Engineering Probability 2018 Exam 2 Solution - Thu 2018-03-29
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:
$$f_X(x)= \begin{cases} a(2-x) & \text{if } 0\le x\le1\\ 0&\text{otherwise}\end{cases}$$
-
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 $0<x<1$.
-
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 $\le$.
-
What is E[X]?
$$E[X] = \int_0^1 x f(x) dx = \int (4/3 x -2/3 x^2)dx = \left. \left(\frac{2 x^2}{3} - \frac{2 x^3}{9}\right)\right|_0^1 = \frac{4}{9}$$
-
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.
-
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.
-
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.
-
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.
-
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
$$f_Y(y) = f_X(y/2)/2 = \begin{cases} \frac{2}{3} - \frac{y}{6} & 0<y<2\\ 0 & \text{otherwise}\end{cases}$$
-
What is $F_Y(y)$?
$$F_Y(y) = F_X(y/2)$$
$$F_Y(y)= \begin{cases} \frac{2}{3} y - \frac{y^2}{12} & 0<y<2\\ 0 & y<0 \\ 1 & y>2 \end{cases}$$
-
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.
-
What is the name of appropriate distribution for the number of hits per second?
Poisson
-
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 $2 e^{-2}$.
-
What is the name of appropriate probability distribution for the time between successive hits?
Exponential
-
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.
-
Using the Markov inequality, what's P[X>3]?
See page 181. $\mu=1$. $P\le 1/3$.
-
Using the Chebyshev inequality, what's P[X>3]?
$\mu=\sigma=1$, P[X<0]=0, so $$P[X>3]= P[|X-1|>2] \le 1/4$$
-
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.
-
P[X>100].
0.5
-
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).