1 Go to the American Time Use Survey web page.the sample standard deviation from

Question

1 Go to the American Time Use Survey web page.the sample standard deviation from youe Research variables of interest to you (or would be of interest to your readers) Determine amount of time Americans who are enrolled in school spend doing the activities you find interesting 2 Use StatCrunch or an online polling site (such as unusual? populaboh survey sample as the population standard 4. Compute probabilities regarding statistics obtained from the study Are any as that you belleve the mean deviation values st th Are ary of the teu 5. Write an article for your newspaper findings. Be sure any interpretations are wr 5 Write an article for your sample of students at your school Write questions to learn how students at your school use their time For example. how are statistically accurate, but rin much time does a student at your school spend attending surveymonkey.com) to conduct a survey of a random dable tor t are statistically accurate, but unders statistically untrained reader CASE STUDY Y sampling Distribution of the Median The exponential distribution is an example of a skewed distribution. It is defined by the density function In our discussion of summary statistics seen that the mean is used as a measure of the data sets whose distribution is (roughly) symm no outliers. In the presence of outliers or skeu used the median as the measure of the center with parameter A.The mean of an exponential distribution isand the median is M- is resistant to extreme values. However, for distributions, it is often difficult to determine measure of center is more useful. In some cases the failure rate of electrical components. For many electrical is more intuitive to the reader within the contex The exponential distribution is used to model the though not resistant, is easier to find than the the failure rate over time appears components, apio othtut. The exponential distribution problem. Here we loo the mean and median for two different opulation!n k at the sampling distribution d to be shaped like a exp is used widely in reliability engineering for the constant hazard rate portion of the bathtub curve, components spend most of their useful life. The portion1. Using statistical software, generate 250 sampe curue owherenormal (ymmetric) and one exponential ckewed, of the curve prior to the constant rate is referred to as the burn-in period, and the portion after the constant rate is referred to as the wear-out period (rows) of size 80 (columns) from a nom distribution with mean 50 and standard deviatin 2. Compute the mean for the first 10 values in each and store the values in C81. You might label this co n10mn" for normal distribution, sample size 10.m 3. Compute the median for the first 10 values in cad row and store the values in C82. You might label column "n10med" for normal distribution, samp Another interesting property of the exponential distribution is the memoryless property. This says that the time to failure of a component does not depend on how long the component has been in use. For example. the probability that a component will fail in the next hour is the same whether the component has been in use for 10, 50, or 100 hours, and so on. size 10, median.

Explanation / Answer

Here we can use R software.

1. Code:

set.seed(1689)
sample <- as.data.frame(matrix(rnorm(250*80,50,10),nrow = 250))
names(sample) <- gsub("V",replacement = "C",x = names(sample))

Output:

> set.seed(1689)
> sample <- as.data.frame(matrix(rnorm(250*80,50,10),nrow = 250))
> names(sample) <- gsub("V",replacement = "C",x = names(sample))

[Note: Your image is not clear so I assumed the standard deviation is 10.If this will change then just put the third element under the rnorm() as the right value of sd]

2. Code:

sample$"C81" <- apply(sample[,1:10], 1, mean)

Output:

> sample$"C81" <- apply(sample[,1:10], 1, mean)

3. Code:

sample$"C82" <- apply(sample[,1:10], 1, median)

Output:

> sample$"C82" <- apply(sample[,1:10], 1, median)

4. Code:

sample$"C83" <- apply(sample[,1:20], 1, mean)
sample$"C84" <- apply(sample[,1:20], 1, median)
sample$"C85" <- apply(sample[,1:40], 1, mean)
sample$"C86" <- apply(sample[,1:40], 1, median)
sample$"C87" <- apply(sample[,1:80], 1, mean)
sample$"C88" <- apply(sample[,1:80], 1, median)

Output:

> sample$"C82" <- apply(sample[,1:10], 1, median)
> sample$"C83" <- apply(sample[,1:20], 1, mean)
> sample$"C84" <- apply(sample[,1:20], 1, median)
> sample$"C85" <- apply(sample[,1:40], 1, mean)
> sample$"C86" <- apply(sample[,1:40], 1, median)
> sample$"C87" <- apply(sample[,1:80], 1, mean)
> sample$"C88" <- apply(sample[,1:80], 1, median)

5.Code:

summary(sample[,81:88])

Output:

> summary(sample[,81:88])
C81 C82 C83 C84
Min. :41.10 Min. :38.41 Min. :43.49 Min. :42.41  
1st Qu.:48.06 1st Qu.:47.63 1st Qu.:48.80 1st Qu.:48.26  
Median :50.21 Median :50.42 Median :50.00 Median :50.19  
Mean :50.22 Mean :50.20 Mean :50.16 Mean :50.15  
3rd Qu.:52.32 3rd Qu.:52.75 3rd Qu.:51.65 3rd Qu.:51.99  
Max. :62.30 Max. :62.19 Max. :56.36 Max. :57.94  
C85 C86 C87 C88
Min. :45.55 Min. :44.55 Min. :46.27 Min. :46.39  
1st Qu.:49.04 1st Qu.:48.67 1st Qu.:49.46 1st Qu.:49.25  
Median :50.37 Median :50.51 Median :50.14 Median :50.22  
Mean :50.23 Mean :50.26 Mean :50.20 Mean :50.20  
3rd Qu.:51.44 3rd Qu.:51.69 3rd Qu.:50.96 3rd Qu.:51.15  
Max. :55.45 Max. :54.93 Max. :53.42 Max. :53.64

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