In models for the lifetimes of mechanical components, one sometimes uses random variables with distribution functions from the so-called Weibull family. Here is an example: F(x) = 0 for x < 0, and F(x) = 1 − e^{−5x^2} for x ≥ 0.
Construct a random variable Z with this distribution from a U(0, 1) variable.

Use Excel to perform the calculations and attach it/screenshot it

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 students using Method 1 produces a testing average of 61 . A sample of 156 students using Method 2 produces a testing average of 64.6 . Assume that the population standard deviation for Method 1 is 18.53 , while the population standard deviation for Method 2 is 13.43 . Determine the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2. Step 2 of 3 : Calculate the margin of error of a confidence interval for the difference between the two population means. Round your answer to six decimal places.

An urn contains 10 red and 12 blue balls. They are withdrawn one at a time without replacement until a total of 4 red balls have been withdrawn. Find the probability that exactly 7 balls withdrawn/

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 68 inches and standard deviation 4 inches.

(a) What is the probability that an 18-year-old man selected at random is between 67 and 69 inches tall? (Round your answer to four decimal places.)


(b) If a random sample of twenty-six 18-year-old men is selected, what is the probability that the mean height x is between 67 and 69 inches? (Round your answer to four decimal places.)


(c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this?

1The probability in part (b) is much higher because the standard deviation is smaller for the x distribution. 2The probability in part (b) is much higher because the mean is larger for the x distribution. 3The probability in part (b) is much lower because the standard deviation is smaller for the x distribution. 4The probability in part (b) is much higher because the standard deviation is larger for the x distribution.5The probability in part (b) is much higher because the mean is smaller for the x distribution.

PLease show work in excell

A major client of your company is interested in the salary distributions of jobs in the state of Minnesota that range from $30,000 to $200,000 per year. As a Business Analyst, your boss asks you to research and analyze the salary distributions. You are given a spreadsheet that contains the following information:

A listing of the jobs by title

The salary (in dollars) for each job

The client needs the preliminary findings by the end of the day, and your boss asks you to first compute some basic statistics.

Background information on the Data

The data set in the spreadsheet consists of 364 records that you will be analyzing from the Bureau of Labor Statistics. The data set contains a listing of several j

obs titles with yearly salaries ranging from approximately $30,000 to $200,000 for the state of Minnesota.

What to Submit

Your boss wants you to submit the spreadsheet with the completed calculations. Your research and analysis should be present within the answers provided on the worksheet.

Income East and West of the Mississippi

For a random sample of households in the US, we record annual household income, whether the location is east or west of the Mississippi River, and number of children. We are interested in determining whether there is a difference in average household income between those east of the Mississippi and those west of the Mississippi.

Incorrect answer iconYour answer is incorrect.

(a) State the null and alternative hypotheses. Your answer should be an expression composed of symbols:

Let group 1 be the households east of the Mississippi River and let group 2 be the households west of the Mississippi River.

(b) What statistic(s) from the sample would we use to estimate the difference?

Let group 1 be the households east of the Mississippi River and let group 2 be the households west of the Mississippi River.

The output voltage of a power supply unit has an unknown distribution. Using a sample size of 36, sixteen samples are taken with the following sample-mean values: 10.35 V, 9.30 V, 10.00 V, 9.96 V, 11.65 V, 12.00 V, 11.25 V, 9.58 V, 11.54 V, 9.95 V, 10.28 V, 8.37 V, 10.44 V, 9.25 V 9.38 V and 10.85 V. Let µ and σ² denote the mean and the variance of the output voltage of the power supply unit. (a) What distribution describes the sample-mean? What are the parameters of the distribution (in terms of µ and σ² )? (b) Test the hypothesis that the population variance σ² = 36 V² at 95% confidence. (c) Construct a two-sided 95% confidence interval for the population’s standard deviation.

