Unformatted text preview: ECSE-2500, Engineering Probability, Spring 2011, Exam 2 Solution (Eng... 1 of 4 http://www.ecse.rpi.edu/Homepages/wrf/pmwiki/EngProbSpring2011/E... To print higher-resolution math symbols, click the
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using the Options pane of the jsMath control panel. ECSE-2500 Engineering Probability, RPI, Spring 2011 Home Page (for grader use only) Finish order _________ Grade _______________ /30
NAME: _ W. Randolph Franklin ____Solution_________________________
EMAIL:[email protected]______________________ RIN:__________________________
There are 5 pages with 4 questions. Answer any 3 questions. (If you answer 4, then we will grade
the first 3.)
We record the order in which you hand in your exam the better to keep track of the exams, and
sometimes to plot scatterplots of order vs grade. (There is never any significant correlation.)
Your order does not affect your grade.
This exam is closed book but a calculator and two 2-sided letter-paper-size note sheets are
allowed. You may not use other computers or communication devices, or share material with
1. In your body, an average of 4400 radioactive potassium atoms decay each second. (ref:
http://en.wikipedia.org/wiki/Becquerel). You're interested in the number of decays in one
a. [_____/1] What's the relevant probability distribution? Poisson
b. [_____/1] What's the value of the relevant parameter? α=4.4
c. [_____/2] What are the mean and standard deviation of the number of decays in one
millisecond? μ=4.4, σ=sqrt(4.4), approx 2.1.
d. [_____/2] What's the probability that there will be zero decays? Write this as a formula. You
don't need to give a number. 0! e− = e−4 4 = 0 01 You don't need to give the 0.01. e. [_____/1] What's the relevant probability distribution for the time between consecutive
decays? Exponential, λ=4.4
f. [_____/1] What's the mean time between consecutive decays? = 1 = 1
44 msec. g. [_____/1] What's the standard deviation of the time between consecutive decays?
= 1 = 414 msec. 4/24/2011 11:06 AM ECSE-2500, Engineering Probability, Spring 2011, Exam 2 Solution (Eng... 2 of 4 http://www.ecse.rpi.edu/Homepages/wrf/pmwiki/EngProbSpring2011/E... h. [_____/1] What's the probability that the time between consecutive decays is greater than
one 1msec? f (x ) = e − x F (x ) = x −x
0e = 1 − e− x Answer = e− = e−4 4 = 0 01 2. I write a digit that is either a 0 or a 1. You have trouble reading my writing. This table shows
the joint probability of what I write and you read. X is the random variable for what I write, Y
the random variable for what you read.
Y=0 .4 .2
Y=1 .1 .3
a. [_____/1] What's the marginal pmf of X? fx (0) = fx (1) = 5
b. [_____/1] What's the mean of X? .5
c. [_____/1] What's the variance of X? E[X2]=.5, Var[X]=.5-.5*.5 = .25
d. [_____/2] What's the mean of XY? .3
e. [_____/2] What's the covariance of X and Y? E[Y]=.4, Cov[X,Y]=E[XY]-E[X]E[Y]=.3-.5*.4=.1
f. [_____/2] What's the correlation coefficient of X and Y?
E[Y2]= .4, Var[Y]= .4-.42=.24, x =5 y = 24 5 XY = Cov[X Y ]
XY 4 . See page 259. g. [_____/1] If you read a 0, what's the probability that you're wrong? Read it off the table: .2/.6
= 1/3 3. [_____/10] Evaluate and give me the answer as an expression perhaps involving
Show your work. and . 2 e−x dx =
Hint: Transform this to something closer to 1
2 e − (x− 2)
2 2 dx = 1 .
Let =0 = 1
2 − 1 2 e−x dx = 1 . Therefore the answer is 2 0 e−x dx = ( )
2 . 4/24/2011 11:06 AM ECSE-2500, Engineering Probability, Spring 2011, Exam 2 Solution (Eng... 3 of 4 http://www.ecse.rpi.edu/Homepages/wrf/pmwiki/EngProbSpring2011/E... 4. You're bicycling the Alaska Highway. Assume that the lifetime of a bicycle tire is exponential
with mean = 3 days. You carry 3 spares.
a. [_____/3] What is the expected number of days before you do not have 2 good tires to bicycle
on? (Hint: You do not need to use any distribution that we didn't study. In fact, there is a
relatively easy way to compute the answer.)
Wrong answer, which we'll accept anyway: That happens when 4 tires fail. Expectations add,
so this is 3*4=12.
Correct answer: Since you're using, and wearing out, two tires at a time, tires are failing at
the rate of one every 1.5 days, on average. Four tires will have failed after 6 days.
b. [_____/1] Let's assume that the answer is a mean of 5 days. (It is not.) Using only that info,
and nothing else about the distribution, give an upper bound on the probability that you're
still bicycling after 10 days.
This is the Markov inequality, page 181. 5/10 = 1/2.
c. [_____/2] Let's further assume that the answer to the first part has a standard deviation of 5
days. (It does not.) Using only that mean and standard deviation info, and nothing else about
the distribution, give an upper bound on the probability that you're still bicycling after 10
This is the Chebyshev inequality. Since P[X<=0]=0, P[X>=10] <=1. I.e., the Chebyshev inequality
is useless here. It would be even better (but not required) to observe that you can still use the
Markov inequality here, and since it's better, use it.
d. [_____/1] Assume that your front tire is two days old, and is still good. What's its expected
The exponential distribution is memoryless, so the expected future lifetime is 3 days.
e. [_____/3] What's the probability that both tires will fail in the first hour? It's ok to write an
expression with an exponential. It's ok to ignore the probability that a replacement tire will
also fail since that's low. Assume that the tires are independent (though in the real world
their failures could be correlated.)
We must assume a certain number of hours per bicycling day, say 8. One tire failure is
exponential with a mean time to failure of 24 hours, so = 24 . P[failure in 1st hour] = 1 − e− 1 24 = 0 04. For two tires, P[both fail in 1st hour]=0.042=0.0016. It's ok to give the expression w/o evaluating it. Nevertheless, it could be computed by hand by using the
1 + x when x is small. (end of exam) Material created by W. Randolph Franklin and others is Copyright © 1994-2010, by the authors, This work is licensed under
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