1999 Prob-Stat I Midterm Answers Page

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45-733 PROBABILITY AND STATISTICS I Midterm Examination Answers

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

1. Suppose we have the bivariate discrete probability distribution


                           æ c(3x²y² - xy)  x = 1, 2
                  f(x,y) = ç                y = 1, 3
                           è      0 otherwise

Find c.


                              y
                           1    3
                         ----------
                       1 | 2   24 |  26
                    x    |        |
                       2 |10  102 | 112
                         |        |
                         ----------
                          12  126 | 138

Therefore, c = 1/138

Find VAR(X) and VAR(Y).

E(X) = å_i=1,n å_j=1,m x_if(x_i, y_j) = å_i=1,n x_if₁(x_i) = 1*(26/138) + 2*(112/138) = 250/138
E(Y) = å_j=1,m å_i=1,n y_jf(x_i, y_j) = å_j=1,m y_jf₂(y_j) = 1*(12/138) + 3*(126/138) = 390/138
E(X²) = å_i=1,n å_j=1,m x_if(x_i, y_j) = å_i=1,n x_if₁(x_i) = 1²*(26/138) + 2²*(112/138) = 474/138
E(Y²) = å_j=1,m å_i=1,n y_jf(x_i, y_j) = å_j=1,m y_jf₂(y_j) = 1²*(12/138) + 3²*(126/138) = 1146/138

VAR(X) = E(X²) - [E(X)]² = (474/138) - (250/138)² = .1529
VAR(Y) = E(Y²) - [E(Y)]² = (1146/138) - (390/138)² = .3176

Are X, Y Independent?

NO. f(1,1) = 2/138 ¹ f₁(1)*f₂(1) = (26/138)*(12/138) » 2.26/138

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

2. The firm you work for has a small office that employs 10 workers in 2 job categories – one job has a high rate of pay, and the other job has a low rate of pay. There are 4 workers in the highly paid job category and 6 in the lower paid job category. Your boss has received complaints that workers have been discriminated against in the office. In particular, 7 of the 10 workers belong to a particular ethnic group. Five of the 7 workers from this ethnic group are in the lower paid job category and only 2 of the 7 are in the higher paying category. Your boss has asked you to check into this and report back. Just based on the numbers, do you think that there is evidence of discrimination?

The probability that we would observe 5 members of the ethnic group in the low paying job and 2 in the higher paying job if job assignment was totally random is:


                         æ7öæ3ö
                         ç ÷ç ÷
                         è5øè1ø      63
                 P(E) = -------- = ---- = .3
                          æ10ö      210
                          ç  ÷
                          è 6ø

This probability is high enough for us to conclude that there probably has not been intentional discrimination.

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

3. Suppose we have the bivariate continuous probability distribution:


                           æ c(4x + y)  0 < x < 1
                  f(x,y) = ç            0 < y < 1
                           è   0 otherwise

Find c.

cò₀¹ ò₀¹ (4x + y)dxdy = cò₀¹ [(4x²/2 + yx)|₀¹]dy = cò₀¹ (2 + y)dy = c(2y + y²/2)|₀¹ = c(5/2)
Hence, c = 2/5

Are X, Y Independent?

f₁(x) = ò_-¥^+¥ f(x,y)dy = ò₀¹ [(8x + 2y)/5]dy = (1/5)[8xy + y²]|₀¹ = (8x + 1)/5
f₂(y) = ò_-¥^+¥ f(x,y)dx = ò₀¹ [(8x + 2y)/5]dx = (1/5)[4x² + 2yx]|₀¹ = (4 + 2y)/5

NO. Clearly f(x,y) ¹ f₁(x)f₂(y)

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

4. The probability that 1 percent of the items produced by a certain process are defective is .8, the probability that 5 percent of the items are defective is .1, and the probability that 10 percent of the items are defective is .1. You draw an item randomly from a large lot and find that it is defective. What is the probability that 5 percent of the items are defective?

Let A₁ = "1% of Items are Defective",
A₂ = "5% of the Items are Defective",
A₃ = "10% of the Items are Defective",
and let B = "Item is Defective".
Graphically, this is:

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

5. You are in the heavy equipment rental business. Let X be the number of rentals per day. Based on long experience, you know that the distribution of daily rentals is:


                         æ .05  x =  0
                         ç
                         ç .20  x =  5
                         ç
                         ç .30  x = 10
                         ç
                  f(x) = ç .25  x = 15
                         ç
                         ç .15  x = 20
                         ç
                         ç .05  x = 25
                         ç
                         è  0 otherwise

Suppose your daily profit, Y, is equal to $5000X – $45,000. Find the mean and the variance of your daily profit.

