ 1, 2/3, 1 < x ≤ 2, 0, otherwise.
Find the corresponding distribution function of X and use it to calculate P (0.5 < X <
1.7).
Answer. Since
If x < 0 then F (x) =∫x
−∞f (t)dt = 0
If 0 < x ≤ 1 then F (x) =∫x
−∞f (t)dt =∫x 0
1
3dt = 3t x
0 = x3
If 1 < x ≤ 2 then F (x) =∫x
−∞f (t)dt =∫1 0
1
3dt +∫x 1
2
3dt = 3t 1
0+ 2t3 x
1
= 13 + 23(x− 1) = 2x3−1
If x > 2 then F (x) =∫x
−∞f (t)dt = 1 therefore
F (x) =
0, x < 0,
x
3, 0 < x ≤ 1,
2x−1
3 , 1 < x ≤ 2, 1, x > 2, and
P (0.5 < X < 1.7) = F (1.7)− F (0.5) = 2(1.7)− 1
3 − 0.5
3 = 19
30 = 0.6¯3.
IE 3101 - IE310 - MAT003 - MBT5001 3
stanbul Kültür University
Que ion 5. 7 + 8 points
Let X be the number on a fair six sided die roll.
(a) Find the moment generating function MX(t).
(b) Use the moment generating function of X to determine µ′1 and µ′2. Answer.
(a) MX(t) = E(etX) =∑6
x=1etxf (x) = 16(et+ e2t+ e3t+ e4t+ e5t+ e6t). (b) µ′1 = µ = MX′ (t)|t=0= 16(et+ 2e2t+ 3e3t+ 4e4t+ 5e5t+ 6e6t)
t=0
= 16(1 + 2 + 3 + 4 + 5 + 6) = 216 = 72, µ′2 = MX′′(t)|t=0= 16(et+ 4e2t+ 9e3t+ 16e4t+ 25e5t+ 36e6t)
t=0
= 16(1 + 4 + 9 + 16 + 25 + 36) = 916.
Que ion 6. 15points
The distribution function of the continuous random variable X is given by
F (x) =
0 for x < 0, x/10, for 0 ≤ x ≤ 10 1, for x > 10.
Find the standard deviation of X.
Answer. We rst need to nd the probability density of X:
f (x) = d
dxF (x) =
d
dx0 = 0 for x < 0,
d dx
(x
10
)= 101 for 0 < x < 10,
d
dx1 = 0 for x > 10.
⇒ f(x) = {1
10 for 0 < x < 10, 0, elsewhere.
Since
E(X2) =
∫ ∞
−∞
x2f (x)dx =
∫ 10 0
x2· 1
10dx = 1 10
x3 3
10
0
= 103
10· 3 = 100 3 , and
E(X) =
∫ ∞
−∞
xf (x)dx =
∫ 10
0
x· 1
10dx = 1 10
x2 2
10
0
= 102
10· 2 = 10 2 , then
σ = √
σ2 =√
E(X2)− (E(X))2 =
√ 100
3 − (10
2 )2
= 5√ 3
3 ≈ 2.8868.
IE 3101 - IE310 - MAT003 - MBT5001 4
