"Radon: The Problem No One Wants to Face" is the title of an article appearing in Consumer Reports. Radon is a gas emitted from the ground that can collect in houses and buildings. At certain levels it can cause lung cancer. Radon concentrations are measured in picocuries per liter (pCi/L). A radon level of 4 pCi/L is considered "acceptable." Radon levels in a house vary from week to week. In one house, a sample of 8 weeks had the following readings for radon level (in pCi/L).

(a) Find the mean, median, and mode. (Round your answers to two decimal places.)

(b) Find the sample standard deviation, coefficient of variation, and range. (Round your answers to two decimal places.)

(c) Based on the data, would you recommend radon mitigation in this house? Explain.

Yes, since the median value is over "acceptable" ranges, although the mean value is not.Yes, since the average and median values are both over "acceptable" ranges.    No, since the average and median values are both under "acceptable" ranges.Yes, since the average value is over "acceptable" ranges, although the median value is not.

Respond to all of the following questions in your posting for this week:

• Describe the characteristics of the F distribution. Provide examples.
• What are we testing when we test for two population variances? Explain your answer and provide an example.
• What are the assumptions of an ANOVA, and when would you use an ANOVA?

Suppose certain coins have weights that are normally distributed with a mean of

5.641 g5.641 g

and a standard deviation of

0.069 g0.069 g.

A vending machine is configured to accept those coins with weights between

5.5215.521

g and

5.7615.761

g.

a. If

300300

different coins are inserted into the vending​ machine, what is the expected number of rejected​ coins?The expected number of rejected coins is

2525.

​(Round to the nearest​ integer.)b. If

300300

different coins are inserted into the vending​ machine, what is the probability that the mean falls between the limits of

5.5215.521

g and

5.7615.761

​g?The probability is approximately

(missing data)

​(Round to four decimal places as​ needed.)

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