1. Let \(X\) be a continuous random variable with cdf \(F(x)=\frac{x^2}{4}\) on \([0,2]\) and let \(Y=F(X)=\frac{X^2}{4}\).
  1. What is \(P(X\leq 1)\)? \(\frac{1}{4}\)
  2. What is \(P(X \leq 1/4)\)? \(\frac{1}{64}\)
  3. What is the pdf for \(X\)? \(F'(x)=f(x)=\frac{x}{2}\)
    \(Y= \frac{(\frac{x^2}{4})^2}{4}=\frac{x^4}{4^3}\)
  4. What is \(P(Y \leq 1)\)? \(\frac{1}{64}\)
  5. What is \(P(Y \leq \frac{1}{4})\)? \(\frac{1}{4^7}\)
  6. What is the cdf for \(Y\)? \[F(Y)=\frac{x^4}{4^3}\]
  7. What is the pdf for \(Y\)? \[F'(Y)=f(Y)=\frac{x^3}{16}\]
  8. \(Y\) has a familiar distribution. What is it? The inverse of an eponential distribution.
  1. 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.
  1. Determine the value of \(k\) that makes \(f\) a pdf.
    \[ \int_{1}^{\infty} \frac{k}{x^4} dx = 1\\ \frac{k}{3} =1\\ k=3 \]
  2. Obtain the cdf of \(X\).
    \[ \int_{1}^{x}\frac{3}{x^4} dx = \frac{-1}{x^3}+1 \]
  3. What is \(E(X)\), and what does it tell you about traffic in this model?
    \[\frac{3}{2}\]
    This tells us that the expected time X of cars consequtively passing averages to 3/2
  4. What is the median of \(X\), and what does it tell you about traffic in this model?
    \[ \int_{1}^{M} \frac{3}{x^4} dx = \frac{1}{2}\\ 1-\frac{1}{M^3} =\frac{1}{2}\\ M = 2^\frac{1}{3} \]
  5. 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?
library(fastR2)
# define the pdf for X
f <- function(x) { 3/x^4 * (1 <= x & x <= Inf) }
integrate(f, 1, Inf)
## 1 with absolute error < 1.1e-14
gf_line(y ~ x, data=data_frame(x = seq(-1, 10, by = 0.001), y=f(x)))

The distribution of X is positively skewed; mean is the most representative measure of center.

  1. 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.
    \[ P(X\geq50)=1-P(X\leq50)\\ P(X\geq50)=1-F(50)\\ P(X\geq50)=1-(1-e^{-.01*50})\\ P(X\geq50)=e^{-.5}\\ P(X\geq50)=0.606 \]

  2. Suppose \(X\)~\(Unif(0,1)\).
  1. What is the mean?
    \[\int_{0}^{1} x*1 dx = \frac{1}{2}\]
  2. What is the standard deviation? \[\int_{0}^{1} x^2*1 dx = \frac{1}{3}\]
    \[\sigma^2=Var(X)=E(X^2)-(E(X))^2=\frac{1}{3} - \frac{1}{2}^2=\frac{1}{12} \]
    \[\sigma=\sqrt{\frac{1}{12}}\]
  3. What percentage of the distribution of \(X\) is within one standard deviation of the mean?
    \[2\sigma *100=.577\]
    57.7% of the distribution of X is within one standard deviation from the mean
  1. 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.
  1. What proportion of eggs laid by these hens are standard? \[\frac{46-50}{5}=\frac{-4}{5}=-.8\]
s=.5-pnorm(-.8)
s
## [1] 0.2881446

28.8% of the eggs laid by these hens are standard
b) What proportion of eggs laid by these hens are medium? \[\frac{56-50}{5}=\frac{6}{5}=-1.2\]

m=pnorm(1.2)-pnorm(-.8)
m
## [1] 0.6730749

67.3% of the eggs laid by these hens are medium
c) What proportion of eggs laid by these hens are large?

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

11.5% of the eggs laid by these hens are large
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 \(P\).

P     |     4-4     |      5-4    |  6-4   
----- | ----------- | ----------- | -----------   
f(F)  | s=0.2881446 | m=0.6730749 | l=0.1150697  
EP=s*(4-4)+m*(5-4)+l*(6-4)
EP
## [1] 0.9032143
  1. 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?
    Setting the target packaging weight at 150 grams, where the the min and max weight are within 5 standard deviations and we can expect over the 99.7% (of 3 standard deviations) of the weights to be within the required range.