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STT441-Spring2011-Midterm2 NAME ID
SOLUTION
The following relates to problems 1 and 2. A class contains 10 girls and 20 boys. A committee of 5 was selected at random and it had one girl and 4 boys. Some people thought that this is discrimination against girls since the girls are 1/3 of the class but only 1/5 of the committee. Let X represent the number of girls in the committee. Problem 1. How is X distributed? (a) Hypergeometric( 30, 10, 5) (b) Geometric (.5) (c) Hypergeometric( 30, 10, 1) (d) Binomial(5, 2/3) (e) Binomial(5, 1/3)
Problem 2. A statistician decided to calculate (^) P ( X 1 )under the assumption of no
discrimination. What did the statistician find?
(1) .61 (2) .27 (3) .39 (4) .49 (5).
Solution:. 45
The following relates to problems 3-7. The average yearly number of earthquakes in California is 2.5. In other words: on average there is an earthquake every 1/ 2.5=0. years. Let X denote the number of earthquakes in California during next year. Let T be the time( in years) between now and the next earthquake. Let U be the time ( in years) between now and the 2nd^ earthquake there.
Problem 3. Find P ( X 2 ).
(1) .61 (2) .71 (3) .39 (4) .49 (5).
Solution: X ~ Poisson ( 2. 5 )
^2.^5 ^2.^5
e ^ e^
P X P X P X
Problem 4. How is T distributed?
(a) exponential( . 5 ) (b) Weibull( ^3 ,^ .^4 )
(c) exponential( 2. 5 ) (d) normal( 2. 5 ,^2 1 )
(e) Gamma( 3 , 2. 5 )
Problem 5. Find P ( T 1 )
(1) e ^2 (2) e ^1 /^2 (3) e ^7 (4) e ^2.^5 (5) e ^5
T~exponential( 2. 5 ) so P ( T 1 ) e ^2.^5 *^1 e ^2.^5
Problem 6. How is U distributed?
(a) exponential( 2. 5 ) (b) Weibull( 2 , . 4 )
(c) normal( 5 , 1 2 ) (d) normal(. 8 , 1 2
(e) Gamma( 2 , 2. 5 )
Problem 7. Find P ( U 1 ). ( Hint: if you like integration then you need to know the
density of U and then integrate appropriately. If you don’t like integration, ask yourself what does the event { U 1 }say about the number of earthquakes in California next
year.) (1) .61 (2) .71 (3) .39 (4) .29 (5).
P ( U 1 ) P ( X 1 ). 29 ( see solution of problem 3)
The following relates to problems 8-10. Let ( X , Y ) be uniformly distributed on the triangle generated by the points (0,0), (1,0) and (1,1). More formally the joint density is given by f ( x , y ) 2 , 0 y x 1.
Problem 8. Complete: f (^) X ( x )= ______ if 0 x 1
a. x b. 2 x c. 2 x d. 3 x e. y
f x dy x
x
y
X ( )^22
0
(^)
if 0 x 1
Problem 9. Find E ( X )
(1) .67 (2) .71 (3) .39 (4) .49 (5).
1
0
E X (^) x xdx
Problem 10. Complete: f (^) Y | X x ( y )= ______ if 0 y x for each 0 x 1.
a. 2 /y b. 3 x c. 1/ x d. 2 ( 1 y ) e. x
Problem 13. Find P ( X 1. 5 ).
(1) .667 (2) .875 (3) .704 (4) .534 (5).
Solution: P(X<1.5)=
- 5 1
(^12) 0
2
- 5
1
1
0
xdx ( 2 x ) dx x / 2 ( 2 x ) / 2.
Problem 14. Find the 70th^ percentile of X.
(1) 1.66 (2) 1.87 (3) 1.77 (4) 1.22 (5) 1. ( Hint: you have to solve for x in the equation P ( X x ). 7 )
We have P(X>x)=.3 and we need to find x. Obviously x >1 and we get
If ( 2 ) / 2. 3 then 1. 22
2
2
2
x x
u du x x
Problem 15. Find Var ( X )
(1) .67 (2) .71 (3) .39 (4) .49 (5).
E ( X^2 ) ( 2 ) / 4 ( 2 / 3 / 4 ) 1. 17
2 1
(^134) 0
4
3
1
2
1
0
x xdx x x dx x x x
E ( X ) 1 due to symmetry of f ( x )about 1.
Var ( X ) 1. 17 12 . 17
The following relates to problems 16-17. Let (X, Y, Z) be the number of the STT 441 finals that will end up being (too easy, just right, too difficult) among the next 4 finals. It is known that P(exam too easy)=.25, P(exam just right)= .45 and P(exam too difficult)=.3.
Problem 16. Find P(X=1, Y=2, Z=1) (1) .16 (2) .18 (3) .25 (4) .32 (5).
Solution. (X, Y, Z) ~Multinomial(n=4; .25, .45, .3)
P(X=1, Y=2, Z=1)=. 25. 45. 3. 18
Problem 17. Find P(X=1) (1) .16 (2) .18 (3) .25 (4) .42 (5).
X~Binomial(n=4, p=.25)
P(X=1)=. 251.^753 1
=.
The following relates to problem 18-20. Let the hazard rate for a random variable be given by ( )
Problem 18. Find ( ) (1) .16 (2) .18 (3) .25 (4) .32 (5).
( ) { ∫ } { }
So (^ )
Problem 19. Find the density (pdf) of
( ) { ∫ }
So ( )^
{ }
(1) (2) (3)
(4) (5)
Problem 20. Everyone gets the points for this problem. Congratulations!
