Download MCS 224 - Introduction to Probability and Statistics FIRST MIDTERM

Çankaya University
Department of Mathematics and Computer Sciences
2010-2011 Spring Semester
MCS 224 - Introduction to Probability and Statistics
FIRST MIDTERM
1) The following data represents the grades of a group of students:
10,17,22,34,44,44,48,49,55,59,60,63,70,81,83,89,90
a) Find the mean
b) Find the median
c) Find the standard deviation
d) How many standard deviation is the top grade above the mean?
2) Gören, Bengisu, Tuncay, Nurşah, Bahadır and Merve are sitting in a row.
a) In how many different ways can they sit if boys and girls alternate?
b) In how many different ways can they sit if Merve and Bahadır are to sit next to each
other but Gören and Nurşah are not to sit next to each other?
3) There are three departments in a Faculty: Math, English and Translation. The Math department has 3 times as many students as the Translation, and 2 times as many students
as the English department. 40% of all students in the Math department, 80% of English
department and 90% of the Translation department are girls. We choose a student from
the Faculty at random. Given that the student is a girl, what is the probability that she’s
a Math student?
4) We pick 3 cards from a deck randomly. We win if they are all the same color (red or black)
or if each of the numbers are less than or equal to 3. Find the probability of winning.
5) X is a continuous random variable with probability density function
f (x) =
k(1 − x)2
0
a) Find k
b) Find the P (0.75 < X).
if
if
06x61
x < 0 or x > 1
Answers
1) a) x̄ = 54
b) x̃ = 55
c) σ = 24.3413
d) 1.4789
2) a) 2 · 3! · 3! = 72
b) 2 · 5! − 2 · 2 · 4! = 144
n
3) Suppose there are n students in Math Dept. Then, there are students in English and
2
11n
n
students in translation departments. the total is
.
3
6
Let M mean the student is from Math Dept. and let G mean the student is a girl.
P (M ∩ G) =
0.4 n
2.4
=
11
11
n
6
n
n
+ 0.9
2
3 = 6.6
11
11
n
6
0.4 n + 0.8
P (G) =
P (M | G) =
4
2.4/11
=
6.6/11
11
4) P (A ∪ B) = P (A) + P (B) − P (A ∩ B)
26
12
6
2
+
−2
3
3
3
=
52
3
= 0.2434
Z
∞
5) a)
f (x) dx = 1
−∞
Z
k
1
(1 − x)2 dx = 1
0
1
(1 − x)3 −k
= 1
3
0
k
= 1
3
⇒
k=3
Z
1
b) P (0.75 < x) =
3(1 − x)2 dx
0.75
= 0.253
=
1
64
Çankaya University
Department of Mathematics and Computer Sciences
2010-2011 Spring Semester
MCS 224 - Introduction to Probability and Statistics
SECOND MIDTERM

 6 − x − y 0 < x < 2, 2 < y < 4
8
1) Given the joint density function f (x) =

0
elsewhere
a) Find marginal distribution of X.
b) Find marginal distribution of Y .
c) Are they independent?

