In agracy2246/MATH4753grac0009: Adam Gracy MATH4753

knitr::opts_chunk$set(echo = TRUE)

Questions{}

Q 1

MS 3.36 pg 105-106

(a)

The probability that an expert will correctly identify the matched prints is p = 92.12%. So, the probability that the expert will not identify the matched prints is q = 100 - p or 100-92.12 = 7.88%

(b)

The probability that a novice will correctly identify the matched prints is p = 74.55%. So, the probability that the novice will not identify the matched prints is q = 100 - p or 100-74.55 = 25.45%

(c)

The parcipant is probably a novice, because each class (expert and novice) has a .5 probability of being the one selected but the novice has the higher fail rate.

Q 2

MS 3.52 - pg 111

(a)

We can solve this by setting it up as a conditional probability problem.

Let + denote positive test results, and user denote a user.

The problem becomes $p(+ | user) = \frac{p(+ \cap user )}{p(user)}$

$p(+ \cap user) = \frac{users\space testing\space positive}{sample \space size} = \frac{50}{1000}$

$p(user) = \frac{users}{sample \space size} = \frac{100}{1000}$

Finally, we get $\frac{p(+ \cap user )}{p(user)} = \frac{\frac{50}{1000}}{\frac{100}{1000}} = .5$ or 50%

(b)

Again, this is a conditional probability problem.

Let - denote negative test results, and nonuser denote a nonuser

The problem becomes $p(- | nonuser) = \frac{p(- \cap nonuser )}{p(nonuser)}$

$p(- \cap nonuser) = \frac{nonusers\space testing\space negative}{sample \space size} = \frac{891}{1000}$

$p(nonuser) = \frac{nonusers}{sample \space size} = \frac{900}{1000}$

Finally, we get $\frac{p(- \cap nonuser )}{p(nonuser)} = \frac{\frac{891}{1000}}{\frac{900}{1000}} = .99$ or 99%

(c)

Let +,-,u,nu* be positive result, negative result, user, nonuser***, respectively.

The problem becomes $p(u | +) = \frac{p(u)p(+|u)}{p(u)p(+|u)+p(nu)p(+|nu)}$

Then,

$p(u) = .1$, $p(nu) = .9$

$p(+|u) = \frac{\frac{50}{1000}}{.1}$, $p(+|nu) = \frac{\frac{9}{1000}}{.9}$

Finally,

$p(u|+) = \frac{(.1)(.5)}{(.1)(.5)+(.9)(.01)} = .8475$

Q 3

Let $A = {x_1,x_2,x_n}, B = {y_1,y_2,y_n}$

Suppose A, B are events with $x_n, y_n$ possible outcomes.

We can compute the amount of total possible outcomes between them by calculating $|A| \times |B| = N$ (cardinalities)

${x_1,y_1},{x_1,y_2},{x_1,y_3},$

${x_2,y_1},{x_2,y_2},{x_2,y_3},$

${x_n,y_1},{x_n,y_2},{x_n,y_n}$

And for n = 3 we can see that we have $|A| \times |B| = 3\times3=9$ possible outcomes.$\square$

Q 4

Let $A = {x_1,x_2,x_n}$ be a set of elements.

Let $N = |A|, n = amount\space of\space elements\space we\space want\space in\space our\space permutations$

Then, $\frac{N!}{(N-n)!}$ will tell us how many permutations we create, using n elements from a set of size N.

Let $A = {x_1,x_2,x_3}$ be a set of elements.

If we let N = 3, and n = 2 then we can calculate how many permutations we'll get using 2 elements from a set of cardinality 3.

$\frac{3!}{(3-2)!}=\frac{6}{1} = 6$

We get $x_1x_2,x_1x_3,x_2x_1,x_2x_3,x_3x_1,x_3x_2\square$

Q 5

Let A be a set of N unique elements and we are going to partition them into k sets. Then we will use $\frac{N!}{n_1!...n_k!}$

Q 6

The proof for the Combinations rules is derived from the proof of the Partition Rule.

