ECSE-2500 Engineering Probability, RPI, Spring 2011 Home Page

# ECSE-2500 Engineering Probability, Exam 2, April 5, 2011

*(for grader use only) Finish order _________ Grade _______________ /30*

NAME: _ W. Randolph Franklin ____Solution_________________________

EMAIL:___wrf@ecse.rpi.edu______________________ 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 other students.

- 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 millisecond.
- [_____/1] What's the relevant probability distribution? Poisson
- [_____/1] What's the value of the relevant parameter? α=4.4
- [_____/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.
- [_____/2] What's the probability that there will be zero decays? Write this as a formula. You don't need to give a number. {$ \frac{\alpha^0}{0!} e^{-\alpha} = e^{-4.4} = 0.01. $} You don't need to give the 0.01.
- [_____/1] What's the relevant probability distribution for the time between consecutive decays? Exponential, λ=4.4
- [_____/1] What's the mean time between consecutive decays? {$ \mu=\frac{1}{\lambda} = \frac{1}{4.4} $} msec.
- [_____/1] What's the standard deviation of the time between consecutive decays? {$ \sigma=\frac{1}{\lambda} = \frac{1}{4.4} $} msec.
- [_____/1] What's the probability that the time between consecutive decays is greater than one 1msec? {$ f(x)=\lambda e^{-\lambda x}, F(x) = \int_0^x e^{-\lambda x_1} dx_1 = 1-e^{-\lambda x}. $} Answer = {$ e^{-\lambda} = e^{-4.4} = 0.01 $}

- 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.
Joint pmf f _{X,Y}X=0 X=1 Y=0 .4 .2 Y=1 .1 .3 - [_____/1] What's the marginal pmf of X? {$ f_x(0) = f_x(1) = .5 $}
- [_____/1] What's the mean of X? .5
- [_____/1] What's the variance of X? E[X
^{2}]=.5, Var[X]=.5-.5*.5 = .25 - [_____/2] What's the mean of XY? .3
- [_____/2] What's the covariance of X and Y? E[Y]=.4, Cov[X,Y]=E[XY]-E[X]E[Y]=.3-.5*.4=.1
- [_____/2] What's the correlation coefficient of X and Y?
E[Y
^{2}]= .4, Var[Y]= .4-.4^{2}=.24, {$ \sigma_x=.5, \sigma_y=\sqrt{.24}\approx .5, \rho_{XY}=\frac{Cov[X,Y]}{\sigma_X\sigma_Y}\approx .4 $}. See page 259. - [_____/1] If you read a 0, what's the probability that you're wrong? Read it off the table: .2/.6 = 1/3

- [_____/10] Evaluate and give me the answer as an expression perhaps involving {$\pi$} and {$\sqrt{\cdot} $}. Show your work. {$$ {\cal A} = \int_{0}^\infty e^{-x^2} dx $$} Hint: Transform this to something closer to {$$ \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx =1 $$} . Let {$ \mu=0, \sigma=\frac{1}{\sqrt{2}} $}, so {$ \int_{-\infty}^\infty \frac{1}{\sqrt{\pi}} e^{-x^2} dx=1 $} . Therefore the answer is {$ \int_{0}^\infty e^{-x^2} dx = \frac{(\sqrt{\pi})}{2} $} .
- You're bicycling the Alaska Highway. Assume that the lifetime of a
bicycle tire is exponential with mean {$\mu=3$} days. You carry 3
spares.
- [_____/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. - [_____/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.
- [_____/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 days. 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.
- [_____/1] Assume that your front tire is two days old, and is still good. What's its expected future lifetime? The exponential distribution is memoryless, so the expected future lifetime is 3 days.
- [_____/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 {$ \lambda=\frac{1}{24} $}. P[failure in 1st hour] = {$ 1-e^{-1/24}=0.04 $}. For two tires, P[both fail in 1st hour]=0.04
^{2}=0.0016. It's ok to give the expression w/o evaluating it. Nevertheless, it could be computed by hand by using the approximation {$ e^x \approx 1+x $} when {$ |x| $} is small.

- [_____/3] What is the expected number of days before you do not have 2
good tires to bicycle on?

*(end of exam)*