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) =

- Find c.

ò_{1}^{3}(c/x^{3})dx = (c/-2)(x^{-2}|_{1}^{3}= (c/-2)(1/9 - 1) = (4/9)c; hence, c = 9/4

- Find
**F(x)**.

**F(x)**= ò_{1}^{x}(9/4t^{3})dt = (-9/8t^{2})|_{1}^{x}= (9/8)(1 - 1/x^{2})

Hence{ 0 x <= 1 { F(x) = { (9/8)[1 - 1/x**2] 1 < x < 3 { { 1 x >= 3

- 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) =

- 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

- Are
**X, Y**independent?

No. f(3, 0) = 12/90 ¹ f_{1}(3)f_{2}(0) = (39/90)(28/90).

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- 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

- What is
**E(Y).**

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

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

- Find
**P(X > 1 Ç Y > 1)**

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

- 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

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?

(.9998x.15)/[(.0002x.99) + (.9998x.15)] = .9987

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

Hence

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)** = ò_{0}^{1}
ò_{0}^{1}
x(4/7)(2 - xy)dydx =
(4/7)ò_{0}^{1}
[(2xy - x^{2}y^{2}/2)|_{0}^{1}]dx =

(4/7)ò_{0}^{1}
(2x - x^{2}/2)dx =
(4/7)[x^{2} - x^{3}/6)|_{0}^{1}] =
(4/7)(1 - 1/6) = 10/21

**E(Y)** = ò_{0}^{1}
ò_{0}^{1}
y(4/7)(2 - xy)dxdy =
(4/7)ò_{0}^{1}
[(2xy - x^{2}y^{2}/2)|_{0}^{1}]dy =

(4/7)ò_{0}^{1}
(2y - y^{2}/2)dy =
(4/7)[y^{2} - y^{3}/6)|_{0}^{1}] =
(4/7)(1 - 1/6) = 10/21

**E(XY)** = ò_{0}^{1}
ò_{0}^{1}
xy(4/7)(2 - xy)dydx =
(4/7)ò_{0}^{1}
[(xy^{2} - x^{2}y^{3}/3)|_{0}^{1}]dx =

(4/7)ò_{0}^{1}
(x - x^{2}/3)dx =
(4/7)[x^{2}/2 - x^{3}/9)|_{0}^{1}] =
(4/7)(1/2 - 1/9) = 2/9

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

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)**.

(3/10)ò

(3/10)ò

(3/10)ò

(3/10)ò

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(X ^{2})** = 0*.1 + 1*.2 + 4*.3 + 9*.4 = 5.0