 8
x3
2) Let X have the density function f (x) =

0
if 2 < x,
if x < 2
a) Find µ b) Find E (X − 2)2
1
. A student has 8
20
examinations this month on separate days. What is the probability that the student will
be ill
a) on none of the exams?
b) on exactly two exams?
b) on more than two exams?
3) The probability that a student is ill on the day of an exam is
4) We examine a shipment of 200 laptops for defects by choosing a sample of 20 at random.
We reject the shipment if there are 2 or more defective ones in the sample. Assume there
are 10 defective ones among the 200. What is the probability that we will
a) reject the shipment?
b) accept the shipment?
5) On average there are 7 snowstorms per year in a certain area. What is the probability
that there are
a) Exactly 7,
b) More than 7,
c) Less than 7
snowstorms in a given year?
6) MCS 224 grades have a normal distribution with (hopefully) µ = 55 and σ = 12. Emre
hoca wants to fail only the bottom 20% of the class. What should the passing grade be?
Answers
1) g(x) =
=
=
4
6−x−y
dy
8
2
4
6x xy y 2 −
− 8
8
16 2
Z
3−x
4
2
6−x−y
dx
8
0
2
6x x2 yx −
− 8
16
8 0
Z
h(y) =
=
5−y
4
=
g(x) h(y) 6= f (x, y)
Z
⇒
No, they are NOT independent.
∞
8
x dx
x3
2
∞
8 = − x 2
2) a) µ =
= 4
2
Z
∞
b) E[(x − 2) ] =
(x − 2)2
2
8
dx
x3
= ∞
3) a) n = 0 :
8
1
1−
= 0.6634
20
b) n = 2 :
2 6
8
1
19
= 0.0515
2
20
20
c) n = 1 :
7
8
1
19
= 0.2793
1
20
20
⇒
n>2:
1 − 0.6634 − 0.2793 − 0.0515 = 0.0058
 
190
10 190
 20
1
19 

4) a) 1 − 
+
 200

200
20
20
= 0.2628
b) 1 − 0.2628 = 0.7372
5) Using the table, we obtain:
a) p(7, 7) − p(6, 7) = 0.149
or, alternatively,
e−7 77
= 0.149
7!
b) 1 − p(7, 7) = 0.4013
c) p(6, 7) = 0.4497
6) Using the table, we see that %80 corresponds to 0.84 therefore %20 is −0.84.
x−µ
σ
x − 55
−0.84 =
12
z=
⇒
x = 44.92
Çankaya University
Department of Mathematics and Computer Sciences
2010-2011 Spring Semester
MCS 224 - Introduction to Probability and Statistics
FINAL

 3 (x2 + y 2 ) 0 6 x 6 1, 0 6 y 6 1
2
1) Given the joint density function f (x) =

0
elsewhere
Compute the probability that 0 6 x 6 1/2, 0 6 y 6 1/2
2) The average number of hurricanes per year is 12. What is the probability that on 3 of the
next four years, there are 8 or fewer hurricanes?
3) Suppose that airplane engines operate independently and fail with probability equal to
0.4. A plane makes safe flight if at least half of its engines run. Determine whether a
4-engine plane or 2-engine plane has higher probability of successful flight.
4) The average time for a trip is 30 minutes with a standard deviation of 4 minutes. Assume
distribution is normal.
a) Find the probability that a trip takes longer than 40 minutes.
b) The longest 10% of the trips take at least how many minutes?
5) The lifetime in weeks of a certain type of transistor √
is known to follow a gamma distribution with mean 10 weeks and standard deviation 50 weeks. What is the probability
that the transistor will last at most 50 weeks?
6) A certain cable is manufactured with mean strength of 78.3 kilograms and standard
deviation 5.6 kilograms. Find the variance of sample mean if sample size is
a) 64
b) 196
Answers
Z
1/2
Z
1/2
3 2
(x + y 2 ) dy dx
2
0
0
1/2
Z 1/2 3 3
y
x2 y +
=
dx
2
3 0
0
Z 1/2 2
3 x
1
=
+
dx
2 2
24
0
1/2
3 3 x3
+
x
=
4 3
48 0
1) P =
=
1
16
2) Using Poisson Distribution Table, probability for one year: P (8, 12) = 0.155.
For 3 of next 4 years:
4
0.1553 (1 − 0.155) = 0.01259
3
3) 2−engine success rate:
2
2
0
2
0.4 0.6 +
0.41 0.61 = 0.84
0
1
4−engine success rate:
4
4
4
0
4
1
3
0.4 0.6 +
0.4 0.6 +
0.42 0.62 = 0.8208
0
1
2
2−engine plane has higher probability of success.
40 − 30
= 2.5
4
Using normal distribution table, P (z > 2.5) = 1 − 0.9938 = 0.0062
4) a) z =
b) P (z > 1.28) = 1 − 0.8997 = 0.1003
⇒
z = 1.28
x = 30 + 1.28 · 4 = 35.12
5) µ = √10 = αβ
√
σ =
50 =
αβ
⇒
α = 2
β = 5
( 1
xe−x/5
f (x) =
25
0
if
x>0
if
x60
Using integration by parts, we can compute the probability as:
Z
p =
50
f (x) dx
0
=
50
1 −x/5
e
(−25 − 5x)
25
0
= 1 − 11 e−10
= 0.9995
σ
6) a) σx̄ = √
n
5.6
= √
64
= 0.7
σ
b) σx̄ = √
n
5.6
= √
196
= 0.4
Çankaya University
Department of Mathematics and Computer Science
MCS 224 - Introduction to Probability and Statistics
24.03.2011
Name-Surname:
ID Number:
CLASSWORK 1
1) Let (X, Y ) have the joint probability function specified in the table.
a) Compute P (X = Y ) b) Compute P (Y > 2) c) Compute P (X > 1, Y 6 3)
2
0 0.04
X 1 0.08
2 0.08
Y
3
0.08
0.08
0.04
4
0.04
0.20
0.08
5
0.08
0.08
0.12
Answer: 0.08, 0.8, 0.28
2) Let (X, Y ) have joint probability function given below. Compute P (X + Y 6 1)
f (x, y) =
6
(x
7
f (x, y) = 0
3
Answer:
14
+ y)2 ,
0 6 x 6 1, 0 6 y 6 1
elsewhere
Çankaya University
Department of Mathematics and Computer Science
MCS 224 - Introduction to Probability and Statistics
31.03.2011
Name-Surname:
ID Number:
CLASSWORK 2
32
(x + 4)3
1) Let X has the density function f (x) =

