STA 4321 / 5325 – Exam 2 – Spring 2007 i



1 x i i !
 e x i



0 k i 
1
1
 k
( 0
 k

1 ) ( a
 b ) n  i n 

0

 n i

 a i b n
 i
The number of deliveries arriving at an industrial warehouse (during work hours) has a
Poisson distribution with a mean of 2.5 per hour.
What is the probability an hour goes by with no more than one delivery.
P ( Y

1 )
 e

2 .
5
2 .
5
0
0 !
 e

2 .
5
2 .
5
1
1 !

3 .
5 e

2 .
5 
.
2873
Give the mean and standard deviation of the number of deliveries during 8-hour work days.
Mean: 8(2.5)=20
Standard Deviation:

(8(2.5)) = 4.47
The moment-generating function for a random variable that follows a Binomial distribution with n trials, and probability of success p is:
M ( t )

( pe t 
( 1
 p )) n
Use this to derive the expected value of the random variable.
M ' ( t )
 n ( pe t 
( 1
 p )) n

1 pe t 
M ' ( 0 )
 np

E ( X )
Used cars can be classified in two categories: peaches and lemons. Sellers know which type the car they are selling is. Buyers do not. Suppose that buyers are aware that the probability a car is a peach is 0.4. Sellers value peaches at $2500, and lemons at $1500.
Buyers value peaches at $3000, and lemons at $2000.
Obtain the expected value of a used car to a buyer who has no extra information.
E(V) = 3000(0.4) + 2000(0.6) = 1200 + 1200 = 2400
Assuming that buyers will not pay more than their expected value for a used car, will sellers ever sell peaches? YES or NO

An actress has a probability of getting offered a job after a try-out of p =0.10. She plans to keep trying out for new jobs until she gets offered. Assume outcomes of try-outs are independent.
How many try-outs does she expect to have to take?
E(Y) = 1/p = 1/0.10 = 10
What is the probability she will need to attend more than 2 try-outs?
P(Y>2) = 1-P(Y≤2) = 1-[p+pq] = 1-[.10+.10(.90)] = 1-0.19 = 0.81
Derive the probability generating function for the geometric distribution: P(t)=E(t
Y
). p ( y )
 pq y

1 y

1 , 2 ,...
P ( t )

E
 y



1 t y pq y

1  pt

 y

1
( tq ) y

1  pt
1
 tq

1
 pt
( 1
 p ) t
A state lottery game draws 3 numbers at random (without replacement) from a set of 15 numbers. Give the probability distribution for the number of “winning digits” you will have when you purchase one ticket.
# of winning digits Probability
0
1
2
3


3
0




12
3




15
3



220
455


3
1




12
2


15
3






3
2




12
1




15
3




3
3




12
0




15
3





198
455
36
455
1
455

A package of 6 fuses are tested where the probability an individual fuse is defective is
0.05. (That is, 5% of all fuses manufactured are defective).
What is the probability one fuse will be defective? p ( 1 )



6
1


( 0 .
05 )
1
( 0 .
95 )
5 
6 ( 0 .
05 )( 0 .
7738 )

.
2321
What is the probability at least one fuse will be defective.
P ( Y

1 )

1
 p ( 0 )

1

(.
95 )
6 
.
2649
What is the probability that more than one fuse will be defective, given at least one is defective.
P ( Y

1 , Y

1 )

P ( Y

1 )

P ( Y

1 )

P ( Y

1 )

.
2649

.
2321

.
0328

P ( Y

1 | Y

1 )

.
0328
.
2649

.
1238
An industrial salesperson has a salary structure as follows:
S

100

50 Y

10 Y 2 where Y is the number of items sold. Assuming sales are independent from one attempt to another, each with probability of success of p = 0.20. Give the expected salaries on days where the salesperson attempts n = 1, 2, and 3 sales by completing the following table.
n E( Y ) E( Y
2
) E( S )
1 .20 .16+(.2) 2 = 0.20 100+10-2 = 108
2 .40 .32+(.4) 2 = 0.48 100+20-4.8 = 115.2
3 .60 .48+(.6) 2 = 0.84 100+30-8.4 = 121.6