Probability and Statistics Problems and Solutions, Exams of Economics

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Department of Economics Fall 2004 Ozlem Eren ECON 413 FINAL ANSWERS

Part I - Choose 2 out of 4

1. (20) Suppose a student dormitory in a college consists of

i. 30% freshmen, of whom 10% own a car. ii. 40% sophomores, of whom 20% own a car. iii. 20% juniors, of whom 40% own a car. iv. 10% seniors, of whom 60% own a car.

a) Find the probability that a student in the dormitory owns a car.

Let F, S, J, Sn denote respectively the set of freshmen, sophomores, juniors, and seniors; and let C denote the set of students who own a car.

F 10% C

30%

40% S 20% C

20%

10% J 40% C

Sn 60% C

We seek P(C). By the Law of Total Probability, or by using the tree diagram, we obtain: ( ) (0.30)(0.10) (0.40)(0.20) (0.20)(0.40) (0.10)(0.60) 0.03 0.08 0.08 0.

P C = + + +

= + + +

=

b) If a student does own a car, find the probability that the student is a junior.

By Bayes’ Formula: ( ) (^ )^ (^ )^ (0.20)(0.40)^8 0. ( ) (0.25) 25

P J C P J P C^ J

P C

= = = =

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2. (20) Let A, B,…,L be 12 given points in the plane R^2 such that no 3 of the points lie on the same line. Find the number n of

a) Lines in R^2 where each line contains two of the points. Each pair of points determines a line, hence n = “12 choose 2” (^12) 12. 66 2 2.

⎛ ⎞ = =

⎜ ⎟

⎝ ⎠

b) Lines in R^2 containing A and one of the other points. We can choose among the 11 remaining points, hence n = 11

c) Triangles whose vertices come from the given points.

Each triple of points determines a triangle, hence n =

⎛ ⎞ =

⎜ ⎟

⎝ ⎠

d) Triangles whose vertices are A and two of the other points.

We can choose two of the remaining points, hence n =

⎛ ⎞ =

⎜ ⎟

⎝ ⎠

(Alternately, there are triangles without A as a vertex; hence

⎛ ⎞ =

⎜ ⎟

⎝ ⎠

220 – 165 = 55 of the triangles do have A as a vertex.)

3. (20) A pair of fair dice is thrown. If the two numbers appearing are different, find the probability p that

a) The sum is 6.

There are 36 ways the pair of dice can be thrown, and 6 of them (1,1), (2,2),…,(6,6) have the same numbers. Thus, the reduced sample space E will consist of 36 – 6 = 30 elements. The sum 6 can appear in 4 ways (1,5),(2,4),(4,2),(5,1). We cannot include (3,3) since the numbers must be different. Thus 4 2 30 15 p = =

Department of Economics Fall 2004 Ozlem Eren

b) An ace appears.

An ace can appear in 10 ways: (1,2),(1,3),…,(1,6) and (2,1),(3,1),…,(6,1). [we cannot include (1,1)] Thus 10 1 30 3

p = =

c) The sum is 4 or less.

The sum will be 4 or less in 4 ways: (3,1),(1,3),(2,1),(1,2).[we cannot include (1,1),(2,2)] Thus 4 2 30 15

p = =

4. (20) A box contains 10 coins where 5 coins are two-headed, 3 coins are two- tailed, and, 2 are fair coins. A coin is chosen at random and tossed.

a) Find the probability that a head appears. Let X, Y, Z denote, respectively, the two-headed coins, the two-tailed coins and the fair coins. Then P(X) = 0.5, P(Y) = 0.3, P(Z) = 0. Note P H ( X )= 1 , that is, a two-headed coin must yield a head. Similarly P H ( Y ) = 0, and , P H ( Z ) = 0.

Tree diagram:

0.5 0.3 0.

X Y Z

1 0 0 1 0.5 0.

H T H T H T

By the Law of Total Probability or by adding the probabilities or the tree paths leading to H, we get P(H) = (0.5)(1) + (0.3)(0) + (0.2)(0.5) = 0.

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b) If a head appears, find the probability that the coin is fair.

By Bayes Rule ( ) ( ) (^) (0.2)(0.5) 1 ( ) 16.7% ( ) (0.6) 6

P Z P H Z

P Z H

P H

= = = =

Part II - Answer ALL the questions

1. (15) Let Y 1 , Y 2 , and Y 3 be random variables, where E(Y 1 ) = 1, E(Y 2 ) = 2, E(Y 3 ) = - Var(Y 1 ) = 1, Var(Y 2 ) = 3, Var(Y 3 ) = 5 Cov(Y 1 , Y 2 ) = -0. Cov(Y 1 , Y 3 ) = 0. Cov(Y 2 , Y 3 ) = 2 Find the expected value and variance of U = Y 1 - 2 Y 2 + Y 3

1 1 2 2 3 3 1 2 3 1 2 3

1 2 1 2 2 2 3 2 1 3 1 3 2 3 2 3

U = a 2 E(U) = E(Y ) - 2E(Y ) + E(Y ) = 1 -2(2) - 1 = - Var(U) = (a ) (Y ) (a ) (Y ) (a ) (Y ) 2a a ov( , ) 2a a ( , )

Y a Y a Y U Y Y Y

Var Var Var C Y Y Cov Y Y

+ +

= − +

+ +

+ +

=

3

2. (15) A point Q is chosen at random within a circle of radius 1 unit. Interpret this to mean that probability is proportional to area. (Reminder: Area of circle is πr^2 )

a) Find the c.d.f. of X the distance from Q to the center of the circle.