We want to choose n elements from a sample of N elements and break them into two groups: the n elements that are being selected and the N-n elements that are not.

So we get $N\choose n$ $= \frac{N!}{n!(N-n)!}$

Q 7

MS 4.2 - pg 138

(a)

.09 + .3 + .37 + .2 + .04 

(b)

$p(x=3 \cup x=4) = p(x=3) + p(x=4) = .20 + .04 = .24$

(c)

$p(2 > x) = p(x=0) + p(x=1) = .09 + .30 = .39 $

Q 8

MS 4.12 - pg 143

(a)

The first rule that 0 <= p(y) <= 1 is met because each p(y) falls within the range The second rule that the sum of all p(y) = 1 is also met, as shown below.

.17+.10+.11+.11+.10+.10+.07+.05+.03+.02+.02+.02+.02+.02+.01+.01+.01+.01+.01+.005+.005

(b)

Just sum p(10) through p(20)

.02+.02+.02+.02+.01+.01+.01+.01+.01+.005+.005

(c)

numbers <- seq(0,20)
mu <- sum(numbers*c(.17,.10,.11,.11,.10,.10,.07,.05,.03,.02,.02,.02,.02,.02,.01,.01,.01,.01,.01,.005,.005))

ySquared = numbers^2

variance <- (numbers - mu)^2 * c(.17,.10,.11,.11,.10,.10,.07,.05,.03,.02,.02,.02,.02,.02,.01,.01,.01,.01,.01,.005,.005)
variance <- sum(variance)
variance <- round(variance, 4)
paste0("The mean is: ", mu)
paste0("The var is: ", variance)

(d)

Chebyshev's says that 75% will lie within k = 2 so (mu - 2s, mu + 2s)

std = sqrt(variance)
paste0("The interval is: ","(",round((mu - (2*std)),4), ", ", round((mu+(2*std)),4), ")")

Q 9

MS 4.34 - pg 154

(a)

Simple binomial probability $\binom{25}{10}.7^{10}.3^{15} = .0013$

Alternatively:

round(dbinom(10,size=25, prob=.7),4)

(b)

$\binom{25}{0}.7^{0}.3^{25} + \binom{25}{1}.7^{1}.3^{24} + \binom{25}{2}.7^{2}.3^{23} +\binom{25}{3}.7^{3}.3^{22} + \binom{25}{4}.7^{4}.3^{21} + \binom{25}{5}.7^{5}.3^{20} = .000000345744$

Alternatively:

pbinom(5,size=25,.7)

(c)

$\mu = np = 25(.7) = 17.5$

$\sigma^2 = npq = 25(.7)(.3) = 5.25$

$\sigma = \sqrt{\sigma^2} = \sqrt{5.25} = 2.2913$

paste0("The mean is: 17.5, the standard deviation is: 2.2913")

(d)

My interpretation: If you sample 25 graduate students, on average 17.5 students will have earned their PhD with a standard deviation of 2.2913.

Q 10

MS 4.46 - pg 158

(a)

$N=50, p_k=.1$

Then the formula is: $p(y_1=5,y_2=5,y_3=5,y_4=5,y_5=5,y_6=5,y_7=5,y_8=5,y_9=5,y_10=5)=$

$\frac{50!}{(5!)^{10}} \times (.1)^{50} = .0000005$

(b)

For each of the 50 experiments we will ask if the train is assigned to Track #1.This is a binomial distribution. We want to know the probability that there will be fewer than two success to that question. In other words, $p(y=0) + p(y=1) = \binom{50}{0}\times.1^0\times.9^{49} + \binom{50}{1}\times.1^1\times.9^{50} = .0338$

Q 11

MS 4.54 - pg 162

(a)

