1.) Exercise 3.10 in the textbook: Let \(X\) be a continuous random variable with cdf \(F(x) = x^2/4\) on \([0,2]\) and let \(Y = F(X) = X^2/4\).

1.a.) \(P(X \leq 1)\)

\[P(X \leq 1) = \frac{x^{2}}{4} \\ = \frac{1}{4}\]

1.b.) \(P(X \leq 1/4)\)

\[P(X \leq 1/4) = \frac{x^{2}}{4} \\ = \frac{(\frac{1}{4})^{2}}{4} \\ = (\frac{1}{4})^3 \\ = \frac{1}{64}\]

1.c.) What is the pdf for \(X\)?

The pdf for \(x \in (0,2)\) is:

\[ f_{pdf}(x) = \frac{d}{dx} f(x)\\ = \frac{d}{dx}{\frac{x^{2}}{4}} \\ = \frac{x}{2} \] Outside of \((0,2)\) \(f_{pdf} = 0\)

1.d.) \(P(Y \leq 1)\) If \(Y = 1\); \(X = 2\)

\[ F(x) = \frac{x^{2}}{4} \\ = 1\]

1.e.) \(P(Y \leq 1/4)\)

If \(Y = \frac{1}{4}\); \(X = 1\)

\[ F(x) = \frac{x^{2}}{4} \\ = \frac{1}{4}\]

1.f.) cdf for \(Y\)?
cdf from \([0,1]\):
\[F_{cdf}(Y) = y\]

1.g.) pdf for \(Y\)?

pdf:

\[F_{pdf}(Y) = 1\]

1.h.) \(Y\) has a familiar distribution, what is it?

Uniform distribution

2.) The time \(X\) between two randomly selected consecutive cars in a traffic flow model is modeled with the pdf \(f(x)\) which equals \(\frac{k}{x^{4}}\) on \([1,\infty)\) and \(0\) otherwise.

2.a.) Determine the value of \(k\) that makes \(f\) a pdf. \[ \int^\infty_1 k * x^{-4} \\ [-\frac{k}{3}*x^{-3}]^\infty_1 \\ -\frac{k}{3 * \infty^3} + \frac{k}{3} \\ 0 + \frac{k}{3} = 1 \\ k = 3\] When \(k=3\) the integral of \(f(x)\) from \([1, \infty)\) equals 1. Thus, \(f\) is a pdf

2.b.) cdf

\[ \mbox{F}(x)=\left\{ \begin{array}{rl} 0 & \mbox{if x<1} \\ 1-x^{-3} & \mbox{from x $\geq$ 1} \end{array} \right. \]

2.c.) What is \(E(X)\), and what does it tell you about traffic in this model?

\[ E(X) = \int^\infty_1 (3 * x^{-4})x \\ = \int^\infty_1 (3 * x^{-3}) \\ = -\frac{3}{2 * \infty^2} + \frac{3}{2} \\ = 0 + \frac{3}{2} \\ = \frac{3}{2}\] The mean of the distribution, \(E(X)\), tells us that expect time in between two consecutive cars in traffic is \(\frac{3}{2}\)

2.d.) What is the median of \(X\), and what does it tell you about traffic in this model? cdf where \(M = x\): \[ 1 - M^{-3} = \frac{1}{2}\\ -M^{-3} = -\frac{1}{2} \\ M^{-3} = \frac{1}{2} \\ M^{3} = 2 \\ M = 2^{1/3} = 1.2599 \\ \]

2.e.) Graph the pdf of \(X\). Is the distribution of \(X\) symmetric, positively skewed, or negatively skewed? Which measure of center (mean or median) is most representative of this data?
pdf:

library(fastR2)
fV <- function(v) {3 / v^(4) * (v >= 1)}
integrate(fV, 1, Inf)
## 1 with absolute error < 1.1e-14
gf_line(y ~ v, data = data_frame(v = seq(0, 10, by = 0.01), y = fV(v)))
## Warning: Removed 1 rows containing missing values (geom_path).

The graph shows positively skewed and the medium would provide a better reprensentation of the graph because the data is skewed.

3.) The time (in seconds) between the arrival of successive vehicles at a point on a country road is exponentially distributed with rate parameter \(\lambda =0.01\) cars per second. An elderly pedestrian takes 50 seconds to cross the road at this intersection. Calculate the probability that, if he sets off as one vehicle passes, he will complete the crossing before the next vehicle arrives.

\[P(X \geq 50) = 1 - P(X \leq 50) \\ = 1 - (1 - e^{-\lambda x}) \\ = e^{-(0.01)(50)} \\ = e^{-.5} \\ = 0.607\]

4.) Suppose \(X\)~Unif\((0,1)\)

4.a.) What is the mean?

\[E(X) = \int^1_0 x \\ = \frac{1}{2}\]

4.b.) What is the standard deviation?

\[ Sd = \sqrt{Var(X)} \\ = \sqrt{E(X^2) - E(x)^2} \\ = \sqrt{\frac{1}{3} - \frac{1}{4}} \\ = \sqrt{\frac{1}{12}} \\ = \frac{\sqrt{3}}{6}\]

4.c.) What percentage of the distribution of \(X\) is within one standard deviation of the mean?

\[ \frac{(\mu + \sigma) - (\mu - \sigma)}{1-0} = 2\sigma = 0.57734 \\ 0.57734 * 100 = 57.734\%\]

5.) Eggs may be classified as standard if they weigh less than 46 grams, medium if they weigh between 46 grams and 56 grams, or large if they weigh over 56 grams. Suppose that eggs laid by a particular breed of hens have a mean weight of 50 grams and that these weights are normally distributed with standard deviation 5 grams.

5.a.) What proportion of eggs laid by these hens are standard?
With the mean and standard deviation the z-score can be found for any x.
\[ z = \frac{(x - mean)}{sd} \\ = \frac{46 - 50}{5} \\ = -0.8\]
Using the z-score chart the proportion of eggs that weighed under 46 grams was 21.19%.
R can also be used.

pnorm(-0.8) - pnorm(-5)
## [1] 0.2118551

5.b.) What proportion of eggs laid by these hens are medium?

pnorm(1.2) - pnorm(-0.8)
## [1] 0.6730749

5.c.) What proportion of eggs laid by these hens are large?

1 - pnorm(1.2)
## [1] 0.1150697

5.d.) Suppose that the selling prices of standard, medium, and large eggs are 4 cents, 5 cents, and 6 cents each, respectively, and the cost of production is 4 cents each, no matter the size. Let P denote the random variable corresponding to the profit per egg sold. Find the pmf and expected value of P.

pdf:

P f(P)
0 0.2118551
1 0.6730749
2 0.1150697

\[E(P) = \sum^n_{k=1}P_k * f(P) \\ = 0 + 0.6730749 + (2) * (0.1150697) \\ = 0.9032143\]

6.) The product specification for packets of sweets states that each packet must weigh between 140 and 160 grams. The packet weights are normally distributed with variance 4 grams squared. If you are the owner of the factory that produces these packets, where would you set the target packing weight, and why?

The target packing weight will represent the best mehttp://127.0.0.1:11661/help/library/ggformula/help/gf_facet_gridan that would guarantee the most products for packets of sweets will weigh between 140 and 160 grams. Thus, the halfway point of 150 grams will be the target packing because 140 and 160 grams are both 5 standard deviations away from 150, making it very unlikely for a produced packet to weigh outside of 140 to 160 grams.