Suppose you have a bag with 18 small colored stones. There are 6 red stones, 7 green stones, and the rest of the stones  are blue.

a. Would reaching into the bag, pulling out a stone, and recording the color be considered a random experiment? Explain your answer.

b. If you reach into the bag and pull out a stone, what is the probability that the stone is blue?

c. What is the probability that you will not pull out a blue stone?

d. Suppose you pull out a red stone. If you do not replace the first stone, what is the probability that you will pull out a blue stone this time?

Consider a new hotel deciding on cleaning staff hiring for the upcoming season. Cleaning times depend on whether it is a stay-over room or a check-out. Suppose that a guest will check-out on a given day with probability 40%. From your experience in similar hotels you estimate that a stay-over room cleaning time is well-described with normal distribution with average 15 minutes and standard deviation 1 minute. Check-out room cleaning time is also normal but with average 30 minutes and standard deviation 10 minutes.

i. Consider an occupied room (stay-over or check-out), what is the average cleaning time for such a room?

ii. Find the variance for the cleaning time for an occupied room.

iii. Suppose that the hotel has 200 rooms, and you estimate that on a given day a room will be occupied with probability 90%. Only occupied rooms need cleaning. Find the average total cleaning time for the hotel. iv. Find the variance of the total cleaning time for the hotel.

Hints: remember var(X) = EX^2 − (EX)^2 .

betting theory on tennis

A bookmaker has quoted odds on a tennis match between players I and II. The match consists of the best two out of three sets, i.e., if a player wins the first two sets, the third set is not played and the bet on it is canceled. The bookmaker is giving odds of 5 to 2 that player I will win the match and odds of 3 to 2 that player I will win each set. A bettor has 100 dollars which he can distribute by betting on either player I or II to win the match and any of the sets. All bets are made before the match starts (if there are only two sets, all bets on the third set are returned to the bettor).
(a) Find a way of placing bets so that no matter what happens the bettor is assured of winning an amount z where z is as large as possible. Formulate this problem as a linear programming and solve it using AMPL.
(b) What if now we have best three out of five sets, i.e., once a player wins three sets, no more sets are played and their corresponding bets are canceled, and everything else keeps the same? Re-solve the problem and compare the answer with part (a).

Researcher conducts a study to decide whether support groups improve academic performance for at-risk high school students. Ten such students are randomly selected to take part in the support group for a semester, while the other 10 at-risk students serve as a control group. At the end of the semester, the improvement in GPA versus the previous semester is recorded for each student.
Support Group: 0.5, 0.8, 0.7, 0.7, -0.1, 0.2, 0.4, 0.4, 0.5, 0.4
Control Group: -0.3, 0.0, -0.1, 0.2, -0.1, -0.2, -0.2, 0.0, -0.1, 0.1

At the 10% level, use R to compare the two groups using a permutation test (with 100,000 randomly generated permutations). You need to write your hypotheses, the test statistic, the pvalue, and the decision/conclusion in the context of the problem.

R code for reference:

SupportGroup <- c(0.5, 0.8, 0.7, 0.7, -0.1, 0.2, 0.4, 0.4, 0.5, 0.4)
ControlGroup <- c(-0.3, 0.0, -0.1, 0.2, -0.1, -0.2, -0.2, 0.0, -0.1, 0.1)

mean(SupportGroup);sd(SupportGroup)
mean(ControlGroup);sd(ControlGroup)

#permutation test on difference of means
choose(20,10)#number of possible permutations
new.dat <- c(SupportGroup,ControlGroup)
obs.mean.diff <- mean(SupportGroup) - mean(ControlGroup)
nsim <- 100000
sim.mean.diff <- rep(NA,length=nsim)
for (i in 1:nsim){
grps <- sample(c(rep(1,10),rep(2,10)),replace=FALSE)
sim.mean.diff[i] <- mean(new.dat[grps==1]) - mean(new.dat[grps==2])
}

hist(sim.mean.diff);abline(v=obs.mean.diff,col="red",lty=2)
length(sim.mean.diff[sim.mean.diff<=obs.mean.diff])/nsim #estimated p-value

A random sample of 49 measurements from one population had a sample mean of 16, with sample standard deviation 3. An independent random sample of 64 measurements from a second population had a sample mean of 18, with sample standard deviation 4. Test the claim that the population means are different. Use level of significance 0.01. (a) What distribution does the sample test statistic follow? Explain.

