#Prob ECSE-2500 Engineering Probability, Spring 2019, Rensselaer Polytechnic Institute (Posts about exam)

Engineering Probability Exam 3 - Sat 2019-05-04

W Randolph Franklin (WRF), RPI — Sat, 04 May 2019 04:00:00 GMT

Name, RPI email:

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Rules:

You have 80 minutes.
You may bring three 2-sided 8.5"x11" papers 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 nine pages.
Each part of a question is worth 5 points.
When answering a question, don't just state your answer, prove it.
You may write FREE as your answer for two questions, and get the 5 points.

Questions:

These few questions are about the population of adult males, which has a mean of 70 inches and a standard deviation of 4 inches.
1. What is the probability that a particular person's height is between 68 and 74?
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2. If we take a sample of 100, what is its mean?
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3. What is its standard deviation?
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These questions are about tossing 3 fair dice and looking at the 3 numbers that show. However, these dice have only 2 faces (to make this question easier). The faces are labeled 1 and 2.
1. What's the expected value of the number showing on the first die?
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2. What's the pmf of the smallest die?
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3. What's the expected value of the smallest die?
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4. What's the probability that the smallest number is 1 given that the first number is 2?
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5. What's the probability that the first number is 2 given that the smallest number is 1?
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6. Are the first number and the smallest number are independent? Prove your answer.
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7. What is the MAP estimator for the smallest number, given that the first number is 2?
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This question is about a continuous probability distribution on 2 variables.

$$f_{XY}(x,y) = \begin{cases} c (x+y) & \text{ if } (0\le x) \ \& \ (0\le y)\ \& \ (0\le x+y \le 1) \\ 0 & \text{ otherwise}\end{cases}$$

The nonzero region is the triangle with vertices (0,0), (1,0) and (0,1).

c is some constant, but I didn't tell you what it is.
1. What is c?
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2. What is $F_{XY}(x,y)$?
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3. What is $f_X(x)$?
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4. Are X and Y independent?
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5. What is $P[X\le Y]$ ?
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6. What is $E[X]$?
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7. What is $COV[X,Y]$?
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8. What is $\rho_{X,Y}$?
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9. What is $f_Y(y|x)$?
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10. What is $E[Y|x]$?
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I'm comparing two types of widgets, red and blue. Assume that the probability of each widget dieing in a small interval dt, given that it was alive at the start, is independent of its age. Assume that the probability of the red widget dieing in the next hour is 0.1%, for the blue, it's 0.01%.
1. Give the pdf for the red widget's lifetime. (You have enough info to do this; there is only one possible probability distribution.)
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2. If you have 100 red widgets, what's the probability that their mean lifetime is within 10% of the population mean?
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3. If you start two widgets at the same time, what's the probability that the red widget will last longer?
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Normal distribution:

x          f(x)      F(x)      Q(x)                x          f(x)      F(x)      Q(x)
-3.0000    0.0044    0.0013    0.9987             0.1000    0.3970    0.5398    0.4602
-2.9000    0.0060    0.0019    0.9981             0.2000    0.3910    0.5793    0.4207
-2.8000    0.0079    0.0026    0.9974             0.3000    0.3814    0.6179    0.3821
-2.7000    0.0104    0.0035    0.9965             0.4000    0.3683    0.6554    0.3446
-2.6000    0.0136    0.0047    0.9953             0.5000    0.3521    0.6915    0.3085
-2.5000    0.0175    0.0062    0.9938             0.6000    0.3332    0.7257    0.2743
-2.4000    0.0224    0.0082    0.9918             0.7000    0.3123    0.7580    0.2420
-2.3000    0.0283    0.0107    0.9893             0.8000    0.2897    0.7881    0.2119
-2.2000    0.0355    0.0139    0.9861             0.9000    0.2661    0.8159    0.1841
-2.1000    0.0440    0.0179    0.9821             1.0000    0.2420    0.8413    0.1587
-2.0000    0.0540    0.0228    0.9772             1.1000    0.2179    0.8643    0.1357
-1.9000    0.0656    0.0287    0.9713             1.2000    0.1942    0.8849    0.1151
-1.8000    0.0790    0.0359    0.9641             1.3000    0.1714    0.9032    0.0968
-1.7000    0.0940    0.0446    0.9554             1.4000    0.1497    0.9192    0.0808
-1.6000    0.1109    0.0548    0.9452             1.5000    0.1295    0.9332    0.0668
-1.5000    0.1295    0.0668    0.9332             1.6000    0.1109    0.9452    0.0548
-1.4000    0.1497    0.0808    0.9192             1.7000    0.0940    0.9554    0.0446
-1.3000    0.1714    0.0968    0.9032             1.8000    0.0790    0.9641    0.0359
-1.2000    0.1942    0.1151    0.8849             1.9000    0.0656    0.9713    0.0287
-1.1000    0.2179    0.1357    0.8643             2.0000    0.0540    0.9772    0.0228
-1.0000    0.2420    0.1587    0.8413             2.1000    0.0440    0.9821    0.0179
-0.9000    0.2661    0.1841    0.8159             2.2000    0.0355    0.9861    0.0139
-0.8000    0.2897    0.2119    0.7881             2.3000    0.0283    0.9893    0.0107
-0.7000    0.3123    0.2420    0.7580             2.4000    0.0224    0.9918    0.0082
-0.6000    0.3332    0.2743    0.7257             2.5000    0.0175    0.9938    0.0062
-0.5000    0.3521    0.3085    0.6915             2.6000    0.0136    0.9953    0.0047
-0.4000    0.3683    0.3446    0.6554             2.7000    0.0104    0.9965    0.0035
-0.3000    0.3814    0.3821    0.6179             2.8000    0.0079    0.9974    0.0026
-0.2000    0.3910    0.4207    0.5793             2.9000    0.0060    0.9981    0.0019
-0.1000    0.3970    0.4602    0.5398             3.0000    0.0044    0.9987    0.0013
      0    0.3989    0.5000    0.5000

End of exam 3, total 100 points.

