Hypergeometric - Probability and Statistics - Solved Exam, Exams of Probability and Statistics

This is the Solved Exam of Probability and Statistics which includes Hypergeometric, Distribution, Binomial, Discrimination etc. Key important points are: Hypergeometric, Random, Represents, Number, Committee, Geometric, Binomial, Distributed, Increase, Committee

Typology: Exams

2012/2013

Uploaded on 02/27/2013

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STT441-Spring2011-Midterm2.5 NAME ID
SOLUTION
The following relates to problems 1- 3. A class contains 10 girls and 20 boys. A
committee of 5 is to be selected at random from the class.
Problem 1. Let X represents the number of girls in the committee when no student can be
selected more than once. 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. Let Y represents the number of girls in the committee when students can be
selected more than once. How is Y 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 3. It was decided to increase the size of the committee to 30 (observe: 30 is also
the size of the class). Find ) )). Hint: What can you say about in this
case?
(1) (6.6, 6.7) (2) (0, 6.7) (3) (6, 6.3) (4) (0, 1.4) (5) None of the
above
Solution: ) so ) . Also,
) ) (
)(
)
____________________________________________________________________
The following relates to problems 4-7. The average number of calls that Jill gets in an
hour is 2.
Problem 4. Let X denotes the number of calls that Jill will get in next hour. Find
)1( XP
.
(1) .69 (2) .86 (3) .79 (4) .49 (5) .25
Solution:
)2(~ PoissonX
86.1
)0(1)1(
2
e
XPXP
Problem 5. Let T be the time( in hours) between now and the next call that Jill will get.
Find
)2( TP
.
pf3
pf4
pf5

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STT441-Spring2011-Midterm2.5 NAME ID

SOLUTION

The following relates to problems 1- 3. A class contains 10 girls and 20 boys. A committee of 5 is to be selected at random from the class.

Problem 1. Let X represents the number of girls in the committee when no student can be selected more than once. 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. Let Y represents the number of girls in the committee when students can be selected more than once. How is Y 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 3. It was decided to increase the size of the committee to 30 (observe: 30 is also the size of the class). Find ) )). Hint: What can you say about in this case?

(1) (6.6, 6.7) (2) (0, 6.7) (3) (6, 6.3) (4) (0, 1.4) (5) None of the

above

Solution: ) so ). Also,

) ) (^) ( ) ( )

____________________________________________________________________

The following relates to problems 4-7. The average number of calls that Jill gets in an hour is 2. Problem 4. Let X denotes the number of calls that Jill will get in next hour. Find P ( X  1 ).

Solution: X ~ Poisson ( 2 )

 ^2 

e^ 

P X P X

Problem 5. Let T be the time( in hours) between now and the next call that Jill will get. Find P ( T  2 ).

(1) e ^2 (2) e ^1 /^2 (3) e ^4 (4) e ^2.^5 (5) e ^5

T~exponential(  2 ) so P ( T  2 ) e ^2 *^2  e ^4

Problem 6. Let U be the time ( in hours) between now and the 3 rd^ call that Jill will get. How is U distributed?

(a) exponential(  2 ) (b) Weibull(  3 , . 4 )

(c) normal(  5 , ^2  1 ) (d) Gamma(  3 ,  2 )

(e) Gamma(  2 ,  2 )

Problem 7. Find P ( U  1 ). (Hint: { } { })

^2  ^2 

e^ 

PU PX PX P X PX

______________________________________________________________________

The following relates to problems 8-10. Let ( X , Y ) be uniformly distributed on the triangle generated by the points (0,0), (0,2) and (2, 2). More formally the joint density is given by f ( x , y ) 1 / 2 , 0  xy  2.

Problem 8. Complete: f X ( x )= ______ if 0  x  2

a. 4 x b. 2 y c. ( 1 / 2 ) x^2 d. 3 ( 1  x ) e. ( ) )

2 f x dy x y x

X ^    

if 0  x  2

Problem 9. Find P ( Y  1 ) (Hint: Look at all ( x , y )from 0  xy  2 where y  1 )

P ( Y  1 )( 1 / 2 )* area ( B )( 1 / 2 )*( 1 / 2 ) 1 / 4

, where B is the triangle generated by the points (0,0), (0,1) and (1, 1).

Problem 10. Complete: f Y | X  x ( y )= ______ if x  y  2 for each 0  x  2.

x

(2) 3 x (3)

2  y

2  x

(5) x

f x x

f x y f y X

Y Xx    2 

(1/2)(2-x)

Problem 15. Find E ( X^2 )

(1)3.67 (2) 2.71 (3) 4.67 (4) 5.49 (5) 2.

E ( X^2 )

4 2

(^234) 0

4

4

2

2

2

0

 x x dx  x x dx x x x

______________________________________________________________________

The following relates to problems 16-18. Let (X, Y, Z) be the number of customers that will end up filling their car with fuel that is(regular, premium, premium+) among the next 5 customers. It is known that P(regular)=.25, P(premium)= .45 and P(premium +)=.3.

Problem 16. Find P(X=1, Y=3, Z=1) (1) .66 (2) .18 (3) .25 (4) .14 (5).

Solution. (X, Y, Z) ~Multinomial(n=5; .25, .45, .3)

P(X=1, Y=3, Z=1)=. 25. 45. 3. 14

Problem 17. How is X distributed? (a) Hypergeometric( 5, 1, 1) (b) Geometric (.25) (c) Poisson( .25) (d) Binomial(5, .75) (e) Binomial(5, .25)

Problem 18. How is distributed? (Hint: Those are the customers who are not using regular)

(a) Hypergeometric( 5, 1, 1) (b) Geometric (.25) (c) Poisson( .25) (d) Binomial(5, .75) (e) Binomial(5, .25)

The following relates to problem 19-20. Let the hazard rate for a random variable be given by )

Problem 19. Find )

(1) (2) (3) (4) (5)

Problem 20. Find the density (pdf) of

(1) (2) (3)