y p(y) y*p(y) Sum

Transcription

1 ISQS 5347 Homework #5 1.A) The probabilities of the number of luxury cars sold in a month, p(y), are greater than zero for all y. The sum of the probabilities equals one: Both requirements for a discrete pdf are therefore fulfilled, so the table displays a valid pdf. B) The graph of the function p(y) and the R code to produce it are given below. y = 0:6 p(y) = c(0.18,0.16,0.14,0.34,0.10,0.05,0.03) plot(y,py,type="h", ylim = c(0,0.35), xlab = "Number of Luxury Cars Sold in a Month", ylab = "Density", yaxs="i") C) The expected value of the number of cars sold can be found by multiplying each y value with it s probability, and summing these values. This was done in the table below, and produced an expected value of 2.29.

2 y p(y) y*p(y) Sum D) According to the Law of Large Numbers, the average of n individual data points produced by this model will approach the true expected value of the distribution, 2.29, as n approaches infinity. E) To find the probability that Y is more than the mean, we can sum the probabilities of all possible y values that are greater than We can use the equation Pr> This means that in about 52 out of 100 months, the number of luxury cars sold in that month will be more than 2, and be greater than the expected number of cars sold in any given month. 2.A) The distribution /9 is always greater than or equal to zero because a squared term is always positive; therefore, the first requirement for a valid continuous pdf is fulfilled. The other requirement can be confirmed by calculating 01; therefore, this is a valid continuous pdf. B) The graph of the function p(y) and the R code to produce it are given below. y = seq(0,3,0.01) py = y^2/9 plot(y,py,type="l", ylab = "Density", yaxs="i", ylim=c(0,1.1))

3 C) The mean of p(y) can be calculated as follows: D) According to the Law of Large Numbers, the average of n data points produced by p(y) will approach 2.25, the true mean of the model, as n approaches infinity. E) To find the probability that Y is more than the mean, we can integrate p(y) over y values greater than We can use the equation Pr> This means that the value of y will be greater than 2.25, the distribution mean, in about 58 out of 100 y values produced by p(y). 3.A) I generated 100,000 values of U and Y using the R code below. u=runif(100000) y=202*u^3-5*u-7

4 B) The histogram of Y values and the R code used to produce it are given below. hist(y, freq=f) The normal q-q plot of the Y values and the R code used to produce it are below. qqnorm(y) qqline(y)

5 Neither the histogram nor the q-q plot suggests that the distribution of Y is approximately normal. From the histogram, it appears that the distribution of Y is extremely skewed relative to the normal; from the q-q plot, it appears that the distribution of Y does not produce as extreme upper or lower values as the normal produces. C) The R code to estimate Pr(Y>9) is given below. B = (y>9) mean(b) This outputs a value of This is the estimated expected value of the random variable B, which is bernoulli. Section 8.5 of the book gives the result that the average of an iid sample of bernoullli random variables approaches Pr(B = 1) as n approaches infinity. The B variable as I defined it in R is a simulation of a sample of 100,000 iid Bernoulli outcomes of 0 s and 1 s. Using the law of large numbers, we can use the average of this n = 100,000 sample to estimate the true parameter for this binomial distribution, Pr(B = 1), because n is large. Due to the way the B random variable was defined, Pr(B = 1) = Pr(Y>9), so the estimate of Pr(Y>9) is A) To construct a bootstrap distribution of the food expense data, we assign a probability of to each of the 20 observations. After adding the probabilities of observations that have the same food expense values, done with the help of the R code prop.table(table(food$food)), this results in the distribution shown in table form below. y p(y)

6 This bootstrap distribution is a very rough estimate of the true distribution of food expense obtained by assuming that 1) the only possible values of food expense are the 10 values that were actually observed and 2) each value of food expense has a probability equal to the number of students who reported that value divided by the total number of students surveyed. B) The bootstrap plug in estimate of the mean works by calculating the expected value of the bootstrap distribution created in 4.A. To do this, each value of y is multiplied by p(y), and then summed. y p(y) y*p(y) Sum The bootstrap estimated mean of the food expense data is therefore 13.2.