Engineering Probability Exam 3 Solutions - Sat 2019-05-04

W Randolph Franklin (WRF), RPI — Sat, 04 May 2019 04:00:00 GMT

Name, RPI email:

WRF solutions

OK to give a formula w/o working it out.

Rules:

You have 80 minutes.
You may bring three 2-sided 8.5"x11" papers 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 nine pages.
Each part of a question is worth 5 points.
When answering a question, don't just state your answer, prove it.

Questions:

These few questions are about the population of adult males, which has a mean of 70 inches and a standard deviation of 4 inches.
1. What is the probability that a particular person's height is between 68 and 74?
  
  68 is mean - std/2
  
  74 is mean + std
  
  Q(-.5) - Q(1) = .69 - .16 = .53
2. If we take a sample of 100, what is its mean?
  
  100.
3. What is its standard deviation?
  
  4/sqrt(100) = .4
These questions are about tossing 3 fair dice and looking at the 3 numbers that show. However, these dice have only 4 faces (to make this question easier).
1. What's the expected value of the number showing on the first die?
  
  2.5
2. What's the pmf of the smallest die?
  
  Enumeration works. 8 cases: 111, 112, 121, 122, 211, 212, 221, 222
  
  Let X = smallest die.
  
  p(X=1) = 7/8, p(X=2)=1/8
3. What's the expected value of the smallest die?
  
  1/8 * 2 + 7/8 * 1 = 9/8
4. What's the probability that the smallest number is 1 given that the first number is 2?
  
  Enumerate. p= 3/4
5. What's the probability that the first number is 2 given that the smallest number is 1?
  
  3/7
6. Are the first number and the smallest number are independent? Prove your answer.
  
  p(1st is 2) = 1/2. p(smallest is 2) = 1/8. p(1st is 2 and smallest is 2) = 1/4
  
  1/2 * 1/8 ne 1/4. Not independent
7. What is the MAP estimator for the smallest number, given that the first number is 2?
  
  P[smallest is 1|first is 2] = 3/4, so MAP = 1
This question is about a continuous probability distribution on 2 variables.

$$f_{XY}(x,y) = \begin{cases} c (x+y) & \text{ if } (0\le x) \ \& \ (0\le y)\ \& \ (0\le x+y \le 1) \\ 0 & \text{ otherwise}\end{cases}$$

The nonzero region is the triangle with vertices (0,0), (1,0) and (0,1).

c is some constant, but I didn't tell you what it is.
1. What is c?
  
  The integral is 1/3.
  
  c=3
2. What is $F_{XY}(x,y)$?
  
  $$F_{XY}(x,y)=\begin{cases} 0 & \text{ if } x\le0 \cup y\le0 \\ 1 & \text{ if } x\ge 1 \cap y\ge1 \\ 3/2 (x^2y+xy^2) & \text{ if } 0\le x \cap 0\le y \cap x+y\le1 \\ (\int_0^x\int_0^{1-x} + \int_0^{1-y}\int_{1-x}^y + \int_{1-y}^x\int_{1-x}^{1-x_0}) (3(x_0+y_0) dy_0dx_0) & \text{ otherwise}\end{cases}$$
  
  The last case above splits the nonzero integration region into two rectangles and a triangle.
  
  It's also acceptable to draw a figure and say something intelligent w/o being explicit about all the details.
3. What is $f_X(x)$?
  
  $f_X(x)= \int_0^{1-x}f_{XY}(x,y) dy = 3/2 (1-x)^2, 0\le x\le1$
  
  Note that $\int_0^1 f_X(x)=1$, which is correct.
4. Are X and Y independent?
  
  $f_X(x)= 3/2 (1-x)^2$
  
  $f_Y(y)= 3/2 (1-y)^2$
  
  $f_{XY}(x,y)= 3(x+y)\ne f_X(x) f_Y(y)$
  
  no.
5. What is $P[X\le Y]$ ?
  
  Integrate: $ \int_0^1 \int_0^{\min{x,1-x)} f(x0,y0) dy0 dx0 = 1/2$
  
  That's reasonable because X and Y are symmetric.
  
  The box around the above expression is a meaningless unwanted artifact.
6. What is $E[X]$?
  
  1/8
7. What is $COV[X,Y]$?
  
  E[XY] = 1/10. E[Y] = E[X]. COV = 1/10 - 1/8 * 1/8 = .085.
8. What is $\rho_{X,Y}$?
  
  OK to write the formula
9. What is $f_Y(y|x)$?
  
  In the nonzero triangle, $f_Y(y|x)=F_{XY}(x,y)/F_X(x)$
10. What is $E[Y|x]$?
  