E(X) = å_i=1,n x_if(x_i) = 0*.05 + 5*.20 + 10*.30 + 15*.25 + 20*.15 + 25*.05 = 12
E(X²) = å_i=1,n x_i²f(x_i) = 0²*.05 + 5²*.20 + 10²*.30 + 15²*.25 + 20²*.15 + 25²*.05 = 182.5

VAR(X) = E(X²) - [E(X)]² = 182.5 - 12² = 38.5

E(Y) = E(5000X - 45000) = 5000*12 - 45000 = $15,000
VAR(Y) = VAR(5000X - 45000) = 5000²VAR(X) = 962,500,000

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

6. Three construction contracts are to be randomly assigned to 3 firms – your firm and your two competitors. If each contract will yield you a profit of $100,000, what is your expected profit?

This is a "boxes and balls" problem.

Let X = "Number of Contracts Received by Your Firm"
P(X = 0) = (2*2*2)/27 = 8/27
P(X = 1) = (1*2*2 + 1*2*2 + 1*2*2)/27 = 12/27
P(X = 2) = (1*1*2 + 1*1*2 + 1*1*2)/27 = 6/27
P(X = 3) = (1*1*1)/27 = 1/27

E(X) = å_i=1,n x_if(x_i) = 0*(8/27) + 1*(12/27) + 2*(6/27) + 3*(1/27) = 1
Hence the expected profit is $100,000.

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

7. You draw 4 cards randomly from a deck of 52 standard playing cards. What is the probability that exactly one suit is missing (e.g., §§ ¨ ª )?


                   æ4öæ3öæ13öæ13öæ13ö
                   ç ÷ç ÷ç  ÷ç  ÷ç  ÷
                   è3øè1øè 2øè 1øè 1ø
            P(E) = ----------------
                         æ52ö
                         ç  ÷
                         è 4ø

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

8. Suppose we have the discrete probability function


                         æ .25  x = 1 
                         ç
                         ç .05  x = 2
                         ç
                  f(x) = ç .20  x = 3 
                         ç
                         ç .50  x = 4 
                         ç
                         è  0 otherwise

Find F(X)


                         æ  0   x < 1
                         ç
                         ç .25  1 £ x < 2
                         ç
                  F(x) = ç .30  2 £ x < 3
                         ç
                         ç .50  3 £ x < 4
                         ç
                         è  1   x ³ 4

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

9. Suppose we have the following circuit with 6 relays. The probability that any relay will work when it is activated is .8 and the failures of the relays are independent. What is the probability that current will flow from A to B when the relays are activated?


                      |--1-----2-------|
                      |                |
                   A--|-----3----------|---B
                      |       ---5---  |
                      |--4----|     |--|
                              ---6---

For current to flow from A to B at least one path has to be complete. Hence
P(Current Flows) = 1 - P(all 3 paths fail)

P(Top Path Fails) = 1 - .8² = .36
P(Middle Path Fails) = .2
P(Bottom Path Fails) = 1 - [.8*(.8 + .8 - .8²)] = .232

P(Current Flows) = 1 - .36*.2*.232 = .9833

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

10. Suppose we have the bivariate discrete probability distribution


                           æ (2x² + y² + 1)/54  x = 0, 1, 2
                  f(x,y) = ç                   y = 0, 1, 2
                           è      0 otherwise

Find P(X > 0 Ç Y > 0).


                               y
                           0   1   2
                         -------------
                       0 | 1   2   5 |  8 
                         |           |    
                    x  1 | 3   4   7 | 14 
                         |           |
                       2 | 9  10  13 | 32 
                         |           |
                         ----------------
                          13  16  25 | 54

P(X > 0 Ç Y > 0) = (4 + 7 + 10 + 13)/54 = 34/54

Find P(X ³ 1 | Y ³ 1)

P(X ³ 1 | Y ³ 1) = P(X ³ 1 Ç Y ³ 1)/P(Y ³ 1) =
(4 + 7 + 10 + 13)/(2 + 5 + 4 + 7 + 10 + 13) = 34/41