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### 45-733 PROBABILITY AND STATISTICS I 1998 Midterm Examination Answers

Probability and Statistics                      Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

1. Suppose we have the continuous probability distribution

f(x) = 1. Find c.

ò13 (c/x3)dx = (c/-2)(x-2|13 = (c/-2)(1/9 - 1) = (4/9)c; hence, c = 9/4

2. Find F(x).

F(x) = ò1x (9/4t3)dt = (-9/8t2)|1x = (9/8)(1 - 1/x2)
Hence
```                           {  0                  x <= 1
{
F(x) = {  (9/8)[1 - 1/x**2]  1 < x < 3
{
{  1                  x >= 3
```

3.

4. Find P(X > 2)

P(X > 2) = 1 - F(2) = 1 - 27/32 = 5/32

Probability and Statistics                      Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

2. You draw 3 cards randomly without replacement from a deck of 52 playing cards. What is the probability that the 3 cards are from different suits? Probability and Statistics                      Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

3. Suppose we have the bivariate discrete probability distribution:

f(x,y) = 1. Find c

```                               y

0   1   2
------------
3 |12  13  14 |39
|           |
x    |           |
|           |
4 |16  17  18 |51
|           |
---------------
28  30  32 |90
```

Hence, c = 1/90
2. Are X, Y independent?

No. f(3, 0) = 12/90 ¹ f1(3)f2(0) = (39/90)(28/90).

Probability and Statistics                      Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

4. Suppose we have two urns. In the first urn there are 10 Red and 10 Black balls. In the second urn there are 5 Red and 15 Black balls. One ball is drawn randomly without replacement from each urn. Let Y equal the number of black balls drawn from the two urns.
1. What is the probability distribution of Y.

P(Y = 0) = (10/20)(5/20) = 50/400
P(Y = 1) = (10/20)(5/20) + (10/20)(15/20) = 200/400
P(Y = 2) = (10/20)(15/20) = 150/400
```                           {  50/400   y = 0
{
f(y) = { 200/400   y = 1
{
{ 150/400   y = 2
```

2. What is E(Y).

E(Y) = 0(50/400) + 1(200/400) + 2(150/400) = 5/4

Probability and Statistics                      Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