  $\int yf_Y(y|x) dy$
I'm comparing two types of widgets, red and blue. Assume that the probability of each widget dieing in a small interval dt, given that it was alive at the start, is independent of its age. Assume that the probability of the red widget dieing in the next hour is 0.1%, for the blue, it's 0.01%.
1. Give the pdf for the red widget's lifetime. (You have enough info to do this; there is only one possible probability distribution.)
  
  exponential.
  
  $f(x) = l e^{-l}$
  
  From the section on reliability, l=001.
2. If you have 100 red widgets, what's the probability that their mean lifetime is within 10% of the mean?
  
  Normal approx works.
  
  Pop variance: $1/l^2$
  
  Sample variance: $1/(100 l^2)$
  
  Sample std: $1/(10 l)$
  
  Pop and sample mean: $1/l$
  
  sample mean w/i 10% of pop mean = w/i one sample std
  
  Prob: Q(-1) - Q(1) = .68
3. If you start two widgets at the same time, what's the probability that the red widget will last longer?
  
  let x be lifetime of a red widget, y blue.
  
  $l_1=0.001, l_2=0.0001$
  
  joint prob: $F(x,y) = l_1 l_2 \exp(-l_1x -l_2y)$
  
  $P[X>Y] = \int_0^\infty \int_0^x F(x,y) dy dx$

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 3, total 70 points.

Engineering Probability 2019 Class 18 and Exam 2 Answers - Thu 2018-03-21

W Randolph Franklin (WRF), RPI — Fri, 22 Mar 2019 04:00:00 GMT

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:

You're designing a circuit with a chip.
1. The chip was manufactured in one of two plants, but you don't know which.
2. The random variable Y says which plant the chip came from. Y=1 or 2.
3. Half the chips come from each plant.
4. Chips from the first plant (Y=1) have a lifetime X that is U[0,10].
5. 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.
1. What are $f(X=x|Y=1)$ and $f(X=x|Y=2)$?
  
  $f(X=x|Y=1) = \begin{cases}0.1 & for& 0\le x\le 10\\ 0 & otherwise\end{cases}$
  
  $f(X=x|Y=2) = \begin{cases}0.05 & for& 0\le x\le 20\\ 0 & otherwise\end{cases}$
2. What are $f(X=x \,\&\, Y=1)$ and $f(X=x \,\&\, Y=2)$?
  
  $P[Y=1] = 1/2$
  
  $f(X=x \,\&\, Y=1) = f(X=x|Y=1) P[Y=1] = \begin{cases}0.05 & for& 0\le x\le 10\\ 0 & otherwise\end{cases}$
  
  $f(X=x \,\&\, Y=2) = \begin{cases}0.025 & for& 0\le x\le 20\\ 0 & otherwise\end{cases}$
3. What is $f_X(x)$ ?
  
  $f_X(x) = f(X=x \,\&\, Y=1) + f(X=x \,\&\, Y=2) \\ = \begin{cases}0.075 & for& 0\le x\le 10\\ 0.025 & for& 10\le x\le 20\\ 0 & otherwise\end{cases}$
4. 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.
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 $
1. What is $c$?
  
  Integrating f(x) from 0 to 1 gives 1/12, so c=12.
2. What is the mean?
  
  $E[X] = \int_0^1 x f(x) = 1/2$
3. What is the variance?
  
  $E[X^2] = \int_0^1 x^2 f(x) = 2/5$
  
  $VAR[X] = 2/5 -(1/2)^2 = 3/20$.
4. 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} $
5. What is the reliability?
  
  $R(x) = 1 - F_X(x) = \begin{cases} 1 & if & x\le0 \\ 1-(3x-6x^2+4x^3) & if & 0\le x\le1\\0 & if & 1\le x \end{cases}$
6. What is the MTTF?
  
  $\int_0^1 R(x) dx = (x- 3/2 x^2 + 2x^3 - x^4)|_0^1 = 0.5$
This question is about a web server with service requests coming independently at an average of 1 per second.
1. 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.
2. Use the Markov inequality to compute an approximation to the above answer.
  
  P <= 3600/3720 = 0.97. (i.e., almost meaningless).
3. 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.
Let X be a normal random variable with mean 1000 and standard deviation 10. Give the following numbers, using the supplied table.
1. P[X>1015].
  
  That will be Q(1.5) in the table = 0.067.
2. P[990<X<1010].
  
  That will be F(1) - F(-1) = 0.68. approx.
3. For what value of x is P[X<x]=.1?

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.

Engineering Probability 2019: Class 18 and Exam 2 - Thu 2018-03-21

W Randolph Franklin (WRF), RPI — Thu, 21 Mar 2019 04:00:00 GMT

Name, RCSID:

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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:

You're designing a circuit with a chip.
1. The chip was manufactured in one of two plants, but you don't know which.
2. The random variable Y says which plant the chip came from. Y=1 or 2.
3. Half the chips come from each plant.
4. Chips from the first plant (Y=1) have a lifetime X that is U[0,10].
5. 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.
1. What are $f(X=x|Y=1)$ and $f(X=x|Y=2)$?
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2. What are $f(X=x \,\&\, Y=1)$ and $f(X=x \,\&\, Y=2)$?
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3. What is $f_X(x)$ ?
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4. What is $P[Y=1|X=5]$ ?
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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 $
1. What is $c$?
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2. What is the mean?
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3. What is the variance?
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4. What is the cdf?
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5. What is the reliability?
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6. What is the MTTF?
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This question is about a web server with service requests coming independently at an average of 1 per second.
1. 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.
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2. Use the Markov inequality to compute an approximation to the above answer.
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3. Use the Chebyshev inequality to compute an approximation to the above answer.
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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.
1. P[X>1015].
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2. P[990<X<1010].
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3. For what value of x is P[X<x]=.1?
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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.

