STA 4321 – Final Exam PRINT Name _____________
1. A game consists of 4 random numbers being selected from the Uniform density over
the range [0,20] independently. That is X1,X2,X3,X4 ~ iid U(0,20), with probability
density function:
1 / 20 0  x  20
f ( x)  
otherwise
 0
a) Give the cumulative distribution function for this distribution and sketch it.
Uniform CDF
1.2
1
F(x)
0.8
0.6
0.4
0.2
0
-0.2-10
0
10
20
30
x
0 x0

1
x

For x  [0,20] F ( x)   f (t )dt   dt 
0  x  20  F ( x)   x / 20 0  x  20
20
20

0

1 x  20

x
x
b) The game pays off the largest of the four values, that is: X(4) =
max(X1,X2,X3,X4). Give the cumulative distribution function and probability
density function for X(4).
G4 ( x)  P( X ( 4)  x)  P( X 1  x, X 2  x, X 3  x, X ( 4 )
0 x0


 x)  [ F ( x)]  [ x / 20]4 0  x  20

1 x  20

  x 3  1 
x3
dG4 ( x)

4

g3 ( x) 
 4[ F ( x)]4 1 f ( x)    20   20  40000
dx

0

4
0  x  20
elsewhere
2. In a large class, on exam 1, the mean and standard deviation of scores were 78 and
20, respectively, For exam 2, the mean and standard deviation were 72 and 15,
respectively. The covariance of the exam scores was 80. Give the mean and standard
deviation of the sum of the two exam scores. Assume all students took both exams.
E ( X1  X 2 )  E ( X 1 )  E ( X 2 )  78  72  150
V ( X 1  X 2 )  V ( X 1 )  V ( X 2 )  2COV ( X 1  X 2 )  (20)2  (15)2  2(80)  785  X 1  X 2  28.0
3. The daily number of arrivals to a rural emergency room is a Poisson random variable
with a mean of 81 people per day. Use the normal approximation to the Poisson
distribution to obtain the approximate probability that 70 or fewer people arrive in a
day.
X ~ Poisson (  81)   X    81,  X    9
Y
X 

~ N (0,1) (Approxima tely)
70  81


P( X  70)  P Z 
 1.22   .5  .3888  .1112
9


4. An airport baggage handling system is made up of a series of n=3 conveyor systems
(each bag travels across conveyor 1, then conveyor 2, etc). Each conveyors’ times
between unplanned maintenance are exponentially distributed with =1200 hours.
Assume times to breakdown of the components are independent. The system must be
shut down when any of the 3 conveyors breaks down. The probability density
function for the smallest order statistic is:
g1 ( x) 
n

e  nx / 
x0
a) Give the mean and standard deviation of the times to breakdown (Hint: what
is this distribution?). Give it in terms of numbers, not symbols.
X (1) ~ Exponentia l( *   / n  1200 / 3  400)
E ( X (1) )  400  X (1)  400
b) What is the probability the machine breaks down in less than 500 hours?
500
P( X (1)  500) 
1
 400 e
0
 x / 400
dx   e x / 400
500
0
 e1.25  (1)  1  e1.25  .7135
5. A store sells washers and dryers. The joint distribution of the numbers of washers
(W) and dryers (D) purchased by households entering the store is given below.
p(w,d)
w=0
w=1
d=0
0.5
0.2
d=1
0.1
0.2
a) Give the marginal distributions of washers sold and dryers sold.
w
0
1
p(w)
0.6
0.4
d
0
1
p(d)
0.7
0.3
b) Give the conditional distribution of dryers sold when 0 washers are sold.
When 1 washer is sold.
d
0
1
sum
p(d|w=0)
0.5/0.6=.833
0.1/0.6=.167
1
p(d|w=1)
0.2/0.4=.5
0.2/0.4=.5
1
c) Are the number of dryers and washers sold independent? Yes or No
6. Based on the Central Limit Theorem:
a) Give the approximate sampling distribution for the following quantity based on
random samples of independent observations:
X
X 1    X 100
100
where E ( X i )  80 V ( X i )  900
900


X ~ N   X  80, X2 
 9  (Approxima tely)
100


b) What is the approximate probability the sample mean will be between 77 and 83?
 77  80 X   X 83  80 
  P(1  Z  1)  .6826
P(77  X  83)  P