5. Suppose we have a discrete bivariate probability distribution

f(x,y) = ```                               y

0   1   2
------------
1 | 2   3   4 | 9
|           |
x  2 | 4   5   6 |15
|           |
3 | 6   7   8 |21
|           |
---------------
12  15  18 |45
```

1. Find P(X > 1 Ç Y > 1)

P(X > 1 Ç Y > 1) = 6/45 + 8/45 = 14/45

2. Find P(X £ 2 ½ Y £ 1)

P(X £ 2 ½ Y £ 1) = P(X £ 2 Ç Y £ 1)/P(Y £ 1)

P(X £ 2 Ç Y £ 1) = 2/45 + 3/45 + 4/45 + 5/45 = 14/45

P(Y £ 1) = 12/45 + 15/45 = 27/45

Hence, P(X £ 2 ½ Y £ 1) = (14/45)/(27/45) = 14/27

Probability and Statistics                      Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

6. Suppose there is a new test for detecting a specific disease. One person in every 5000 is known to have the disease. If a person has the disease the probability of a positive test is .99. If a person does not have the disease the probability of a positive test is .15. A person is randomly drawn from a large population and the test is applied to the person. If the test is positive what is the probability that the person does not have the disease?

A = Person Has Disease, B = Postive Test Result
P(A) = .0002, P(B | A) = .99, P(B | Ac) = .15

P(Ac | B) = P(Ac)P(B | Ac)/[P(A)P(B | A) + P(Ac)P(B | Ac)] =
(.9998x.15)/[(.0002x.99) + (.9998x.15)] = .9987

Probability and Statistics                      Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

7. An urn contains 6 red and 4 white balls. Three balls are randomly drawn without replacement from the urn. Find the probability that all 3 of the removed balls are red if it is known that at least 1 of the removed balls is red.

Let X = The number of Red balls. Then
P(X = 3 | X ³ 1) = P(X = 3 Ç X ³ 1) = P(X = 3)/P(X ³ 1) = P(X = 3)/[1 - P(X = 0)]

P(X = 3) = [6 choose 3]/[10 choose 3]
P(X = 0) = [4 choose 3]/[10 choose 3]

Hence P(X = 3 | X ³ 1) = [6 choose 3]/{[10 choose 3] - [4 choose 3]}

Probability and Statistics                      Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

8. Suppose we have the bivariate continuous probability function

f(x,y) = Find COV(X, Y)

E(X) = ò01 ò01 x(4/7)(2 - xy)dydx = (4/7)ò01 [(2xy - x2y2/2)|01]dx =
(4/7)ò01 (2x - x2/2)dx = (4/7)[x2 - x3/6)|01] = (4/7)(1 - 1/6) = 10/21

E(Y) = ò01 ò01 y(4/7)(2 - xy)dxdy = (4/7)ò01 [(2xy - x2y2/2)|01]dy =
(4/7)ò01 (2y - y2/2)dy = (4/7)[y2 - y3/6)|01] = (4/7)(1 - 1/6) = 10/21

E(XY) = ò01 ò01 xy(4/7)(2 - xy)dydx = (4/7)ò01 [(xy2 - x2y3/3)|01]dx =
(4/7)ò01 (x - x2/3)dx = (4/7)[x2/2 - x3/9)|01] = (4/7)(1/2 - 1/9) = 2/9

COV(X, Y) = E(XY) - E(X)E(Y) = 2/9 - (10/21)(10/21)

Probability and Statistics                      Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

9. Suppose we have the continuous bivariate probability distribution Find VAR(X) and VAR(Y).

E(X) = ò01 ò12 x(3/10)(x2 + 2y)dydx = (3/10)ò01 [(x3y + xy2)|12]dx =
(3/10)ò01 (x3 + 3x)dx = (3/10)[x4/4 + 3x2/2)|01] = (3/10)(1/4 + 3/2) = 21/40

E(X2) = ò01 ò12 x2(3/10)(x2 + 2y)dydx = (3/10)ò01 [(x4y + x2y2)|12]dx =
(3/10)ò01 (x4 + 3x2)dx = (3/10)[x5/5 + x3)|01] = (3/10)(1/5 + 1) = 18/50

E(Y) = ò12 ò01 y(3/10)(x2 + 2y)dxdy = (3/10)ò12 [(yx3/3 + 2y2x)|01]dy =
(3/10)ò12 (y/3 + 2y2)dy = (3/10)[y2/6 + 2y3/3)|12] = (3/10)(3/6 + 14/3) = 93/60

E(Y2) = ò12 ò01 y2(3/10)(x2 + 2y)dxdy = (3/10)ò12 [(y2x3/3 + 2y3x)|01]dy =
(3/10)ò12 (y2/3 + 2y3)dy = (3/10)[y3/9 + y4/2)|12] = (3/10)(7/9 + 15/2) = 149/60

VAR(X) = 18/50 -(21/40)2 = .0844

VAR(Y) = 149/60 -(93/60)2 = .0808

Probability and Statistics                      Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Midterm
Keith Poole

(10 Points)

10. The manager of a stockroom in a factory has constructed the following probability distribution for the daily demand (number of times used) for a particular tool.

```                           {  .1   x = 0
{
{  .2   x = 1
{
f(x) = {  .3   x = 2
{
{  .4   x = 3
{
{  0 otherwise
```

It costs the factory \$10 each time the tool is used. Find the mean and the variance of the daily cost of the tool.

Let Y = 10X

E(X) = 0*.1 + 1*.2 + 2*.3 + 3*.4 = 2

E(X2) = 0*.1 + 1*.2 + 4*.3 + 9*.4 = 5.0

E(Y) = E(10X) = 10E(X) = 10*2 = 20

VAR(Y) = VAR(10X) = 100VAR(X) = 100(5 - 22) = 100