0
Find the expected value of X.


x>0
elsewhere
Answer: 4
2) The random variable X represents the number of defective items in a lot of 1000. It has
the probability distribution given in the table. Find the variance of X.
0
1
2
3
4
x
f (x) 0.02 0.08 0.3 0.45 0.15
Answer: 0.8131
Çankaya University
Department of Mathematics and Computer Science
MCS 224 - Introduction to Probability and Statistics
07.04.2011
Name-Surname:
ID Number:
CLASSWORK 3
If the joint density function of X and Y is given by:

 2 (x + 2y) 0 < x < 1, 1 < y < 2
7
f (x) =

0
elsewhere
find the expected value of g(X, Y ) =
Answer: 0.122
X2
Y3
Çankaya University
Department of Mathematics and Computer Science
MCS 224 - Introduction to Probability and Statistics
14.04.2011
Name-Surname:
ID Number:
CLASSWORK 4
We consider purchasing a welding machine. An efficient machine is successful at 99%
of the jobs. An inefficient machine is successful at 95% of the jobs. We perform 100 welds
with each machine and purchase the ones that are successful at 97 or more welds.
a) What is the probability that we will reject an efficient machine?
b) What is the probability that we will accept an inefficient machine?
Answer: 0.01837, 0.2578
Çankaya University
Department of Mathematics and Computer Science
MCS 224 - Introduction to Probability and Statistics
21.04.2011
Name-Surname:
ID Number:
CLASSWORK 5
1) Two chess players A and B will meet in 9 matches. The one to win 5 matches wins
the tournament. The probability that A will win in a single match is 0.7. What is the
probability that B wins the tournament in 9 matches?
Answer: 0.0408
2) On average, there are 6 misprints per a report. What is the probability that a given report
a) has no misprints?
b) has more than 5 misprints?
Answer: 0.5543
Çankaya University
Department of Mathematics and Computer Science
MCS 224 - Introduction to Probability and Statistics
05.05.2011
Name-Surname:
ID Number:
CLASSWORK 6
MCS 224 grades have a normal distribution with µ = 53 and σ = 15. Emre hoca wants
to give CC to middle 30% of the class. What should the limits for CC be?
Answer: 47.15, 58.85
Prepared by: Dr. Emre Sermutlu
(2011)