Numerical proof of Var(X + Y ) = σ 2 X + σ 2 Y + 2Cov(X, Y ):

***Please use R commands, that is where my confusion lies***

2.1 State how you create a dependent pair of variables (X, Y), give Var(X), Var(Y ), Cov(X, Y ), and Var(X + Y ).

2.2 Choose a sample size n and generate (xi , yi), i = 1, . . . , n according to the (X,Y) distribution. Give the sample variances of the x-sample and y-sample, the sample covariance of the (x,y)-sample, and the sample variance of the x+y sample.

2.3 Turn in all R commands and the output showing the results.

If a variable is Normal (µ = 10, σ = 1.2)
to. Find the probability that X is between 10 and 12. (10 points)
b. Calculate the X corresponding to the 80% percentile (10 points)
c. Find the probability that X is greater than 9 (10 points)
d. If a sample of 15 data is taken, calculate the probability that the average is between 9.2 and 10. (10 points)
and. Calculate the 90% percentile of the average of X if the sample is 15 data. (10 points)
F. If you sample 10 data, find the probability that the Total is between 105 and 120. (10 points)

D3 a product is either rectangular or triangular. What is the probability that, in a random sample of 10 units, there are more than 8 rectangular units of product? Round your answer to 4 decimal places. Assume that the population proportion of rectangular products is the same as the sample proportion of rectangular products

Fruit flies, like almost all other living organisms, have built-in circadian rhythms that keep time even in the absence of external stimuli. Several genes have been shown to be involved in internal timekeeping, including per (period) and tim (timeless). Mutations in these two genes, and in other genes, disrupt timekeeping abilities. Interestingly, these genes have also been shown to be involved in other time-related behavior, such as the frequency of wingbeats in male courtship behaviors. Individuals that carry particular mutations of per and tim have been shown to copulate for longer than individuals that have neither mutation. But do these two mutations affect copulation time in similar ways? The following table summarizes some data on the duration of copulation for flies that carry either the tim mutation or the per mutation (Beaver and Giebultowicz 2004):

b) Do the populations carrying these mutations have different variances in copulation duration? Enter the calculated F-value.

c) Do the populations carrying these mutations have different variances in copulation duration? Finish this hypothesis test and describe your conclusion.

A sample of 44 observations is selected from a normal population. The sample mean is 24, and the population standard deviation is 3. Conduct the following test of hypothesis using the 0.05 significance level.

H0: μ ≤ 24 H1: μ > 24

a) Is this a one- or two-tailed test?

b) What is the decision rule?

c) What is the value of the test statistic?

d) What is your decision regarding H0?

e) What is the p-value?

f) Interpret the p-value?

. Define the stochastic process Xt=2A+3Bt where PA=2=PA=-2=PB=2=PB=-2=0.5. Find PXt≥0 | t  

A sample of parts provided the following contingency table data on part quality by production shift.

Shift Number Good Number Defective

First 368 32

Second 285   15

Third 176 24

Use a=.05 and test the hypothesis that part quality is independent of the production shift. What is your conclusion? Please explain the computation in Excel all the needed functions to resolve the solution so that I can understand the concept of Excel procedures. Explain the computation of the chi-square test statistics for determining whether the column variable independent of the row variable. As well as area in the upper tail and degree of freedom. Thank you very much for your support and help

Problem Binomial Distribution: A consumer advocate claims that 80 percent of cable television subscribers are not satisfied with their cable service. In an attempt to justify this claim, a randomly selected sample of cable subscribers will be polled on this issue.

Suppose that the advocate’s claim is true, and suppose that a random sample of 25 cable subscribers is selected. Assuming independence, find:

(2) The probability that more than 20 subscribers in the sample are not satisfied with their service. Minitab instructions: Go to Calc > select Probability Distributions > select Binomial > select Cumulative probability > Number of Trials insert 25 > Event Probability insert “.8” > Input Constant insert “20.” Click OK.Paste your Minitab results here and then show your work to calculate P(x > 20) = 1 – P(x ≤ 20) to get your final answer:

(3) The probability that between 20 and 24 (inclusive) subscribers in the sample are not satisfied with their service.

(A) Miniab instructions: Go to Calc > select Probability Distributions > select Binomial > select Cumulative probability > Number of Trials insert 25 > Event Probability insert “.8” > Input Constant insert “19.” Click OK.