Engineering Probability Class 12 and Exam 1 - Thu 2019-02-21

W Randolph Franklin (WRF), RPI — Thu, 21 Feb 2019 05:00:00 GMT

Name, RCSID:

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Rules:

You have 80 minutes.
You may bring one 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 six pages.
Do any 14 of the 17 questions or subquestions. Cross out the 3 that you don't do.
When answering a question, don't just state your answer, prove it.

Questions:

You are trying to pass your driving test to get a driving license. You can take the test only once a month. For each time you try, you pass with probability 1/3. The random variable is the number of months until you pass for the first time.
1. (5 pts) What's the relevant probability distribution?
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2. (5 pts) What's the expected number of months until you pass?
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3. (5 pts) What's the standard deviation?
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In this year 2019, which has 365 days, the set of outcomes is the set of days of the year, from Jan 1 to Dec 31. Event A is that the day is Monday. Event B is that the day is in Jan. Here is a calendar for Jan:
```
    January 2019
Su Mo Tu We Th Fr Sa
       1  2  3  4  5
 6  7  8  9 10 11 12
13 14 15 16 17 18 19
20 21 22 23 24 25 26
27 28 29 30 31
```
1. (5 pts) What is the probability of B?
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2. (5 pts) Are A and B independent? Prove your answer (don't just state it).
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You are scanning a B&W page and transmitting it over a noisy channel. A black bit is coded as 1 and a white bit as 0. Event A is that a random bit is black. P(A) = .01. Event B is that the bit is received as black. Sometimes the bit is changed during transmission. 1% of the white bits are changed to black. But 10% of the black bits are changed to white.
1. (5 pts) What is the probability of B?
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2. (5 pts) What is P(A and B))?
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3. (5 pts) What is P(A' and B')?
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4. (5 pts) What is the probability that the bit arrived correct?
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5. (5 pts) What is P(A|B), the probability that 1 was transmitted, if you received 1?
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6. (5 pts) What is P(A' | B'), the probability that 0 was transmitted, if you received 0?
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7. (5 pts) Now you take that received page and transmit it a second time over the same noisy channel. Let event C be that you receive a black bit the 2nd time. Compute P(C).
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(5 pts) An LCD display has 1000 * 1000 pixels. A display is accepted if it has 15 or fewer faulty pixels. The probability that a pixel is faulty coming out of the production line is 1e-5. Find the proportion of displays that are accepted.
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Pretend that there are only 28 students in this class. This is about their birthdays. Assume that they are uniformly distributed over the year (although this is actually false.)
1. (5 pts) What's the probability that no one has a birthday on Feb 29? (Year 2000 was a leap year.) As always a reasonable formula is ok.
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2. (5 pts) Now pretend that no one has a birthday on Feb 29. What's the probability everyone has a different birthday?
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3. (5 pts) Assume that RPI has 7000 students, and their birthdays are uniformly distributed and no one is on Feb 29. Use a reasonable approximate distribution to compute the probability that exactly 20 students' birthday is today, Feb 21.
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(5 pts) Able and Baker take turns tossing a coin until one gets Head. The winner is the person who made that toss. So, if Able gets a head on the first toss, he wins. If Able tosses Tail and then Baker tosses Head, Baker wins. And so on. What's the probability that Able eventually wins the game?
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End of exam 1, total 70 points.

Engineering Probability Class 12 and Exam 1 Answers - Thu 2019-02-21

W Randolph Franklin (WRF), RPI — Thu, 21 Feb 2019 05:00:00 GMT

Name, RCSID:

WRF answers

Rules:

You have 80 minutes.
You may bring one 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 six pages.
Do any 14 of the 17 questions or subquestions. Cross out the 3 that you don't do.
When answering a question, don't just state your answer, prove it.

Questions:

You are trying to pass your driving test to get a driving license. You can take the test only once a month. For each time you try, you pass with probability 1/3. The random variable is the number of months until you pass for the first time.
1. (5 pts) What's the relevant probability distribution?
  
  geometric
2. (5 pts) What's the expected number of months until you pass?
  
  p=1/3, mean=1/p. So 3.
3. (5 pts) What's the standard deviation?
  
  sqrt(1-p)/p = 3 sqrt(2/3) = 2.4
In this year 2019, which has 365 days, the set of outcomes is the set of days of the year, from Jan 1 to Dec 31. Event A is that the day is Monday. Event B is that the day is in Jan. Here is a calendar for Jan:
```
    January 2019
Su Mo Tu We Th Fr Sa
       1  2  3  4  5
 6  7  8  9 10 11 12
13 14 15 16 17 18 19
20 21 22 23 24 25 26
27 28 29 30 31
```
1. (5 pts) What is the probability of B?
  
  31/365 = .0849
2. (5 pts) Are A and B independent? Prove your answer (don't just state it).
  
  One year is 52 weeks plus one day. In 2019, the extra day is Tues.
  
  P(A) = 52/365=.1424
  
  P(A and B) = 4/365 = .0109. That does not equal P(A)P(B)=.0120
  
  no independent.
You are scanning a B&W page and transmitting it over a noisy channel. A black bit is coded as 1 and a white bit as 0. Event A is that a random bit is black. P(A) = .01. Event B is that the bit is received as black. Sometimes the bit is changed during transmission. 1% of the white bits are changed to black. But 10% of the black bits are changed to white.
1. (5 pts) What is the probability of B?
  