(^22)

( ) ( )

, 0 1

F x P X x area of circle of radius x area of circle of radius

π x x x

π

= ≤ =

= = < <

b) Find the c.d.f. of Y = 12X (the distance in inches) 2 ( ) ( ) 12 144 F y = P X ≤^ y^ = y

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3. (15) A fair die is tossed 180 times. Determine the probability p that the face 6 will appear

a) Between 29 and 32 times inclusive. This is a Binomial experiment B(n,p) with n = 180, p = 1/6, q = 1-p = 5/6. Then

2

180. 1 5. 25, 5

np

npq

μ

σ σ

= = =

= = = =

Let X denote the number of times the faces 6 appears. We seek BP (29 ≤ X ≤ 32), or assuming continuity NP (28.5 ≤ X ≤32.5). Converting X values to standard units we get NP ( 0.3− ≤ Z ≤ 0.5) =0.

b) Between 31 and 35 times inclusive.

We seek BP (31 ≤ X ≤35), or assuming continuity, the Normal Probability NP (30.5 ≤ X ≤ 35.5) = NP (0.1 ≤ Z ≤ 1.1) =0.

c) Less than 22 times. We seek the one-sided probability P(X < 22) or approximately NP X ( ≤22.5) NP Z ( ≤ −1.5) =0. Remark: Since Binomial variable is never negative, we should actually replace BP X ( <22)by BP (0 ≤ X ≤ 22) ≈ NP ( 0.5− ≤ X ≤ 22.5) = NP ( 6.1− ≤ Z ≤ −1.5) However P Z ( ≤ −6.1)is negligible.

4. (15) Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected, and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is x = 11.6, and the population standard deviation is σ = 4.1. (Suppose we know σ). Estimate μ, the mean number of unoccupied seats per flight during the past year, using a 90% Confidence Interval.

2 11.6 (1.645). 4.1 11.6 4. 225 90% : (11.15,12.05)

x z x

CI

α σ

⎛ ⎞ =

⎜⎝ ⎟⎠

∓

∓ ∓

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Y

Bonus Questions – You are allowed to answer AT MOST 2 from this section.

1. (10) Show that

Cov X Y ( , ) = ∑ ( xi − μ x )( y j − μ Y ) f ( xi , y j ) = ∑ x y fi j ( xi , yj )−μ x μ

Note that the last term is E XY ( ) − μ X μ Y

Proof is given for the case that X and Y are discrete and finite:

,

,

,

, ,

, ,

,

( , ) ( )

( , ) ( )

( , )

( )( ) ( , ) ( ) ( ,

( , ) ( , )

( , ) ( , )

i j j^ i^ j^ j j^ j^ Y i i j i i X i j i

i j i^ j

i j i^ X^ j^ Y^ i^ j^ i j i^ j^ X^ j^ Y^ i^ X^ Y^ i^ j

i j i^ j^ i^ j^ X^ i j j^ i^ j Y (^) i j i i j X Y (^) i i j

y f x y y g y

x f x y x f x

f x y

Therefore x y f x y x y y x f x )

x y f x y y f x y

y

x f x y f x y

μ

μ

μ μ μ μ μ μ

μ

μ μ μ

= =

= =

− − = − − +

= −

− +

∑ ∑

∑ ∑

∑

∑ ∑

∑ ∑

∑ ,

,

,

( , )

( , )

j i j i j X Y X Y X Y i j

i j i^ j^ i^ j^ X^ Y

x y f x y

x y f x y

μ μ μ μ μ μ

μ μ

= − − +

= −

∑

∑

∑

2. (10) Z has p.d.f

2 2 , 0 ( ) 0 ,

e z z f z otherwise

⎧ − >

= ⎨

⎩

What is the p.d.f of T^1 z

=?

