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- A population consists of the following 6 values: {10, 20, 30, 40, 50, 60}. A sample of size 3 will be chosen from this population without replacement and the median of the sample will be computed: { (0.5 (£n/2 + £n/2+1) m = x(n+1)/2 n odd n even In other words, in a sample of size n = 3 ordered from smallest to largest the median is the 2nd number. The median of a discrete random variable, μ, is defined as the value of such that and P(X M) ≤ 0.5 Determine the median of the population. Think of the population as an equally likely sample space. Determine the sampling distribution of the sample median, m for this population by (1) identifying all possible samples (how many are there?), (2) calculating the median in each sample and (3) using that to determine the probability of each median. Calculate E (m) Determine the standard error or the sample medianThe assets (in billions of dollars) of the four wealthiest people in a particular country are 41, 25, 19, 14. Assume that samples of size n = 2 are randomly selected with replacement from this population of four values. each sample, construct a table representing the sampling distribution of the sample mean. In the table, values of the sample mean that are the same have been combined. X 41 33 30 27.5 25 (Type integers or fractions.) Probability The mean of the population, sample means, (Round to two decimal places as needed.) b. Compare the mean of the population to the mean of the sampling distribution of the sample mean. means is The sample means 22 19.5 19 16.5 14 Probability c. Do the sample means target the value of the population mean? In general, do sample means make good estimates of population means? Why or why not? the mean of the the population mean. In general, sample make good estimates of population means because the mean is estimator.5. Prove that the mean of Binomial Distribution is np and the variance is np(1 – p).
- Suppose that a simple random sample of n = 2 numbers will be selected from the list of values 10, 12, 14, 16, 18. (a) There are ten possible equally likely samples of n 2 numbers that could be selected (10 and 12, 10 and 14, etc.). Calculate the sample mean for each sample. = Sample 10, 12 Sample 12, 16 I 10, 14 12, 18 10, 16 14, 16 10, 18 12, 14 14, 18 16, 18 (b) Summarize the results of part (a) into a table showing the sampling distribution of possible sample means. To do this, list each possible value for the sample mean along with the probability the value would occur. (Enter your answers from smallest to largest value for the sample mean.) ProbabilityWhich one of the followings is true about the variance of the sample mean of a random sample of size greater than one? a.)It is equal to the variance of the population from which the sample is drawn b.)It is smaller than the variance of the population from which the sample is drawn c.)It is greater than the variance of the population from which the sample is drawn d.)Its value cannot be compared with the variance of the population from which the sample is drawn e.)None of the aboveA population consists of the two numbers (7, 8). Consider all possible samples ofsize 2 that can be drawn with replacement from this population. Find the following:a. The mean of the sampling distribution of meansb. The variance of the sampling distributionc. The standard deviation of the sampling distribution of means.d. Illustrate the probability histogram of the sampling distribution of themeans. The average number of liters of water that a person consumes in a month is 12 liters. Assume that the standard deviation is 4.5 liters and the distribution isapproximately normal.a. Find the probability that a person selected at random consumes morethan 13 liters per month.b. If a sample of 20 individuals is selected, find the probability that mean ofthe sample will be more than 13 liters. Identify the region under the t-distribution corresponding to the t-value which is 1.440 and the area is on both sides of the t-distribution. Identify the region under the t-distribution corresponding to…
- 5. Suppose X is the number of successes in a random sample of size n from a large population with a proportion of success p. What are the mean and the standard deviation of the sampling distribution of X in terms of n and p? The mean is p(l-pland the standard deviation is p(1-p) b. The mean is npand the standard deviation is ap(1-p). The mean is p(1-pland the standard deviation is pp. C. d. The mcan is 7pand the standard deviation is rp.Four-year-olds in China average 3 unsupervised hours per day. Most of the unsupervised children live in rural areas, considered safe. Suppose that the amount of unsupervised time is normally distributed with standard deviation 0.5 . A Chinese 4-year-old is randomly selected from a rural area. We are interested in the amount of time the child spends alone per day. In each appropriate box you are to enter either a rational number in "p/q" format or a decimal value accurate to the nearest 0.01 a. (3%) In words, the random variable X in which we are interested is defined by X = (pick one) b. (3%) Find the probability that the child spends less than one hour per day unsupervised. P(X 10) = d. (3%) The 70th percentile of the distribution of unsupervised time is given by P(X < = 0.70 .SITUATION: A population of PWD learners in 5 schools in a certain municipality of Cavite City are (x) 9, 5, 6, 12 and 15. Suppose that 2 schools were selected as samples, determine the mean and variance of the sampling distribution of sample mean List all possible samples and their corresponding means SAmples od size 2 Mean Construct the sampling distribution of the sample means SAMPLING DISTRIBUTION OF SAMPLE MEANS Sample means Frequency Probability Total (f) Compute the mean of the sampling distribution of the sample means. Sample mean Probability S.Mean * Probability