  P(B)=P(B|A)P(A)+P(B|A ' )P(A ' ) = .9*.01+.01*.99 = .0189
2. (5 pts) What is P(A and B))?
  
  P(A and B) = P(B|A) P(A) = .9 .01 = .009
3. (5 pts) What is P(A' and B')?
  
  P(A' and B') = P(B' | A' ) P(A') = .99 * .99 = .9801
4. (5 pts) What is the probability that the bit arrived correct?
  
  P(A and B) + P(A' and B') = .9891
5. (5 pts) What is P(A|B), the probability that 1 was transmitted, if you received 1?
  
  P(A|B) = P(A and B)/P(B) = .009/.0189 = .4761
6. (5 pts) What is P(A' | B'), the probability that 0 was transmitted, if you received 0?
  
  P(A' | B') = P(A' and B') / P(B') = .9801 / .9811 = .9989
7. (5 pts) Now you take that received page and transmit it a second time over the same noisy channel. Let event C be that you receive a black bit the 2nd time. Compute P(C).
  
  P(C|B) = P(B|A), P(C|B') = P(B|A')
  
  P(C) = P(C|B) P(B) + P(C|B') P(B') = .9 * .0189 + .01 * .9811 = .0268
  
  That is, with the noise, as you repeatedly retransmit, black bits become more likely.
(5 pts) An LCD display has 1000 * 1000 pixels. A display is accepted if it has 15 or fewer faulty pixels. The probability that a pixel is faulty coming out of the production line is 1e-5. Find the proportion of displays that are accepted.

This would be a Poisson distribution with the random variable being the number of bad pixels in a display. The parameter, a=1e6 * 1e-5 = 10. We want P(X<=15). That is

$\sum_{k=0}^{15} e^{-10} 10^k / k! = .9513$
Pretend that there are only 28 students in this class. This is about their birthdays. Assume that they are uniformly distributed over the year (although this is actually false.)
1. (5 pts) What's the probability that no one has a birthday on Feb 29? (Year 2000 was a leap year.) As always a reasonable formula is ok.
  
  (365/366)^28 = .9244
2. (5 pts) Now pretend that no one has a birthday on Feb 29. What's the probability everyone has a different birthday?
  
  1st person is always ok.
  
  P(2nd person's birthday is different) = 364/365
  
  P(3rd is different from previous 2) = 363/365
  
  P(k-th student is different from 1st k-1) = (366-k)/365
  
  P(all different) = product = 365!/338!/365^28 = .3455
3. (5 pts) Assume that RPI has 7000 students, and their birthdays are uniformly distributed and no one is on Feb 29. Use a reasonable approximate distribution to compute the probability that exactly 20 students' birthday is today, Feb 21.
  
  Use Poisson. The expected number of students with a birthday today is a=7000/365= 19.17.
  
  The probability of exactly 20 is $e^{-19.17} 19.17^{20}/20!$ = .0873
(5 pts) Able and Baker take turns tossing a coin until one gets Head. The winner is the person who made that toss. So, if Able gets a head on the first toss, he wins. If Able tosses Tail and then Baker tosses Head, Baker wins. And so on. What's the probability that Able eventually wins the game?

P(Able wins) = P(Able wins on 1st toss) + P(wins on 3rd) + P(5th) + ...

= 1/2 + 1/8 + 1/32 + ... = 2/3

End of exam 1, total 70 points.

Engineering Probability 2018 Exam 3 - Tues 2018-05-08

W Randolph Franklin (WRF), RPI — Mon, 07 May 2018 04:00:00 GMT

Name, RCSID:

.



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Rules:

You have 80 minutes.
You may bring three 2-sided 8.5"x11" papers 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 nine pages.
Each part of a question is worth 5 points.
You may cross out three question parts, which will not be graded.
When answering a question, don't just state your answer, prove it.

Questions:

You toss two coins. Each comes up heads half of the time. However, for some funny reason, they both come up heads together, or both come up tails together. Intuitively, they not independent. This question is to prove that from the definition of independence.
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This time, you toss three coins, A, B, and C. These are the probabilities:

P[TTT] = P[THH] = P[HTH] = P[HHT] = 0

P[TTH] = P[THT] = P[HTT] = P[HHH] = 1/4

My notation is that TTH means that coin A is tails, coin B tails, and coin C heads. Etc.
1. Are the individual coins fair (i.e., heads half the time)?
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2. Are coins A and B independent?
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3. Are all 3 coins independent?
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This question is about a continuous probability distribution on 2 variables.

$$f_{XY}(x,y) = \begin{cases} c x y & \text{ if } (0\le x) \ \& \ (0\le y)\ \& \ (0\le x+y \le 1) \\ 0 & \text{ otherwise}\end{cases}$$

The nonzero region is the triangle with vertices (0,0), (1,0) and (0,1).

c is some constant, but I didn't tell you what it is.
1. What is c?
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2. What is $F_{XY}(x,y)$?
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3. What is $f_X(x)$?
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4. Are X and Y independent?
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5. What is $P[X\le Y]$ ?
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6. Define a new random variable $Z=X+Y$. What is $F_Z(z)$?
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7. What is $E[X]$?
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8. What is $COV[X,Y]$?
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9. What is $\rho_{X,Y}$?
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10. What is $f_Y(y|x)$?
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11. What is $E[Y|x]$?
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Compute $$\int_0^\infty e^{-x^2} dx$$ .
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What is the legal range for a correlation coefficient?
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What is the variance of the sum of 100 independent variables, each of which is N(0,1)?
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You have 10 independent random variables. Each is uniform on [0,1]. What is the expected value of the max?
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You toss 10 independent fair coins, one after the other.
1. What is the expected total number of heads, given that the first 5 coins came up heads?
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2. What is the probability of the total number of heads being 10, given that the first 5 coins came up heads?
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End of exam 3, total 90 points (considering that 3 questions aren't graded).

