Engineering Probability 2019 Class 18 and Exam 2 Answers - Thu 2018-03-21
Name, RCSID: W. Randolph Franklin answers
Rules:
- You have 80 minutes.
- You may bring two 2-sided 8.5"x11" paper with notes.
- You may bring a calculator.
- You may not share material with each other during the exam.
- No collaboration or communication (except with the staff) is allowed.
- Check that your copy of this test has all seven pages.
- Do all questions.
- Each part of a question is worth 5 points.
- When answering a question, don't just state your answer, prove it.
Questions:
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You're designing a circuit with a chip.
- The chip was manufactured in one of two plants, but you don't know which.
- The random variable Y says which plant the chip came from. Y=1 or 2.
- Half the chips come from each plant.
- Chips from the first plant (Y=1) have a lifetime X that is U[0,10].
- For chips from the second plant (Y=2), X is U[0,20].
You pick a random chip, power it up, and it's still working at time X=5.
Answer the following questions.
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What are $f(X=x|Y=1)$ and $f(X=x|Y=2)$?
$f(X=x|Y=1) = \begin{cases}0.1 & for& 0\le x\le 10\\ 0 & otherwise\end{cases}$
$f(X=x|Y=2) = \begin{cases}0.05 & for& 0\le x\le 20\\ 0 & otherwise\end{cases}$
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What are $f(X=x \,\&\, Y=1)$ and $f(X=x \,\&\, Y=2)$?
$P[Y=1] = 1/2$
$f(X=x \,\&\, Y=1) = f(X=x|Y=1) P[Y=1] = \begin{cases}0.05 & for& 0\le x\le 10\\ 0 & otherwise\end{cases}$
$f(X=x \,\&\, Y=2) = \begin{cases}0.025 & for& 0\le x\le 20\\ 0 & otherwise\end{cases}$
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What is $f_X(x)$ ?
$f_X(x) = f(X=x \,\&\, Y=1) + f(X=x \,\&\, Y=2) \\ = \begin{cases}0.075 & for& 0\le x\le 10\\ 0.025 & for& 10\le x\le 20\\ 0 & otherwise\end{cases}$
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What is $P[Y=1|X=5]$ ?
There are different ways to do this. Here's one, which works with $P[5\le X\le5+d]$ for some small positive $d$.
$P[Y=1|X=5] = \lim_{d\rightarrow 0} P[Y=1|5\le X\le5+d] $
$P[Y=1|5\le X\le5+d] = P[Y=1 \& 5\le X\le5+d] / P[5\le X\le5+d]$
$= .05d / ( .075d) = 2/3 $
Note that this is different from asking what are the probabilities for the two source for chips that are still alive at time 5. This is asking about chips whose lifetime is 5, not 5 or more.
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Assume that a bicycle tire's pdf for failing at time x is: $f(x) = c(1/4 - x + x^2), \,\, 0\le x\le1 $
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What is $c$?
Integrating f(x) from 0 to 1 gives 1/12, so c=12.
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What is the mean?
$E[X] = \int_0^1 x f(x) = 1/2$
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What is the variance?
$E[X^2] = \int_0^1 x^2 f(x) = 2/5$
$VAR[X] = 2/5 -(1/2)^2 = 3/20$.
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What is the cdf?
$F_X(x) = \begin{cases} 0 & if& x\le0 \\ 3x-6x^2+4x^3&if&0\le x\le1\\1 &if&1\le x \end{cases} $
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What is the reliability?
$R(x) = 1 - F_X(x) = \begin{cases} 1 & if & x\le0 \\ 1-(3x-6x^2+4x^3) & if & 0\le x\le1\\0 & if & 1\le x \end{cases}$
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What is the MTTF?
$\int_0^1 R(x) dx = (x- 3/2 x^2 + 2x^3 - x^4)|_0^1 = 0.5$
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This question is about a web server with service requests coming independently at an average of 1 per second.
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What is the probability that in one hour there are more than 3720 hits? Here I want a number, not just a formula. The table at the end of this exam is useful.
The thinking part of this question is this: The distribution of the number of hits in one hour is Poisson with a=3600. So, the mean is 3600 and the std 60. The Gaussian is an excellent approximation near the mean.