(B) Minitab instructions: Go to Calc > select Probability Distributions > select Binomial > select Cumulative probability > Number of Trials insert 25 > Event Probability insert “.8” > Input Constant insert “24.” Click OK. Paste each of your Minitab results here and then show your work to calculate P(20 ≤ x ≤ 24) = P( x≤ 24) – P(x≤ 19) to get your final answer:

(4)  The probability that exactly 24 subscribers in the sample are not satisfied with their service.Paste your Minitab results here:

c) Suppose that when we survey 25 randomly selected cable television subscribers, we find that 15 are actually not satisfied with their service. Using a probability you found in this exercise as the basis for your answer, do you believe the consumer advocate’s claim? Explain your answer here:

Answer the following questions using R (include both the input and output for each question).

I) A random variable P has a Poisson distribution with a mean of 10. Solve for the probability that random variable P is greater than 8.

II) What is the probability that in 30 tosses of a fair coin, the head comes up 10 or 15 times?

III) What is the probability that a normal random variable is less than 40, assuming that it has a mean of 33 and a variance of 36?

Please very important to explain all the transactions answers

Chapter 12 Monte Carlo Simulation and Risk Analysis

3. A professional football team is preparing its budget for the next year. One component of the budget is the revenue that they can expect from ticket sales. The home venue, Dylan Stadium, has five different seating zones with different prices. Key information is given below. The demands are all assumed to be normally distributed.

Seating Zone    Seats available Ticket price    Mean Demand    Standard Deviation

First Level Sideline    15,000    $100.00 14,500 750

Second Level    5,000 $90.00    4,750    500

First Level End Zone    10,000    $80.00    9,000 1250

Third Level Sideline 21000 $70.00 17,000 2500

Third Level End Zone    14000 $60.00 8,000 3000

Determine the distribution of total revenue under these assumptions using an Excel data table with 50 simulated trials. Summarize your results with a histogram.

Suppose that a study shows that an electric bike is driven 23,500 km/year on average and has a standard deviation of 3900 km. Assume that the measurements’ distribution is approximately normal.

a) Calculate the 3rd quartile.

b) Solve for the probability that a randomly selected electric bike is driven at most 15,000 km per year, and then solve for the probability that a randomly selected electric bike is driven between 10,000 and 20,000 km per year.

c) If the average cost of leasing an electric bike is $300 per month with an SD of $110. You want to buy an electric bike so you randomly sample 20 people who are currently leasing. Solve for the probability that the sample mean, amongst the randomly selected 20 people, is less than $250.

d) Solve for the probability that the sample mean is greater than $300, amongst the randomly selected 20 people sampled in part c)

Answer the following questions using R (include both the input and output for each question).

I) A random variable P has a Poisson distribution with a mean of 10. Solve for the probability that random variable P is greater than 8.

II) What is the probability that in 30 tosses of a fair coin, the head comes up 10 or 15 times?

III) What is the probability that a normal random variable is less than 40, assuming that the variable has a mean of 33 and a variance of 36?

Question 1

Both F and Chi Square distributions are unimodal and symmetric

1. True
2. False

Question 2

Both F and Chi Square distributions vary with df

1. True
2. False

Question 3

Both F and Chi Square distributions can be used to compare multiple sample means

1. True
2. False

Question 4

For Chi Square tests, all tests discussed can be used for both experiments and surveys

1. True
2. False

Question 5

For ANOVA: the null hypothesis is: Mu1 = Mu2 = Mu3......etc

1. True
2. False

Question 6

For ANOVA: the alternative hypothesis is: Mu1 LaTeX: \ne≠Mu2 LaTeX: \ne≠Mu3......etc.

1. True
2. False

Question 7

Chi Square tests have an assumption of normality

1. True
2. False

Question 8

Both ANOVA and Chi square tests are always right tail tests

1. True
2. False

Question 9

For a Chi Square test of Association, you ask if mouse hair color (black, grey or white) is associated with whisker length (short, medium or long). The df for this analysis would be: Give answer as X

Question 10

For ANOVA, you only complete a Tukey's means comparisons test if the overall F* value was significant (i.e., p<0.05)

1. True
2. False

The 22,000 students at NCC have mean mileage on their vehicles of µ = 54,000 miles

       with a standard deviation of s = 3,125 miles. Assuming a normal distribution

   a) what is the probability that a randomly selected student has a car with mileage between 55,000
           and 60,000 miles?

b) what percent of student vehicles have mileage above 60,000 miles?

c) how many students have cars with mileage below 50,000?