İ 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
MCB1007
Introduction to Probability and Statistics Final Exam
Fall 2014-2015
Number:
Name:
Department:
Section:
– 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. The answer key to this exam will be posted on Department of Mathematics and Computer Science board after the exam.
Good luck!
M. Fatih Uçar, PhD.
Emel Yavuz Duman, PhD.
Arzu Yemişçi, PhD.
Question 1. Question 4.
Question 2. Question 5.
Question 3. Question 6.
TOTAL
8+8 points A fair coin is tossed 120 times. Using normal approximation, find the probability of obtaining
(a) between 48 and 72 heads.
Answer. Since this is a normal approximation to the binomial distribution we have μ = nθ = 120 × 12 = 60 and σ =
nθ(1 − θ) =
120 × 12 × 12 = 5.48. Thus
P (48 ≤ X ≤ 72) = P (47.5 < X < 72.5)
≈ P
47.5 − 60
5.48 < Z < 72.5 − 60 5.48
= P (−2.28 < Z < 2.28) = 2 × 0.4887
= 0.9774.
(b) at least 75 heads.
Answer. Using the same arguments in (a) we obtain P (X ≥ 75) = P (X > 74.5)
≈ P
Z > 74.5 − 60 5.48
= P (Z > 2.65) = 0.5 − 0.4960 = 0.004.
5+5+6 points
There are several types of trees in a forest. In a random sample of 240 trees, they have average height 35 mt with a standard deviation 7 mt. For the mean of the population sampled,
(a) Construct a 95% confidence interval.
Answer. Since 1 − α = 0.95 we have zα/2 = 1.96, and n = 240, ¯x = 35 and σ = 7.
Substituting these values in to the formula ¯x − zα/2√σ
n < μ < ¯x + zα/2√σ
n we obtain 35 − 1.96√2407 < μ < 35 + 1.96√2407 , or 35 − 0.8856 < μ < 35 + 0.8856, thus
34.1144 < μ < 35.8856.
(b) What should be the minimum sample size, if error does not exceed 0.97?
Answer. Since the error E is less than zα/2√σ
n with the given degree of the confidence we have
1.96 × √7
n ≤ 0.97 ⇒ n ≥ 200.06.
So, the minimum sample size should be201.
(c) With what degree of confidence could we say that the average height of trees are 35 ∓ 0.794?
Answer.Since zα/2√σn = 0.794, we have
zα/2× √7
240 = 0.794 ⇒ zα/2 = 1.76.
The corresponding area for this zα/2 = 1.76 value is 0.4608. So, the degree of confidence should be2 × 0.4608 = 0.9216.
MCB1007 - Int. to Prob. and Statistics 2 Final Exam
14 points The following table shows the Mathematics exam scores of 57 student. Find the mean and median scores of the exam?
Score 21 ≤ x < 30 31 ≤ x < 40 41 ≤ x < 50 51 ≤ x < 60 61 ≤ x < 70 71 ≤ x < 80
Num. of Stud. 2 11 17 20 5 2
Answer. The mean number of the grouped data is given by
¯x = 25.5 × 2 + 35.5 × 11 + 45.5 × 17 + 55.5 × 20 + 65.5 × 5 + 75.5 × 2
57 = 49.18
Median is the 29th value. In this case it lies in the41 ≤ x < 50 class interval. The 16th value in the interval is needed. Using the continuity correction we obtain that median is equal to
40.5 + 50.5 − 40.5
17 × 16 = 49.91
5+5+8 points
A continuous random variable X has the following probability density function f (x) =
ke−|x2|, 0 < x < ∞, 0, elsewhere.
(a) Find k and μ.
Answer.
1 =
∞
−∞f (x)dx =
∞
0 ke−x/2dx = k lim
b→∞−2e−x/2b
0
= −2k lim
b→∞(e−b/2− e0) = 2k =⇒ k = 1/2.