This will be a Geometric Probability distribution. So, we first need to find p. What percentage of customers give a reason that is NOT because of label or packaging? $p = \frac{12+6+4+18}{100} = .4$ and since q = 1-p, this tells us q = .6

Now, the formula will be $p(y) = (p)(q)^{y-1}$

(b)

Since $\mu =E(y) = \frac{1}{p}$ we get $\frac{1}{.4} = 2.5$

(c)

The probability that it takes exactly 1 consumer to get an answer that is other than "labeling or packaging" is $p(y) = (.4)(.6)^{1-1} = .4$

(d)

I'll first find p(y <= 2) $p(1) + p(2) = (.4)(.6)^{1-1} + (.4)(.6)^{2-1} = .64$ and since $p + p^c = 1$ then $p(y > 2) = .36$

Q 12

MS 4.66 - pg 168

(a)

Asking for the expected value or mean which is $\mu = \frac{nr}{N} = \frac{80}{209} = .3828$ This means that the expected maximum is 1 facility that treats on site.

(b)

$\frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}} = .00017$

Q 13

MS 4.78 - pg 173

(a)

By definition: $\mu = \sigma^2$ so $\sigma^2 = .03$

(b)

For the assumptions to hold, the experiment will need to meet the three required characteristics:

• Experiment consists of counting the number of time Y a particular (rare) event occurs during a given unit of time or in a given area or volume.

• The probability that an event occurs in a given unit of time, area, or volume is the same for all the units. Also, units are mutually exclusive.

• The number of events that occur in one unit of time, area, or volume is independent of the number that occur in other units.

(c)

We simply plug y=0 into the function $\lambda =.03;\quad y(0) = \frac{.03^0\times e^{-.03}}{0!} = .9704$

Q 14

MS 5.2 - pg 191

(a)

$f(y) = 2c-yc ;\quad Solve\space \int_0^1f(y)dy = 1$

$2c \int_0^1 dy - c \int_0^1 y dy = 1$

$\quad\quad 2c\space-\space\frac{c}{2} = 1$

$\quad\quad c=\frac{2}{3}$

(b)

The Cdf is the integral of f(y):

$F(y) = \int_ {-\infty}^yf(x)dx = \int_0^yf(x)dx$

$\frac{2}{3}\int_0^y(2-y) = \frac{2}{3}(2\int_0^ydx - \int_0^yxdx)$

$= \frac{2}{3}(2x\rvert_0^y -\frac{x^2}{2}\rvert_0^y)$

$= \frac{2}{3}(2y-\frac{y^2}{2})$

Thus, the Cdf is $F(y) = 0\quad y\leq0; \quad\quad \frac{2}{3}(2y-\frac{y^2}{2})\quad 01$

(c)

$y = .4\quad$ Since $0 < y \leq 1$ holds we use $\frac{2}{3}(2y-\frac{y^2}{2})$

$F(.4)= \frac{2}{3}(2(.4)-\frac{.4^2}{2}) = 0.48$

(d)

Being that the values of Y will be $0 < y \leq 1$ throughout the given interval we use $F(y) = \frac{2}{3}(2y-\frac{y^2}{2})$ and calculate $F(.6) - F (.1)$

$F(.1 \leq y \leq .6) = F(.6) - F(.1) = \frac{2}{3}(2(.6)-\frac{.6^2}{2}) - \frac{2}{3}(2(.1)-\frac{.1^2}{2}) = 0.55$

Q 15

MS 5.10 - pg 196

(a)

Distributing we get $f(y) = \frac{3}{20} -\frac{3y^2}{500}$

To find the mean: $\mu = E(Y) = \int_{-5}^{5}y(\frac{3}{20} -\frac{3y^2}{500})dy=\int_{-5}^{5}(\frac{3y}{20} -\frac{3y^3}{500})dy = \frac{3}{20}\int_{-5}^{5}ydy - \frac{3}{500}\int_{-5}^5y^3dy = 0$

