Question 1. Question 4.

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 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! Emel Yavuz Duman, PhD.

Question 1. Question 4.

Question 2. Question 5.

Question 3. 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.

PPPGenderPPPPPP

GPA Under 2.0 2.0 − 3.0 Over 3.0

Male 0.05 0.25 0.10

Female 0.10 0.30 0.20

One student is selected at random. Let we define two random variables X and Y as

X=

0, student is a male

1, student is a female, Y =

⎧⎪

⎨

⎪⎩

0, student has GPA under 2.0

1, student has GPA between 2.0 and 3.0 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

HHx HHHH

y 0 1 2 g(x)

0 0.05 0.25 0.10 0.4 1 0.10 0.30 0.20 0.6 h(y) 0.15 0.55 0.30 1

The conditional distribution of Y given X = 1 is w(y|1) = f(1, y)

g(1) = f(1, y) 0.60 for y= 0, 1, 2. That is, that

w(0|1) = f(1, 0) g(1) = 1

6, w(1|1) = f(1, 1) g(1) = 3

6, w(2|1) = f(1, 2) g(1) = 2

6. On the other hand

E[Y²|1] =

2 y=0

y²w(y|1) = 0²· 1

6+ 1²·3

6 + 2² ·2 6 = 11

6 ,

E[Y |1] =

2 y=0

yw(y|1) = 0 ·1

6 + 1 · 3

6 + 2 · 2 6 = 7

6. Since the conditional variance of Y given that X = 1 is

σ_{Y |1}² = E[Y²|1] − (E[Y |1])² we find that

σ_{Y |1}² = 11 6 −

7 6

₂

= 17 36.

 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,

f(0, 0) = P (T T T ) = 1

8, f(0, 1) = P (T T H) = 1

8, f(0, 2) = 0, f(1, 0) = P (HT T ) = 1

8, f(1, 1) = P (T HT ) = P (HT H) = 2

8, f(1, 2) = P (T HH) = 1 8, f(2, 0) = 0, f(2, 1) = P (HHT ) = 1

8, f(2, 2) = P (HHH) = 1 8. Thus we have the following table

HHx HHHH

y 0 1 2 g(x)

0 1/8 1/8 0 2/8

1 1/8 2/8 1/8 4/8

2 0 1/8 1/8 2/8

h(y) 2/8 4/8 2/8 1 On the other hand, since

μ_X = E(X) =²

x=0

xg(x) = 0 · g(0) + 1 · g(1) + 2 · g(2) = 0 · 2 8+ 14

8 + 2 · 2 8 = 1, μ_Y = E(Y ) =

2 y=0

yh(y) = 0 · h(0) + 1 · h(1) + 2 · h(2) = 0 ·2 8 + 14

8 + 2 · 2 8 = 1,

μ^_XY = E(XY ) =²

x=0

2 y=0

xyf(x, y) = 1 · 1 · 2

8+ 1 · 2 · 1

8+ 2 · 1 · 1

8+ 2 · 2 · 1 8 = 10

8

then we have

Cov(X, Y ) = σXY = E(XY ) − E(X)E(Y ) = 10

8 − 1 · 1 = 2 8.

 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

b(2; 120, 0.03) =

120 2

0.03²(1 − 0.03)¹¹⁸= 120!

2! · 118!× 0.03²× 0.97¹¹⁸ = 0.1766059636.

(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 = ¹⁸₅ = 3.6 into the formula for Poisson distribution, we get

p(2; 3.6) = 3.6²× e^−3.6

2! = 0.1770577215.

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

h(3; 20, 250, 12) = ₁₂

3

₂₃₈

17

 ₂₅₀

20

 = 12!

3! · 9! × 238!

17! · 221!

250!

20! · 230!

.

(b) How many defective chips would you expect to select?

Answer. The mean of the hypergeometric distribution is μ = ^nM_N . So the expected number of defective chips is

μ= nM

= 20 · 12 = 24 = 0.96.

 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

b^∗(12; 10, 0.80) =

11 9

0.8¹⁰(1 − 0.8)² = 11!

9!2!0.8¹⁰0.2² ≈ 0.236.

(b) Substituting x = 6, and θ = 0.8 into the formula for the geometric distribution, we get

g(6; 0.8) = 0.8(1 − 0.8)⁶⁻¹= 0.8 × 0.2⁵ = 4

15625 = 2.56 × 10⁻⁴.

 10 + 7 points

(a) Find the moment generating function of the random variable whose probability density is given by

f(x) =

2e^−2x, for x > 0, 0, elsewhere.

Answer. The moment generating function of the random variable X is M_X(t) = E[e^tX] =

 _∞

0

e^tx2e^−2xdx= 2  _∞

0

e^x(t−2)dx= 2 lim

c→∞

 _c

0

e^x(t−2)dx

= 2

t− 2 lim

c→∞e^{x(t−2) c}

0 = 2

t− 2 lim

c→∞(e^c(t−2)− e⁰) = 2

t− 2(0 − 1)

= 2

2 − t for t <2

(b) Use the moment generating function to determine μ^₁ and μ^₂ for f(x) given in (a).

Answer. Since μ^_r = d^rM_X(t) dt^r

t=0

for r= 1, 2 we obtain

μ^₁ = dM_X(t) dt

t=0

= d

dt(2(2 − t)⁻¹) t=0

= (2(2 − t)⁻²)

t=0 = 1 2, μ^₂ = d²M_X(t)

dt²

t=0

= d

dt(2(2 − t)⁻²) t=0

= (4(2 − t)⁻³)

t=0 = 1 2.