The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.     The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.

(b) State the hypotheses.

H₀: μ₁ ≠ μ₂; H₁: μ₁ = μ₂H₀: μ₁ = μ₂; H₁: μ₁ ≠ μ₂     H₀: μ₁ = μ₂; H₁: μ₁ < μ₂H₀: μ₁ = μ₂; H₁: μ₁ > μ₂

(c) Compute

x₁ − x₂.

x₁ − x₂ =

Compute the corresponding sample distribution value. (Test the difference μ₁ − μ₂. Round your answer to three decimal places.)

An eating disorders clinic would like to assess the efficacy of their 10-week mindfulness training program with clients who have Binge Eating Disorder (BED). The clinic researchers first measured the number of binges in the previous week for 16 clients through self-report. One month after the mindfulness training sessions were conducted, the clients were again asked to report the number of binges in the last week. The data are listed in the table below. The clinic researchers have set the significance level at α = .05. # of Binges per week Subject Before Training After Training

Part I. (25 points total) a) Identify the outcome (dependent) variable and the independent variable (that differentiates the two populations being compared). What are the “samples” in this paired-samples t test? (Or, what are the “means” in this dependent-means t test?) (1 point) b) The clinic researchers predict the number of binges per week will decrease after the mindfulness training. In other words, the researchers believe the mindfulness training will be helpful in reducing binge eating. What would be the null and alternative hypotheses in both words and symbol notations? c) Calculate the difference scores by subtracting the “before” scores from the “after” scores. (In other words, set up the columns to calculate after minus before.) Create a table below for “difference score.” d) Calculate the mean from the sample of the difference scores. e) Estimate the standard deviation of the comparison population (that represents the null hypothesis). f) Calculate the standard error (standard deviation of the sampling distribution). g) Calculate the t statistic for the sample. h) Because the hypotheses are directional, a one-tailed test can be performed. Determine the critical t value based on the degrees of freedom and the preset alpha level. Compare the t statistic with the critical t value. Is the calculated t statistic more extreme or less extreme than the critical t value? Then make a decision about the hypothesis test, stating explicitly “reject” or “fail to reject” accordingly. i) Interpret the result in 1-2 sentences (you may restate the hypothesis accepted or explain it in your own words). ( a) Calculate the raw and standardized effect size of this hypothesis test. The clinic researchers could also set up the hypothesis to see if there are any differences (increases or decreases) in binge eating behavior after mindfulness training. a) What would be the null and alternative hypotheses for this alternative analysis? Compose them in symbol notations only. b) Since a non-directional hypothesis is examined with a two-tailed test, determine the critical t values for the two-tailed test using the same alpha level and degree of freedom. c) Compare the t statistic with the critical t values. Is the calculated t statistic more extreme or less extreme than the critical t value? What is the decision of the hypothesis test now? d) Was the two-tailed test result (Part II) different from the one-tailed test result (from Part I)? Why or why not?

This is the first question and I know how to solve this one, but I am confused by the second one (The admissions office of a small, selective liberal-arts college will only offer admission to applicants who have a certain mix of accomplishments, including a combined SAT score of 1,300 or more. Based on past records, the head of admissions feels that the probability is 0.58 that an admitted applicant will come to the college. If 500 applicants are admitted, what is the probability that 310 or more will come? Note that “310 or more” means the set of values {310, 311, 312, …, 500}. )

The following is the second question

Consider the admissions office in the previous problem. Based on financial considerations, the college would like a class size of 310 or more. Find the smallest n, number of people to admit, for which the probability of getting 310 or more to come to the college is at least 0.95.