STA 4321 – Final Exam PRINT Name _____________
1. A game consists of 3 random numbers being selected from the Uniform density over
the range [0,10] independently. That is X1,X2,X3 ~ iid U(0,10), with probability
density function:
1 / 10 0  x  10
f ( x)  
otherwise
 0
a) Give the cumulative distribution function for this distribution and sketch it.
0 x0

1
x

For x  [0,10] F ( x)   f (t )dt   dt 
0  x  10  F ( x)   x / 10 0  x  10
10
10

0
 1 x  10

x
x
Uniform CDF
1.2
1
F(x)
0.8
0.6
0.4
0.2
0
-0.2 -10
-5
0
5
10
15
20
x
b) The game pays off the largest of the three values, that is: X(3) =
max(X1,X2,X3). Give the cumulative distribution function and probability
density function for X(3).
G3 ( x)  P( X ( 3)
0 x0


 x)  P( X 1  x, X 2  x, X 3  x)  [ F ( x)]3  [ x / 10]3 0  x  10

1 x  10

  x 2  1  3x 2
dG3 ( x)

3 1
g 3 ( x) 
 3[ F ( x)] f ( x)  3 10   10   1000
dx

0

0  x  10
elsewhere
2. In a large class, on exam 1, the mean and standard deviation of scores were 72 and
15, respectively, For exam 2, the mean and standard deviation were 68 and 20,
respectively. The covariance of the exam scores was 120. Give the mean and standard
deviation of the sum of the two exam scores. Assume all students took both exams.
E ( X 1  X 2 )  E ( X 1 )  E ( X 2 )  72  68  140
V ( X1  X 2 )  V ( X 1 )  V ( X 2 )  2COV ( X1  X 2 )  (15)2  (20)2  2(120)  865  X 1  X 2  29.4
3. The daily number of arrivals to a rural emergency room is a Poisson random variable
with a mean of 100 people per day. Use the normal approximation to the Poisson
distribution to obtain the approximate probability that 112 or more people arrive in a
day.
X ~ Poisson (  100)   X    100,  X    10
Y
X 

~ N (0,1) (Approxima tely)
112  100


P( X  112)  P Z 
 1.12   .5  .3849  .1151
10


4. An airport baggage handling system is made up of a series of n=4 conveyor systems
(each bag travels across conveyor 1, then conveyor 2, etc). Each conveyors’ times
between unplanned maintenance are exponentially distributed with =1000 hours.
Assume times to breakdown of the components are independent. The system must be
shut down when any of the 4 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  1000 / 4  250)
E ( X (1) )  250  X (1)  250
b) What is the probability the machine breaks down in less than 200 hours?
200
P( X (1)  200) 
1
 250 e
0
 x / 250
dx   e  x / 250
200
0
 e 0.8  (1)  1  e 0.8  .5507
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
0.6
0.1
d=1
0.1
0.2
a) Give the marginal distributions of washers sold and dryers sold.
w
0
1
p(w)
0.7
0.3
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)
0.6/0.7=.857
0.1/0.7=.143
1
p(d|w=1)
0.1/0.3=.333
0.2/0.3=.667
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 )  100 V ( X i )  400
400


X ~ N   X  100, X2 
 4  (Approxima tely)
100


b) What is the approximate probability the sample mean will be between 96 and 104?
 96  100 X   X 104  100 
  P(2  Z  2)  .9544
P(96  X  104)  P




2

2
X


7. A game is divised where each of the following prize amounts are placed at random in
cases, and you choose one case.
1
10
100
1000
10000
a) What is your Expected winnings after you select the case?
1
1
1
1
1
E (W )  (1)   (10)   (100)   (1000)   (10000)   2222.2
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 1, 10. Then Your expected winnings are now (conditional on
seeing that 1, 10 are not in yours): (100+1000+10000)/3 = 3700
Lowest: You open 1000, 10000. Conditional Expected winnings: (1+10+100)/3=37
8. A designer buys fabric from 2 suppliers. Supplier A provides 60% of the fabric,
supplier B provides the remaining 40%. Supplier A has a defect rate of 5%. Supplier
B has a defect rate of 8%. Given a batch of fabric is defective, what is the probability
it came from Supplier A?
P( A)  .60 P( D | A)  .05  P( AD)  .030
P( B)  .40 P( D | B)  .08  P( BD )  .032
P( D)  P( AD)  P( BD )  .062
P( A | D) 
P( AD) .030

 .4839
P( D) .062
9. The moment-generating function for a Binomial random variable is:
M(t)=[pet+(1-p)]n
Use this to derive the mean of a random variable X that is binomial with n trials, and
probability of success p.
M ' (t )  n[ pet  (1  p)]n 1 ( pet )  M ' (0)  n[ p(1)  1  p]n 1 p(1)  np
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 > 1.5
1
0.9
0.8
0.7
0.6
0.5
By Geometry:
P(X1+X2>1.5)=(1/2)(1/2)(1/2)=1/8
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
1
By Calculus : P( X 1  X 2  1.5) 

1
1
1
1dx2dx1   x2
0.51.5  x1
0.5
1
 1  (1.5  x )dx1
dx1 
1.5  x1
1
0.5
1
 x12

1
 1
 1
  ( x1  0.5)dx1    0.5 x1     0.5     0.25  
 8
 8
 2
 0.5  2
0.5
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) for each case. Assume outcomes of individual trials are independent
with constant probability of success (e.g. Bernoulli trials).
a) A missile designer will launch missiles until one successfully reaches its target. The
underlying probability of success is 0.40.
Distribution: Geometric
p(y)= (0.60) y 1 (0.40)
y  1,2,....
b) A clinical trial enrolls 20 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 0.60.
Distribution: Binomial
 20 
p(y)=  (0.60) y (0.40) 20 y
 y
y  0,1,...,20
Have a great summer!!!