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EF 507
REVIEW FOR MIDTERM EXAM
FALL 2007
Chapter 5
The following table displays the joint probability distribution of two random variables
X and Y.
Y
0
1
2
0
0.13
0.16
0.07
X
1
0.08
0.14
0.04
2
0.21
0.03
0.13
Determine the marginal probability distribution of X.
ANSWER:
x
0
1
2
P(x)
0.36
0.26
0.38
Calculate the mean of X.
ANSWER:
 X   xP  x   1.02
Calculate the standard deviation of X.
ANSWER: Since  X2   x2 P  x    X2  1.78  1.02  0.7396 , then  X  0.86 .
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Determine the marginal probability distribution of Y.
ANSWER:
y
0
1
2
P(y)
0.43
0.33
0.24
Calculate the mean of Y.
ANSWER:
Y   yP  y   0.81
Calculate the standard deviation of Y.
ANSWER:  Y2   y 2 P  y   Y2  1.29   0.81  0.6339 . Hence,  Y  0.7962
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Calculate the covariance between X and Y.
ANSWER: Cov(X, Y) =  xyP  x, y    X Y = 0.80 -(1.02)(0.81) = - 0.0262
x
y
Calculate the correlation between X and Y.
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ANSWER:
  Corr  X , Y  
Cov  X , Y 
 X Y

0.262
 0.0383
 0.86 0.7961
Chapter 6
1. A normal random variable x has an unknown mean  and standard deviation
  2.5 If the probability that x exceeds 7.5 is 0.8289, find  .
ANSWER:
It is given that x is normally distributed with  = 2.5 but with unknown mean  , and
that P (x > 7.5) = 0.8289. In terms of the standard normal random variable z, we can
write
P( X  7.5)  P[ Z  (7.5   ) / 2.5]  0.8289
Since the area to the right of (7.5   ) / 2.5 is greater than 0.5, then (7.5   ) / 2.5 must
be negative, and that P[ (7.5   ) / 2.5 < z < 0] = 0.3289. Hence,
(7.5   ) / 2.5 = -.095. This implies  = 9.875.
2. Suppose that the time between successive occurrences of an event follows an
exponential distribution with mean 1/  minutes. Assume that an event occurs.
a) Show that the probability that more than 4 minutes elapses before the occurrence of
the next event is e 4  .
ANSWER:
P(X > 4) = 1 – P(X  4) = 1 – F(4) = 1 - [1 - e 4  ] = e 4 
b) Show that the probability that more than 8 minutes elapses before the occurrence of
the next event is e 8 .
ANSWER:
P(X > 8) = 1 – P(X  8) = 1 – F(8) = 1 - [1 - e 8  ] = e 8 
c) Using the results of (a) and (b), show that if 4 minutes have already elapsed, the
probability that a further 4 minutes will elapse before the next occurrence is e
Explain your answer in words.
4 
.
ANSWER: P(X > 8 | X >4) = P(X > 8) / P(X > 4) = e 8 / e 4  = e 4 
The probability of an occurrence within a specified time in the future is not related to
how much time has passed since the most recent occurrence.
3. A random variable X is normally distributed with mean of 50 and variance of 50,
and a random variable Y is normally distributed with mean of 100 and variance of
200.
Given the random variables X and Y have a correlation coefficient equal to -0.50, find
the mean and variance of the random variable W = 4X+3Y.
ANSWER: X  50,  X2  50, Y  100,  Y2  200, Corr( X ,Y )  0.50
W  4 X  3Y  (4)(50) + (3)(100) = 500
2
W2  (4)2 X2  (3)2 Y2  2(4)(3)Corr( X , Y ) X Y
= (16)(50)+(9)(200) + (24)(-0.50)(7.071)(14.142) = 1,400.023
4. You are the Webmaster for your firm’s Website. From your records, you know that
the probability that a visitor will buy something from your firm is 0.23. If the number
of visitors in one day is 952, what is the probability that less than 200 of them will
buy something from your firm? Use the normal approximation to the binomial
without the continuity correction.
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ANSWER: n = 952, P = 0.23,  =E(X) = 218.96,  =Var(X) = nP(1-P) = 168.5992,
 = 12.985
P(X<200) =P(Z<-1.46) = 0.0721
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