Engineering Probability 2018 Exam 3 solution - Tues 2018-05-08

W Randolph Franklin (WRF), RPI — Mon, 07 May 2018 04:00:00 GMT

Name, RCSID: W. Randolph Franklin, frankwr

OK to give the formulas w/o working them out.

Rules:

You have 80 minutes.
You may bring three 2-sided 8.5"x11" papers 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 nine pages.
Each part of a question is worth 5 points.
You may cross out three question parts, which will not be graded.
When answering a question, don't just state your answer, prove it.

Questions:

You toss two coins. Each comes up heads half of the time. However, for some funny reason, they both come up heads together, or both come up tails together. Intuitively, they not independent. This question is to prove that from the definition of independence.

P[HT] = 0. However P[A=H] = P[B=T] = 1/2, so P[HT] != P[A=H]P[B=T]. That's the def of not independent.
This time, you toss three coins, A, B, and C. These are the probabilities:

P[TTT] = P[THH] = P[HTH] = P[HHT] = 0

P[TTH] = P[THT] = P[HTT] = P[HHH] = 1/4

My notation is that TTH means that coin A is tails, coin B tails, and coin C heads. Etc.
1. Are the individual coins fair (i.e., heads half the time)?
  
  P[A=H] = 0+0+1/4+1/4 = 1/2 so fair. Ditto B and C.
2. Are coins A and B independent?
  
  P[A=H,B=H] = 1/4, good. P[A=H,B=T]=1/4, good. P[TH] = 1/4, good. P[TT] = 1/4, good. Yes independent.
3. Are all 3 coins independent?
  
  P[HHH] = 1/4 != P[A=H]P[B=H]P[C=H] = 1/8. Not independent.
This question is about a continuous probability distribution on 2 variables.

$$f_{XY}(x,y) = \begin{cases} c x y & \text{ if } (0\le x) \ \& \ (0\le y)\ \& \ (0\le x+y \le 1) \\ 0 & \text{ otherwise}\end{cases}$$

The nonzero region is the triangle with vertices (0,0), (1,0) and (0,1).

c is some constant, but I didn't tell you what it is.
1. What is c?
  
  $\int_0^1\int_0^{1-x} xy\ dy\ dx = 1/24$ so $c=24$
2. What is $F_{XY}(x,y)$?
  
  $$F_{XY}(x,y)=\begin{cases} 0 & \text{ if } x\le0 \cup y\le0 \\ 1 & \text{ if } x\ge 1 \cap y\ge1 \\ 6x^2y^2 & \text{ if } 0\le x \cap 0\le y \cap x+y\le1 \\ (\int_0^x\int_0^{1-x} + \int_0^{1-y}\int_{1-x}^y + \int_{1-y}^x\int_{1-x}^{1-x_0}) (12x_0y_0 dy_0dx_0) & \text{ otherwise}\end{cases}$$
  
  The last case above splits the nonzero integration region into two rectangles and a triangle.
  
  It's also acceptable to draw a figure and say something intelligent w/o being explicit about all the details.
3. What is $f_X(x)$?
  
  $f_X(x)= \int_0^{1-x}f_{XY}(x,y) dy = 12x(1-x)^2$
  
  Note that $\int_0^1 f_X(x)=1$, which is correct.
4. Are X and Y independent?
  
  $f_X(x)=12x(1-x)^2,f_Y(y)=12y(1-y)^2,f_X(x)f_Y(y)\ne f_{XY}(x,y)$
  
  No.
5. What is $P[X\le Y]$ ?
  
  $\int_0^1\int_0^{\min(x,1-x)} f_{XY}(x,y) dy\ dx$. However, since $f_{XY}(x,y) = f_{XY}(y,x)$, $P[X\le Y]=1/2$
6. Define a new random variable $Z=X+Y$. What is $F_Z(z)$?
  
  $f_Z(z) = \int_0^z f_{XY}(x,z-x) dx = 24\int_0^z x(z-x)dx$ for $0\le z\le 1$
  
  $F_Z(z) = \int f_Z(z)dz$
7. What is $E[X]$?
  
  $\int_0^1 xf_X(x)dx = 2/5$
8. What is $COV[X,Y]$?
  
  E[XY] = $\int_0^1\int_0^{1-x}xyf_{XY}dx dy=8\int_0^1x^2(1-x)^4 dx, E[X]=E[Y]=2/5$, COV[X,Y]=E[XY]-E[X]E[Y]. You don't have to work through the math.
9. What is $\rho_{X,Y}$?
  
  $\sigma_X=\sigma_Y= E[X^2]-E[X]^2$
  
  $\rho_{X,Y}=COV[X,Y]/(\sigma_X\sigma_Y)$
10. What is $f_Y(y|x)$?
  
