İstanbul Kültür University
Faculty of Engineering
MCB1007
Introduction to Probability and Statistics
Final Exam
Fall 2015-2016
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 may
use a calculator. During the exam, please turn off your cell phone(s). You can’t
use any book or lecture note. You have one page “cheat-sheet”. 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.
M. Fatih Uçar, PhD.
Arzu Yemişçi, PhD.
Question 1.
Question 2.
Question 3.
Question 4.
Question 5.
Question 6.
TOTAL
Question 1.
7 + 7 points
A fair die is tossed and its outcome is denoted by X. After that, X independent fair
coins are tossed together and the number of heads obtained by Y .
(a) P (Y = 4) =?
Answer.
4 1 4 2 4
0
5
1
6
1
4
1
1
1
1
1
29
+
+
=
4
4
6
4
2
2
2
2
2
2
384
(b) P (X = 5|Y = 4) =?
Answer.
P (Y = 4|X = 5)P (X = 5)
P (X = 5, Y = 4)
=
P (Y = 4)
P (Y = 4)
5 1 4 1 1
1
× 4 2
10
2
= 6
=
29
29
384
P (X = 5|Y = 4) =
Question 2.
6 + 6 + 6 points
A study of the annual growth of certain cactus showed that, 64 of them, selected at
random in a desert region, grew on the average 52.80 mm with a standart deviation of
4.5 mm.
(a) Construct a 99% confidence interval for the true average annual growth of the
cactus.
Answer.
4.5
52.8 ∓ 2.575 √ = 52.8 ∓ 1.4484375
64
(b) What should be the minimum sample size if the error does not exceed 1.2?
Answer.
4.5
1.2 ≥ 2.575 √ ⇒ n ≥ 93.2 ⇒ n = 94
n
(c) With what degree of confidence could we say that the mean of annual growth of all
64 cactus is (51.5, 54.1)?
Answer.
4.5
(51.5, 54.1) = 52.8 ∓ 1.3 ⇒ 1.3 = zα/2 √ ⇒ zα/2 = 2.31
64
⇒ confidence interval = 2 × 0.4896 = 0.9792 ≈ 98%
MCB1007 - Int. to Prob. and Statistics
2
Final Exam
Question 3.
7 + 10 points
An egg carton contains 10 eggs, 2 of which are cracked. A sample of 5 eggs is randomly
selected from the carton.
(a) Find the probability distribution for X, the number of cracked eggs in the sample.
Answer. Hypergeometric Distribution with parameters n = 5, M = 2, N = 10
210−2
10
10!
x
5−x
h(x; 5, 2, 10) = 10 , x = 0, 1, 2,
=
= 15
5
5!5!
5
28
P (X = 0) =
0
5
15
2
= , P (X = 1) =
9
28
1
4
15
5
= , P (X = 2) =
9
28
2
3
15
=
2
9
(b) What are the mean and variance of X?
Answer.
nM
5×2
μ=
=
=1
N
10
5 × 2(10 − 2)(10 − 5)
4
nM(N − M)(N − n)
=
= = 0.4̄
σ2 =
2
2
N (N − 1)
10 (10 − 1)
9
Question 4.
8 + 8 + 8 points
A fair coin is tossed 5 times. Find the probability of getting 2, 3 or 4 heads by using
(a) Binomial distribution.
Answer. Paremeters x = 2, 3, 4, n = 5, θ = 0.5
P (2 ≤ X ≤ 4) = b(2; 5, 0.5) + b(3; 5, 0.5) + b(4; 5, 0.5)
5
5
5
2
3
3
2
=
0.5 (1 − 0.5) +
0.5 (1 − 0.5) +
0.54 (1 − 0.5)1
2
3
4
5
5
25
5
+
+
=
= 0.78125
=
16 16 32
32
(b) The normal approximation to the
binomial distribution.
√
Answer. μ = nθ = 5 × 0.5 = 2.5, σ = nθ(1 − θ) = 5 × 0.5(1 − 0.5) = 5/2
4.5 − 2.5
1.5 − 2.5
√
<Z< √
P (2 ≤ X ≤ 4) = P (1.5 < X < 4.5) ≈ P
5/2
5/2
= P (−0.89 < Z < 1.79) = 0.3133 + 0.4633 = 0.77660
(c) The Poisson approximation to the binomial distribution.
Answer. λ = nθ = 5 × 0.5 = 2.5
P (2 ≤ X ≤ 4) = p(2; 2.5) + b(3; 2.5) + b(4; 2.5)
2.52 e−2.5 2.53 e−2.5 2.54 e−2.5
+
+
=
2!
3!
4!
= 0.60388
MCB1007 - Int. to Prob. and Statistics
3
Final Exam
Question 5.
7 + 8 points
From the set of numbers {3, 5, 7}, a random sample of size 2 will be selected with
replacement.
(a) Determine the sampling distribution of the mean.
Answer.
(x1 , x2 )
x̄
(3,3) (3,5) (3,7) (5,3) (5,5)
3
4
5
4
5
x̄
f (x̄)
(5,7) (7,3) (7,5) (7,7)
6
5
6
7
3
4
5
6
7
1/9 2/9 3/9 2/9 1/9
(b) Calculate the mean and standard deviation for the population distribution and for
the distribution of X̄.
Answer.
distribution:
For the polulation
3+5+7
μX = x xf (x) = 3 = 5.
(3−5)2 +(5−5)2 +(7−5)2
2
= 83 = 1.633.
σX =
x (x − μX ) f (x) =
3
For the sampling distribution of the mean:
μX̄ = μX = 5.
√
σX̄ =
σX
√
n
=
8
√3
2
=
√2
3
= 1.155.
Question 6.
12 points
Estimate the mean, median and modal class of the following data:
Number of Orders
10 − 12
13 − 15
16 − 18
19 − 21
Frequency
4
12
20
14
Answer.
Class Interval
10 − 12
13 − 15
16 − 18
19 − 21
Mid-Point
11
14
17
20
Totals
Frequency
4
12
20
14
50
Midpoint × Frequency
44
168
340
280
832
832
= 16.64,
50
18.5−15.5
× 9.5 = 677
20
40
Mean= x̄ =
Median= 15.5 +
= 16.925,
Modal Class= 16 − 18.
MCB1007 - Int. to Prob. and Statistics
4
Final Exam