Engineering Probability Exam 3 Solutions - Sat 2019-05-04
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:
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These few questions are about the population of adult males, which has a mean of 70 inches and a standard deviation of 4 inches.
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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
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If we take a sample of 100, what is its mean?
100.
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What is its standard deviation?
4/sqrt(100) = .4
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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 4 faces (to make this question easier).
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What's the expected value of the number showing on the first die?
2.5
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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
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What's the expected value of the smallest die?
1/8 * 2 + 7/8 * 1 = 9/8
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What's the probability that the smallest number is 1 given that the first number is 2?
Enumerate. p= 3/4
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What's the probability that the first number is 2 given that the smallest number is 1?
3/7
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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
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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
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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.
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What is c?
The integral is 1/3.
c=3
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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.
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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.
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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.
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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.
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What is $E[X]$?
1/8
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What is $COV[X,Y]$?
E[XY] = 1/10. E[Y] = E[X]. COV = 1/10 - 1/8 * 1/8 = .085.
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What is $\rho_{X,Y}$?
OK to write the formula
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What is $f_Y(y|x)$?
In the nonzero triangle, $f_Y(y|x)=F_{XY}(x,y)/F_X(x)$
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What is $E[Y|x]$?
$\int yf_Y(y|x) dy$
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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%.
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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.
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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
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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: $F(x,y) = l_1 l_2 \exp(-l_1x -l_2y)$
$P[X>Y] = \int_0^\infty \int_0^x F(x,y) dy dx$
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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 3, total 70 points.