Note : If pdf of X is given as f(x) and another RV is defined as Y = r(x) The distribution function G(y) of Y is

G y ( ) = P Y ( ≤ y ) = P r x ( ( ) ≤ y ) = ∫ f ( ) x dx

Pdf of Y, g(y), can be obtained from: g y ( ) dG y (^ ) dy

=

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In this question, let

1/ (^2 2) 1/ 2 1/ 2 / 0 0 0 2 /

( ) ( ) ( 1 )

(1 )

( 1 )

( ) 1 ( 1 )

( ) 1 2 1 2 1 1 1

( )

t (^) z z t (^) z t t

t

G t P T t P t z P tz P z t G t P z Distribution Function t G t e dz e e e

G t e

− − − −

−

= ≤ = ≤

= ≤

= ≥

= − ≤

= − = − = + = + − =

−

=

∫ e −2 /^ t

We need pdf. 2 / 2 / 2 2 / 2

( ) ( )^ (^ )^2

2 , 0

( )

0 ,

t t

t

g t dG t^ d e e dt dt t e (^) t g t (^) t otherwise

− −

−

= = = −

−

⎧

= ⎨⎪ >

⎪⎩

3. (15) 1 Let X be the (uniform) continuous r.v. with distribution UNIF(a,b), that is, whose p.d.f, f, is a constant k^1 b a

=

−

on the interval I = [a,b], and zero elsewhere. Prove: i.

2

2 2

( ) (^ )

( ) ( )

2( )

2( ) 2( )

b a b

a

E X a^ b

E X xf x dx x dx b a x b a b a b a b a a b

μ

∞ −∞

= +

= = =

−

= ⎡^ ⎤

⎢ − ⎥

⎣ ⎦

= −

− −

= +

∫ ∫

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ii. ( )

[ ]

2

2 2 2

3 3 3 2

2 2 2 2 2 2

2

( )

( ) ( )

3( ) 3( ) 3( ) 3

( ) ( ) ( )

( )

b a b

a

b a Var X

E X x f x dx x dx b a x b a b ab^2 b a b a b a Var X E X E X b ab a a ab b

b a

∞ −∞

−

=

= =

−

= ⎡^ ⎤^ = − = +^ + a ⎢ (^) − ⎥ − − ⎣ ⎦ = −

= +^ +^ − +^ +

= −

∫ ∫

iii. Cumulative distribution F(x) is equal to: 0 , ( ) (^ ) , ( ) 1 ,

for x a F x x^ a for a x b b a for x b

⎧ <

⎪⎪ −

= ⎨ ≤

⎪ −

⎪⎩ >

≤

We have 3 cases:

.

( ) ( ) 0 0

.

( ) ( )^1

. ( ) ( ) ( ) 1 ( ) ( ) 1 ( ) 1

x x

x x x

a

i For x a

F x f t dt dt

ii For a x b

F x f t dt dt t^ x^ a b a b a b a iii For x b Since F x is a cumulative distribution function F x F b But F x P X x Hence F x

−∞ −∞

−∞ −∞

<

= = =

≤ ≤

= = = ⎡^ ⎤ = −

− ⎢⎣^ − ⎥⎦ −

>

≥ =

= ≤ ≤

=

∫ ∫

∫ ∫

Department of Economics Fall 2004 Ozlem Eren

4. (15) 1 Suppose that a radioactive particle is randomly located in a square with sides of unit length. That is, if two regions of equal area are considered, the particle is equally likely to be in either. Let Y 1 and Y 2 denote the coordinates locating the particle. A reasonable model for the relative frequency histogram for Y 1 and Y 2 would be the bivariate analogue of the univariate uniform distribution.

1 2 1 2

1, 0 1 ; 0 1

( , )

0,

y y f y y elsewhere

⎧^ ≤^ ≤^ ≤^ ≤

= ⎨

⎩

a) Sketch the probability density surface.

f ( y 1 (^) , y 2 )= z

1

0.4 0.

y 2 y 1

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b) Find F(0.2, 0.4), and show on graph. 0.4 0. 1 2 1 2

1 2

0.4 0. 1 2 0 0 0.4 (^) 0. 1 0 2 0

(0.2, 0.4) ( , ) ,

0 0.2; 0 0.4. ( )

0.

(0.2, 0.4) (1)

0.

F f y y dy dy corresponds to the volume under F where

y and y Shaded region The desired volume equals

F dy dy

y dy

−∞ −∞

=

≤ ≤ ≤ ≤

=

=

=

∫ ∫

∫ ∫

∫

c) Find P (0.1 ≤ Y 1 (^) ≤ 0.3 , 0 ≤ Y 2 ≤ 0.5),and show on graph.

0.5 0. 1 2 1 2 1 2 0 0. 0.5 0. 1 2 0 0.

1 2

(0.1 0.3, 0 0.5) ( , )

(1)

0.

0.1 0.3 , 0 0.

( 1

P Y Y f y y dy dy

dy dy

Corresponds to the volume under F over the region y y Density or height of the probability is equal to and hence the desired proba

≤ ≤ ≤ ≤ =

=

=

≤ ≤ ≤ ≤

∫ ∫

∫ ∫

bility ( volume ) is P (0.1 ≤ Y 1 (^) ≤ 0.3, 0 ≤ Y 2 ≤ 0.5) = (0.2)(0.5)(1) =0.1)