  $f_Y(y|x)=f(x,y)/f(x) = 12xy/(4x^3) = 2\frac{y}{x^2}$
11. What is $E[Y|x]$?
  
  Integrate the above over $y$ to get $x^{-2}$
Compute $$\int_0^\infty e^{-x^2} dx$$ .

Consider $\sigma=1/\sqrt{2}$. Working a little, this will give $\int_0^\infty e^{-x^2} dx=\sqrt{\pi}/2=0.886$. It was also ok just to use a calculator.
What is the legal range for a correlation coefficient?

-1 to 1
What is the variance of the sum of 100 independent variables, each of which is N(0,1)?

100.
You have 10 independent random variables. Each is uniform on [0,1]. What is the expected value of the max?

Let $W=\max(X_i). F_W(w)=w^{10}. f_W(w)=10w^9.E[W]=10/11.$
You toss 10 independent fair coins, one after the other.
1. What is the expected total number of heads, given that the first 5 coins came up heads?
  
  The last 5 coins do not depend on the first 5. So the expectation is 7.5.
2. What is the probability of the total number of heads being 10, given that the first 5 coins came up heads?
  
  1/32

End of exam 3, total 90 points (considering that 3 questions aren't graded).

Engineering Probability 2018 Exam 2 - Thu 2018-03-29

W Randolph Franklin (WRF), RPI — Thu, 29 Mar 2018 04:00:00 GMT

Name, RCSID:

.



.

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 eleven pages.
Each part of a question is worth 5 points.
You may cross out two questions, which will not be graded.
When answering a question, don't just state your answer, prove it.

Questions:

Consider this probability distribution:

$$f_X(x)= \begin{cases} a(2-x) & \text{if } 0\le x\le1\\ 0&\text{otherwise}\end{cases}$$
1. What is $a$?
```
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2. What is $F_X(x)$?
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3. What is E[X]?
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4. What is the reliability, R[x]?
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5. What is the MTTF?
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6. What is the failure rate, r(x)?
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7. What is $f_X(x|x>.5)$?
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Define a new r.v. Y=2X, where X is the r.v. in the previous question.
1. What is $f_Y(y)$?
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2. What is $F_Y(y)$?
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3. What is E[Y]?
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Your web server gets on the average 1 hit per second. The possible clients are independent of each other.
1. What is the name of appropriate distribution for the number of hits per second?
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2. What is the probability that it gets exactly one hit in the next two seconds?
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3. What is the name of appropriate probability distribution for the time between successive hits?
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4. What is the probability that the time between two successive hits is less than two seconds?
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Let X be an exponential random variable with mean 1.
1. Using the Markov inequality, what's P[X>3]?
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2. Using the Chebyshev inequality, what's P[X>3]?
```
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3. What's the exact P[X>3]?
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Let X be a normal random variable with mean 100 and standard deviation 10. Give the following numbers, using the supplied table.
1. P[X>100].
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2. P[80<X<100].
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You're tossing 10000 fair coins. What's the probability of getting between 5000 and 5100 heads? Use the table.
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Evaluate $$\int_0^\infty e^{-2 x^2} dx$$
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Let $f_X(x) = 1$ and $f_Y(y)=2y$, both in the range $0\le x, y\le1$.

Let Z=max(X,Y).

What is E[Z]?
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```

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

Normal distribution:

x          f(x)      F(x)      Q(x)
      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 1, total 100 points (considering that 2 questions aren't graded).

Engineering Probability 2018 Exam 2 Solution - Thu 2018-03-29

W Randolph Franklin (WRF), RPI — Thu, 29 Mar 2018 04:00:00 GMT

Name, RCSID:

WRF solution

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 eleven pages.
Each part of a question is worth 5 points.
You may cross out two questions, which will not be graded.
When answering a question, don't just state your answer, prove it.

Questions:

Consider this probability distribution:

$$f_X(x)= \begin{cases} a(2-x) & \text{if } 0\le x\le1\\ 0&\text{otherwise}\end{cases}$$
1. What is $a$?
  
  We require that $\int_0^1 f(x) = 1$.
  
  So, $a=2/3$. You don't have to write it, but this gives $f_X(x) = \frac{4}{3}-\frac{2}{3}x$ if $0<x<1$.
2. What is $F_X(x)$?
  
  $$F(x) = \int_0^x f(w) dw = \begin{cases} 0 & \text{if } x\le0 \\ \frac{4}{3}x - \frac{x^2}{3} & 0<x<1 \\ 1 & 1\le x \end{cases} $$
  
  You can use any placeholder variable (other than $x$) in place of $w$.
  
  It doesn't matter where you say $<$ versus $\le$.
3. What is E[X]?
  
  $$E[X] = \int_0^1 x f(x) dx = \int (4/3 x -2/3 x^2)dx = \left. \left(\frac{2 x^2}{3} - \frac{2 x^3}{9}\right)\right|_0^1 = \frac{4}{9}$$
4. What is the reliability, R[x]?
  
  $$R(x) = 1 - F(x) = \begin{cases} 1 - \frac{4}{3}x + \frac{x^2}{3} & 0<x<1 \\ 0 & 1\le x \end{cases} $$
  
  Lifetimes are nonnegative, so I deleted the case for $x<0$, but it doesn't matter.
5. What is the MTTF?
  
  MTTF = $$\int_0^1 R(x) dx = \int \left(1 - \frac{4}{3}x + \frac{x^2}{3}\right) dx = \frac{4}{9}$$
  
  MTTF=E[X]. The integral goes up to 1 because R(x) is 0 when x>1.
6. What is the failure rate, r(x)?
  