 3


3
X


7. A game is divised where each of the following prize amounts are placed at random in
cases, and you choose one case.
2
20
200
2000
20000
a) What is your Expected winnings after you select the case?
1
1
1
1
1
E (W )  (2)   (20)   (200)   (2000)   (20000)   4444.4
5
5
5
5
5
b) You then choose 2 cases (that you have not selected). What is the highest amount
your expected winnings can be now? The lowest?
Note: Due to confusion over this part, no points were taken off, but here’s answer:
Highest: You open 2, 20. Then Your expected winnings are now (conditional on
seeing that 2, 20 are not in yours): (200+2000+20000)/3 = 7400
Lowest: You open 2000, 20000. Conditional Expected winnings: (2+20+200)/3=74
8. A designer buys fabric from 2 suppliers. Supplier A provides 70% of the fabric,
supplier B provides the remaining 30%. Supplier A has a defect rate of 3%. Supplier
B has a defect rate of 8%. Given a batch of fabric is defective, what is the probability
it came from Supplier B?
P( A)  .70 P( D | A)  .03  P( AD)  .021
P( B)  .30 P( D | B)  .08  P( BD )  .024
P( D)  P( AD)  P( BD )  .045
P( B | D) 
P( BD ) .024

 .5333
P( D) .045
9. The moment-generating function for a Gamma() random variable is:
M(t)=[1-t]
Use this to derive the mean of a random variable X~Gamma().
M ' (t )  [1  t ] 1 (  )  M ' (0)  [1   (0)] 1 (  )  
10. Suppose that X1 and X2 have the following joint probability distribution:
1 0  x1  1 0  x 2  1
f ( x1 , x 2 )  
otherwise
0
Give the probability that X1+X2 > 1.2
1
By Calculus : P( X 1  X 2  1.2) 
1
1
1
1
 1dx dx   x
2
0.21.2  x1
1
dx1 
2
0.2
1.2  x1
 1  (1.2  x )dx1
1
0.2
1
1
 x2

1

  ( x1  0.2)dx1   1  0.2 x1     0.2   .02  0.04  0.32

 2
 0.2  2
0.2
By Geometry:
P(X1+X2>1.2)=(1/2)(.8)(.8)=.32
1
0.9
0.8
x2
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
x1
0.8
1
11. The central limit theorem states that as we increase our sample size towards infinity,
the distribution of sampled measurements tends to be approximately normal,
regardless of the shape of the population of individual measurements.
True or False
Assuming independent observations from distribution with finite mean and
variance, the SAMPLE MEAN has a sampling distribution that is approximately
normal
12. Let X1,…,Xn be independent and identically distributed Uniform(0,1). Let Yn be the
minimum of X1,…,Xn. Then Yn has cumulative distribution function:
0 x0


G1 ( x)  P(Yn  x)  1  (1  x) n 0  x  1

1 x 1

Show that Yn converges in Probability to 0 by showing the limiting behavior of
lim P| Yn  0 |   for arbitrary   0
n 
A sequence of rv’s X1,...,Xn converges in probability to a constant c if:
lim P| X n  c |    1   0
n 
Yn  min( X 1 ,..., X n ) c  0
1
 1

P(| Yn  0 |  )  P(Yn   )  
n
1  (1   ) 0    1
Considerin g non - trivial case : 0    1 :
lim P(| Yn  0 |  )  lim 1  (1   ) n  1  lim (1   ) n  1  0  1
n 
n 
n 
P
 Yn  0
13. Which of the following sampling plans is binomial and which is geometric? Give
P(Y=y)=p(y) as a function of y for each case. Assume outcomes of individual trials
are independent with constant probability of success (e.g. Bernoulli trials).
a) A clinical trial enrolls 15 children with a rare disease. Each child is given an
experimental therapy, and the number of children showing marked improvement is
observed. The true underlying probability of success is 0.40.
Distribution: Binomial
15 
p(y)=  (0.40) y (0.60)15 y
 y
y  0,1,...,15
b) A missile designer will launch missiles until one successfully reaches its target. The
underlying probability of success is 0.20.
Distribution: Geometric
p(y)= (0.80) y 1 (0.20)
y  1,2,....
Have a great summer!!!