To find the variance: $E(Y^2) -\mu^2$, so we first need to find $E(Y^2) = \int_{-5}^{5}y^2(\frac{3}{20} -\frac{3y^2}{500})dy = 5$

Now use the $E(Y^2) -\mu^2 = 5 - 0 = 5$

Thus, $\mu = 0,\sigma^2 = 5$

(b)

Being that the random variable Y is in minutes, we can easily convert this in to hours by doing $\frac{Y}{60}$ and redoing the steps from (a). Since the expected value rules hold for continuous random variables, we can simply do $E(\frac{1}{60}Y) = \frac{1}{60}E(Y) = 0$ so $\mu = 0$ still.

For the variance we do the same, but c is squared: $E(\frac{1}{60^2}Y^2) = \frac{1}{60^2}E(Y^2) = \frac{5}{60^2}= .0014$

(c)

We will take the exact same steps but multiply by 60 instead of dividing.

$\mu = E(60Y) = 60E(Y) = 60\times0 = 0$

For the variance: $E(60Y^2) = 60^2E(Y^2) = 60^2\times5 = 18000$

Q 16

MS 5.36 - pg 205

(a)

We have two methods of doing this (*Note: Since Y is continuous, the boundaries we use are not critical as they were with discrete random variables):

• Knowing $p(Y > 45) + p(Y <= 45) = 1$ and then finding 1-pnorm(45, mean=50, sd=3.2)

round(1-pnorm(45, mean=50, sd=3.2),4)

• Knowing that $p(Y>45)$ corresponds to $p(z > \frac{Y - \mu}{\sigma} = \frac{45-50}{3.2}=-1.5625$ and then computing 1- pnorm(-1.5625, mean=0, sd=1)

round(1-pnorm(-1.5625, mean=0, sd=1),4)

And we see that using both methods the answer is $0.9409$

(b)

Again we have two methods of doing this. But since Y is an upperlimit, we only need to find the lower tail.

• We can find $p(Y < 55)$ by calculating $pnorm(55, mean= 50, sd=3.2)$

round(pnorm(55, mean=50, sd=3.2),4)

• Alternatively, we can use our knowledge that $p(Y < 55)$ corresponds to $P(z < \frac{Y - \mu}{\sigma} = \frac{55-50}{3.2} = \frac{5}{3.2} = 1.5625)$ and calculate it using $pnorm(1.5625,mean=0,sd=1$

round(pnorm(1.5625, mean=0, sd=1),4)

Once again we can see that both methods return the same result of $0.9409$ (Note: Again, since Y is continuous, it is not important that we use $<$ or $\leq$)

(c)

Using the same strategy from (a) and (b) we can calculate $p(52) - p(51)$ to get the area in between

• $p( 51 < Y < 52)$ can be found by calculating $pnorm(52, mean=50,sd=3.2) - pnorm(51,mean=50, sd=3.2$

round(pnorm(52, mean=50,sd=3.2) - pnorm(51,mean=50, sd=3.2),4)

• Alternatively, we can do the same thing using $pnorm(z=\frac{52-50}{3.2}, mean=0, sd=1) - pnorm(z=\frac{51-50}{3.2},mean=0,sd=1)$

round(pnorm((52-50)/3.2, mean=0, sd=1) - pnorm((51-50)/3.2,mean=0,sd=1),4)

As you can see, both methods give the same answer of: $0.1113$

Q 17

MS 5.38 - pg 205

(a)

round(pnorm(700, mean = 605, sd=185)-pnorm(500, mean = 605, sd=185),4)

(b)

round(pnorm(500, mean = 605, sd=185)-pnorm(400, mean = 605, sd=185),4)

(c)

round(pnorm(850, mean = 605, sd=185),4)

(d)

round(1-pnorm(1000,mean=605, sd=185),4)

(e)

y <- 605 + qnorm(.9, mean=0, sd=1) * 185
y

agracy2246/MATH4753grac0009 documentation built on April 26, 2020, 9:39 a.m.