μ = E[X] =
∞
−∞xf (x)dx =
∞
0
1
2xe−x/2dx = lim
b→∞−xe−x/2b
0
=0 by L’Hôpital
+
∞
0 e−x/2dx
2
= 2
(b) Find a lower bound for P (|x − μ| < 4) using Chebyshev’s theorem.
Answer. We need to find σ2 value first which is equal to E[X2] − (E[X])2: E[X2] =
∞
−∞x2f (x)dx =
∞
0
1
2x2e−x/2dx = lim
b→∞−x2e−x/2b
0
=0 by L’Hôpital
+
∞
0 2xe−x/2dx = 4μ = 8
So
σ2 = E[X2] − (E[X])2 = 8 − 4 = 4.
Since
P (|x − μ| < 4) = P (|x − 2| < 4) = 1 − P (|x − 2| ≥ 4) we obtain that
P
|x − 2| ≥ 4
ε
≤ σ2 ε2 = 4
16 = 1
4 ⇒ 1 − P (|x − 2| ≥ 4) ≥ 1 − 1 4 = 3
4
MCB1007 - Int. to Prob. and Statistics 3 Final Exam
10+10 points (a) Eight students will select a topic for graduate thesis among topics A, B, C and D.
What is the probability that at most 3 students will select topic A?
Answer. Let X be the number of students who selected topic A. Then P (X ≤ 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
=
8
0
38 48 +
8
1
37 48 +
8
2
36 48 +
8
3
35
48 = 0.8862.
(b) Five pawns are placed at random on a 8 × 8 chessboard, no more than one pawn to a cell. What is the probability that no more than one pawn in the same row or column?
Answer. The total number of outcomes of the sample space is 64
5
. Noting that you can choose five different rows in 8
5
ways, it follows that the number of choices of five different cells for which no row or column contains more than one dot is 8
5
× 8 × 7 × 6 × 5 × 4. This gives the result
8
5
× 8 × 7 × 6 × 5 × 464
5
= 0.0494.
16 points
Suppose that a pair fair dice is rolled together. Let X denotes the number of dice that come up prime number and Y denotes the number of dice that come up bigger than or equal to 5. Find the correlation coefficient of X and Y ?
Hint. In Statistics, correlation coefficient, denoted by ρ(x, y) is a measure of the linear relationship between two random variables X and Y , giving a value −1 < ρ < 1, where 1 is total positive correlation, 0 is no correlation, and −1 is total negative correlation.
It is defined by ρ(x, y) = √ Cov(X,Y )
V ar(X)V ar(Y ).
Answer. All possible pairs and related probabilities are listed below:
1 2 3 4 5 6
1 (0,0) (1,0) (1,0) (0,0) (1,1) (0,1) 2 (1,0) (2,0) (2,0) (1,0) (2,1) (1,1) 3 (1,0) (2,0) (2,0) (1,0) (2,1) (1,1) 4 (0,0) (1,0) (1,0) (0,0) (1,1) (0,1) 5 (1,1) (2,1) (2,1) (1,1) (2,2) (1,2) 6 (0,1) (1,1) (1,1) (0,1) (1,2) (0,2)
HHy HHHH
x 0 1 2 h(y)
0 4/36 8/36 4/36 16/36 1 4/36 8/36 4/36 16/36 2 1/36 2/36 1/36 4/36 g(x) 9/36 18/36 9/36 1
Since
Cov(X, Y ) = E(XY ) − E(X)E(Y ) =
1 · 1 · 8
36+ 1 · 2 · 2
36+ 2 · 1 · 4
36 + 2 · 2 · 1 36
−
1 ·18
36+ 2 · 9 36
1 ·16
36 + 2 · 4 36
= 2
3− 1 · 2 3 = 0 we deduce that
ρ(x, y) = Cov(X, Y )
V ar(X)V ar(Y ) = 0
V ar(X)V ar(Y ) = 0.
MCB1007 - Int. to Prob. and Statistics 4 Final Exam