3720=m+2s. P=Q(2)=0.023.
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Use the Markov inequality to compute an approximation to the above answer.
P <= 3600/3720 = 0.97. (i.e., almost meaningless).
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Use the Chebyshev inequality to compute an approximation to the above answer.
P[| X-3600 | >= 120] <= 60^2/120^2 = 1/4.
Both tails are equally probable because we're quite far from the origin. So
P[X>=3720] <= 1/8.
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Let X be a normal random variable with mean 1000 and standard deviation 10. Give the following numbers, using the supplied table.
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P[X>1015].
That will be Q(1.5) in the table = 0.067.
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P[990<X<1010].
That will be F(1) - F(-1) = 0.68. approx.
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For what value of x is P[X<x]=.1?
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Look for F(x)=.1 in the table. That's about -1.3.
Our x will be 1000+10(-1.3) = 987.
Normal distribution:
x f(x) F(x) Q(x)
-3.0000 0.0044 0.0013 0.9987
-2.9000 0.0060 0.0019 0.9981
-2.8000 0.0079 0.0026 0.9974
-2.7000 0.0104 0.0035 0.9965
-2.6000 0.0136 0.0047 0.9953
-2.5000 0.0175 0.0062 0.9938
-2.4000 0.0224 0.0082 0.9918
-2.3000 0.0283 0.0107 0.9893
-2.2000 0.0355 0.0139 0.9861
-2.1000 0.0440 0.0179 0.9821
-2.0000 0.0540 0.0228 0.9772
-1.9000 0.0656 0.0287 0.9713
-1.8000 0.0790 0.0359 0.9641
-1.7000 0.0940 0.0446 0.9554
-1.6000 0.1109 0.0548 0.9452
-1.5000 0.1295 0.0668 0.9332
-1.4000 0.1497 0.0808 0.9192
-1.3000 0.1714 0.0968 0.9032
-1.2000 0.1942 0.1151 0.8849
-1.1000 0.2179 0.1357 0.8643
-1.0000 0.2420 0.1587 0.8413
-0.9000 0.2661 0.1841 0.8159
-0.8000 0.2897 0.2119 0.7881
-0.7000 0.3123 0.2420 0.7580
-0.6000 0.3332 0.2743 0.7257
-0.5000 0.3521 0.3085 0.6915
-0.4000 0.3683 0.3446 0.6554
-0.3000 0.3814 0.3821 0.6179
-0.2000 0.3910 0.4207 0.5793
-0.1000 0.3970 0.4602 0.5398
0 0.3989 0.5000 0.5000
0.1000 0.3970 0.5398 0.4602
0.2000 0.3910 0.5793 0.4207
0.3000 0.3814 0.6179 0.3821
0.4000 0.3683 0.6554 0.3446
0.5000 0.3521 0.6915 0.3085
0.6000 0.3332 0.7257 0.2743
0.7000 0.3123 0.7580 0.2420
0.8000 0.2897 0.7881 0.2119
0.9000 0.2661 0.8159 0.1841
1.0000 0.2420 0.8413 0.1587
1.1000 0.2179 0.8643 0.1357
1.2000 0.1942 0.8849 0.1151
1.3000 0.1714 0.9032 0.0968
1.4000 0.1497 0.9192 0.0808
1.5000 0.1295 0.9332 0.0668
1.6000 0.1109 0.9452 0.0548
1.7000 0.0940 0.9554 0.0446
1.8000 0.0790 0.9641 0.0359
1.9000 0.0656 0.9713 0.0287
2.0000 0.0540 0.9772 0.0228
2.1000 0.0440 0.9821 0.0179
2.2000 0.0355 0.9861 0.0139
2.3000 0.0283 0.9893 0.0107
2.4000 0.0224 0.9918 0.0082
2.5000 0.0175 0.9938 0.0062
2.6000 0.0136 0.9953 0.0047
2.7000 0.0104 0.9965 0.0035
2.8000 0.0079 0.9974 0.0026
2.9000 0.0060 0.9981 0.0019
3.0000 0.0044 0.9987 0.0013
End of exam 2, total 80 points.