  For $0<x<1$,
  
  $$r(x) = \frac{-R'(x)}{R(x)} = \frac{\frac{4}{3}-\frac{2x}{3}}{1 - \frac{4}{3}x + \frac{x^2}{3}} = \frac{4-2x}{3 - 4 x + x^2}$$
  
  You don't need to simplify it.
  
  Note that $r(x)$ grows to infinity as $x$ approaches 1.
7. What is $f_X(x|x>.5)$?
  
  $P[x>.5] = 1-F(.5) = 5/12$.
  
  $$f_X(x|x>.5) = f(x)/P[x>.5] = \begin{cases} \frac{8}{5}(2-x) & \text{if } 0.5\le x\le1\\ 0&\text{otherwise}\end{cases}$$
  
  As a check, you can see that $\int f(x|x>.5) dx = 1$.
Define a new r.v. Y=2X, where X is the r.v. in the previous question.
1. What is $f_Y(y)$?
  
  The nonzero domain for $f_X(x)$ is $0<x<1$, and Y=2X.
  
  So the nonzero domain for $f_Y(y)$ will be $0<y<2$.
  
  $dy/dx = 2$, so
  
  $$f_Y(y) = f_X(y/2)/2 = \begin{cases} \frac{2}{3} - \frac{y}{6} & 0<y<2\\ 0 & \text{otherwise}\end{cases}$$
2. What is $F_Y(y)$?
  
  $$F_Y(y) = F_X(y/2)$$
  
  $$F_Y(y)= \begin{cases} \frac{2}{3} y - \frac{y^2}{12} & 0<y<2\\ 0 & y<0 \\ 1 & y>2 \end{cases}$$
3. What is E[Y]?
  
  $$ E[Y] = \int_0^2 y\left(\frac{2}{3} - \frac{y}{6}\right) dy = \left.\left(y^2/3-y^3/18\right)\right|_0^2 = \frac{8}{9}$$
  
  Alternatively, E[Y] = 2 E[X].
Your web server gets on the average 1 hit per second. The possible clients are independent of each other.
1. What is the name of appropriate distribution for the number of hits per second?
  
  Poisson
2. What is the probability that it gets exactly one hit in the next two seconds?
  
  That r.v. is Poisson with $\alpha=2$ so $$P[X=1] = \frac{2^1 e^{-2}}{1!} = 2 e^{-2} = .27$$
  
  full points for $2 e^{-2}$.
3. What is the name of appropriate probability distribution for the time between successive hits?
  
  Exponential
4. What is the probability that the time between two successive hits is less than two seconds?
  
  Mean: $1/\lambda= 1$. $F(x) = 1-e^{-x}$. $F(2) = 1-e^{-2}=.14=.86$.
  
  full points for $1-e^{-2}$.
Let X be an exponential random variable with mean 1.
1. Using the Markov inequality, what's P[X>3]?
  
  See page 181. $\mu=1$. $P\le 1/3$.
2. Using the Chebyshev inequality, what's P[X>3]?
  
  $\mu=\sigma=1$, P[X<0]=0, so $$P[X>3]= P[|X-1|>2] \le 1/4$$
3. What's the exact P[X>3]?
  
  $F(x) = 1-e^{-x}$ so $P[X>3] = 1-F[3] = e^{-3} = .05$
  
  full points for $e^{-3}$.
Let X be a normal random variable with mean 100 and standard deviation 10. Give the following numbers, using the supplied table.
1. P[X>100].
  
  0.5
2. P[80<X<100].
  
  Converted to $\mu=0,\ \sigma=1$, this is P[-2<Y<0] = F(0)-F(-2) = .5 - .02 = .48.
You're tossing 10000 fair coins. What's the probability of getting between 5000 and 5100 heads? Use the table.

This is a Bernoulli r.v. with $\mu=5000,\ \sigma=\sqrt{npq}=50$.

Use a normal approximation; the answer is F(2)-F(0) = .98-.5 = .48.
Evaluate $$\int_0^\infty e^{-2 x^2} dx$$

For $\mu=0$, what value of $\sigma$ would make $f(x) = c e^{-2 x^2}$?

$$f(x) = \frac{1}{\sqrt{2\pi} \cdot \sigma} \exp\left(\frac{-x^2}{2\sigma^2}\right)$$

so let $\sigma=1/2$ and

$$f(x) = \sqrt{\frac{2}{\pi} } e^{\left(-2 x^2\right)}$$

So $$\int_\infty^\infty e^{-2 x^2} = \sqrt{\frac{\pi}{2} }$$

and $$\int_0^\infty e^{-2 x^2} dx$$ = $$\sqrt{\frac{\pi}{8} }$$

which could be written various ways.
Let $f_X(x) = 1$ and $f_Y(y)=2y$, both in the range $0\le x, y\le1$.

Let Z=max(X,Y).

What is E[Z]?

$F_X(x) = \int f_X(x) dx = x$,

$F_Y(y) = y^2$.

$F_Z(z) = F_X(z) F_Y(z) = z^3$,

$f_Z(z) = 3 z^2$,

$$E[Z] = \int_0^1 3 z^3 dz = 3/4 z^4|_0^1 = \frac{3}{4}$$.

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

Normal distribution:

x          f(x)      F(x)      Q(x)
      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 1, total 100 points (considering that 2 questions aren't graded).