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45-733 PROBABILITY AND STATISTICS I Practice Midterm

Probability and Statistics
Name__________________________

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Midterm

Keith Poole

(10 Points)

Probability and Statistics
Name__________________________

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Midterm

Keith Poole

(10 Points)

æ (3x - y)/5 1 < x < 2 f(x,y) = ç 1 < y < 3 è 0 otherwise

- Find
**f**and_{1}(x)**f**_{2}(y)

**f**_{1}(x) = ò_{-¥}^{+¥}f(x,y)dy = ò_{1}^{3}[(3x - y)/5]dy = (1/5)[3xy - y^{2}/2]|_{1}^{3}= (1/5)[(9x - 9/2) - (3x - 1/2)] = (6x - 4)/5

**f**_{2}(y) = ò_{-¥}^{+¥}f(x,y)dx = ò_{1}^{2}[(3x - y)/5]dx = (1/5)[3x^{2}/2 - yx]|_{1}^{2}= (1/5)[(6 - 2y) - (3/2 - y)] = (9/2 - y)/5

Therefore:**æ (6x - 4)/5 1 < x < 2 f**_{1}(x) = ç è 0 otherwise æ (9/2 - y)/5 1 < y < 3 f_{2}(y) = ç è 0 otherwise - Are
**X, Y**independent?

NO. Clearly**f(x,y) ¹ f**_{1}(x)f_{2}(y)

Probability and Statistics
Name__________________________

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Midterm

Keith Poole

(10 Points)

æ 13 öæ 39 ö ç ÷ç ÷ è5 4 3 1øè8 9 10 12ø P(E) = ----------------- æ 52 ö ç ÷ è13 13 13 13ø

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Midterm

Keith Poole

(10 Points)

4. Suppose we have a continuous probability distributionæ 2(1 - x) 0 < x < 1 f(x) = ç è 0 otherwise

- Find
**F(x)**.

**F(x) = P(X £ x) = ò**_{0}^{x}2(1 - t)dt = (2t - 2t^{2}/2)|_{0}^{x}= 2x - x^{2}

Hence:**æ 0 x < 0 ç F(x) = ç 2x - x**^{2}0 £ x £ 1 ç è 1 x > 1

- Find
**P(1/2 < X < 1)**

**P(1/2 < X < 1) = F(1) - F(1/2) = 1 - 3/4 = 1/4**

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Midterm

Keith Poole

(10 Points)

5. Suppose we have the discrete bivariate probability distribution

æ c(3y - x) x=2,3 f(x,y) = ç y=2,3,4 è 0 otherwise

- Find
**c**

Therefore,**y 2 3 4 ------------ 2 | 4 7 10 | 21 x | | 3 | 3 6 9 | 18 | | --------------- 7 13 19 | 39****c**= 1/39

- Find
**E(X)**and**E(Y)**.

**E(X) = å**= 2*(21/39) + 3*(18/39) = 96/39_{i=1,n}å_{j=1,m}x_{i}f(x_{i}, y_{j}) = å_{i=1,n}x_{i}f_{1}(x_{i})

**E(Y) = å**= 2*(7/39) + 3*(13/39) + 4*(19/39) = 129/39_{j=1,m}å_{i=1,n}y_{j}f(x_{i}, y_{j}) = å_{j=1,m}y_{j}f_{2}(y_{j})

- Are
**X, Y**Independent.

NO.**f(2,2) = 4/39 ¹ f**_{1}(2)*f_{2}(2) = (21/39)*(7/39)

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Midterm

Keith Poole

(10 Points)

Let

and let

Graphically, this is:

(.5*.1)/(.5*.1 + .3*.6 + .2*.3)= .05/.29 = 5/29

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Midterm

Keith Poole

(10 Points)

Findæ .1 x = 0 ç ç .2 x = 1 ç ç .3 x = 2 ç f(x) = ç .2 x = 3 ç ç .1 x = 4 ç ç .1 x = 5 ç è 0 otherwise

Hence,

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Midterm

Keith Poole

(15 Points)

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

Clearly,

P(top path fails) = P(at least one fails)

Hence,

æ .005149 x = 0 ç ç .082143 x = 1 ç f(x) = ç .381267 x = 2 ç ç .531441 x = 3 ç è 0 otherwise

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Midterm

Keith Poole

(10 Points)

Let

[(1/2)

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Midterm

Keith Poole

(10 Points)

Let

(.65*.3)/(.65*.3 + .35*.2) = .195/(.195 + .07) » .736