Suppose that a sample of size n-3 is selected from a population of four value 4, 8, 0, 12. (1) Find the population distribution. (2) Find the mean and standard deviation of the population distribution. (3) Suppose that the sampling is without replacement. (a) Find the sampling distribution of the sample mean. (b) Find the mean of this sampling distribution. (c) Find the standard deviation of this sampling distribution. (Calculate
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Given Information:
A population consists of four values: 4, 8, 0, 12
A sample of size n=3 is selected from the population.
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- Which of the following is true regarding the sampling distribution of the mean for a large sample size? O It has a normal distribution with the same mean and standard deviation as the population. O It has the same shape, mean, and standard deviation as the population. O It has the same shape and mean as the population, but has a smaller standard deviation. O It has a normal distribution with the same mean as the population but with a smaller standard deviation.Let a population consist of the values 7 cigarettes, 8 cigarettes, and 19 cigarettes smoked in a day. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? Calculate the mean absolute deviation for each possible sample of size 2 from the population. Sample Mean Absolute Deviation {7,7} {7,8} 0 0.5 6 {7,19} {8,7} 0.5 {8,8} 0 {8,19} 5.5 {19,7} 6 {19,8} 5.5 {19,19} 0 (Type integers or decimals rounded to one decimal place as needed.) Calculate the mean of the sample mean absolute deviations. The mean of the sample mean absolute deviations is (Type an integer or decimal rounded to one decimal place as needed.) 2 9The annual salary for one particular occupation is normally distributed, with a mean of about $122,000 and a standard deviation of about $22,000. Random samples of 34 are drawn from this population, and the mean of each sample is determined. Find the mean and standard deviation of the sampling distribution of these sample means. Then, sketch a graph of the sampling distribution.
- Use the central limit theorem to find the mean and standard error of the mean of the indicated sampling distribution. Then sketch a graph of the sampling distribution. The per capita consumption of red meat by people in a country in a recent year was normally distributed, with a mean of 117 pounds and a standard deviation of 39.8 pounds. Random samples of size 20 are drawn from this population and the mean of each sample is determined. = μx ox (Round to three decimal places as needed.) = Sketch a graph of the sampling distribution. Choose the correct graph below. A. B. O C. O D. A A A -108.1 8.9 125.9 90.3 117 143.7 -342.1 8.9 359.9 99.2 117 134.8Use the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution. The amounts of time employees of a telecommunications company have worked for the company are normally distributed with a mean of 5.90 years and a standard deviation of 1.80 years. Random samples of size 14 are drawn from the population and the mean of each sample is determined. Round the answers to the nearest hundredth.Let a population consist of the values 7 cigarettes, 21 cigarettes, and 22 cigarettes smoked in a day. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? Calculate the mean absolute deviation for each possible sample of size 2 from the population. Sample Mean Absolute Deviation {7,7} __ {7,21} __ {7,22} __ {21,7} __ {21,21} __ {21,22} __ {22,7} __ {22,21} __ {22 ,22}
- Use the central limit theorem to find the mean and standard error of the mean of the indicated sampling distribution. Then sketch a graph of the sampling distribution. The per capita consumption of red meat by people in a country in a recent year was normally distributed, with a mean of 117 pounds and a standard deviation of 39.4 pounds. Random samples of size 15 are drawn from this population and the mean of each sample is determined. 4-0 0 = (Round to three decimal places as needed.) Sketch a graph of the sampling distribution. Choose the correct graph below. O A. A 96.7 117 137.3 Q O B. A 340.8 10.2 361.2 Q Q ww ^ A -106.8 10.2 127.2 86.5 117 147.5 O C. O D. QA variable of two populations has a mean of 30 and a standard deviation of 18 for one of the populations and a mean of 30 and a standard deviation of 32 for the other population. Complete parts (a) through (c). a. For independent samples of size 9 and 16, respectively, find the mean and standard deviation of X₁ -X₂. (Assume that the sampling is done with replacement or that the population is large enough.) The mean of X₁ -X₂ is. (Type an integer or a decimal. Do not round.) The standard deviation of X₁-X₂ is (Round to four decimal places as needed.) b. Must the variable under consideration be normally distributed on each of the two populations for you to answer part (a)? Choose the correct answer below. OA. No, the variable does not need to be normally distributed for the formulas for the mean and standard deviation of x₁-x₂ to hold as long as the sample sizes are large enough, as long as the sampling is done with replacement. B. No, the formulas for the mean and standard deviation of…Use the central limit theorem to find the mean and standard error of the mean of the indicated sampling distribution. Then sketch a graph of the sampling distribution. The per capita consumption of red meat by people in a country in a recent year was normally distributed, with a mean of 113 pounds and a standard deviation of 37.7 pounds. Random samples of size 20 are drawn from this population and the mean of each sample is determined.
- Let a population consist of the values 11 cigarettes, 12 cigarettes, and 22 cigarettes smoked in a day. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? Calculate the mean absolute deviation for each possible sample of size 2 from the population. Mean Absolute Deviation Sample {11,11} {11,12} {11,22} {12,11} {12,12} {12,22} {22,11} {22,12} {22,22} (Type integers or decimals rounded to one decimal place as needed.) Budget Pi nd Pow....P Enter your answer in the edit fields and then click Check Answer. Clear All Check Answer partsUse the Central Limit Theorem to find the mean and standard error of the mean of the sampling distribution. Then sketch a graph of the sampling distribution. The mean price of photo printers on a website is $227 with a standard deviation of $68. Random samples of size 21 are drawn from this population and the mean of each sample is determined.Use the central limit theorem to find the mean and standard error of the mean of the indicated sampling distribution. Then sketch a graph of the sampling distribution. The per capita consumption of red meat by people in a country in a recent year was normally distributed, with a mean of 109 pounds and a standard deviation of 38.3 pounds. Random samples of size 15 are drawn from this population and the mean of each sample is determined. μx=nothing σx=nothing (Round to three decimal places as needed.) Sketch a graph of the sampling distribution. Choose the correct graph below. A. 109 89.2128.8 x A normal curve is over a horizontal x overbar axis labeled from 89.2 to 128.8 and is centered on 109. B. 9.9 -99.1118.9 x A normal curve is over a horizontal x overbar axis labeled from negative 99.1 to 118.9 and is centered on 9.9. C. 9.9 -317.1336.9 x A normal…