Average American Watches Television

average american watches television
statistical problems please help?

Some people have an average of 519 and a standard deviation of 34. Several samples are size 60 randomly selected and the calculated half. (A) What value do you expect to find for the average of all these sample means? (B) What is the value you expect to find for the sample standard deviation of all these means? An article in April 2004, HearTheIssues.com said that Americans watch an average of 4.2 hours of television per person per day. If the standard deviation for the many hours of television watched per day is 3.3 and a random sample of 234 U.S. selected, the average of the sample belongs to a sampling distribution. (B) What is the average of the sampling distribution? (C) What is the deviation this sampling distribution standard?

Question 1) The central limit theorem tells us that for any sample if the sample size is sufficiently large, the average will be followed the normal distribution with mean equal to the sample mean and variance equal to the variance of the sample divided by the sample size. In other words, if we have a sample mean and variance μ ² σ then the sample mean, Xbar, follow the normal distribution with μ mean and variance σ ² / N, or standard deviation σ / √ n (). Note: If you do not have stock or υ parameters σthen them closer to the sample values Xbar and S. In this edition, Xbar ~ Normal (μ = 519, σ ² = 1156-1160) Xbar ~ Normal (μ = 519, σ ² = 19.26667) Xbar ~ Normal (μ = 519, σ = 34 / sqrt (60)) Xbar ~ Normal (μ = 519, σ = 4.389381) Question 2) Xbar ~ Normal (μ = 4.2, σ ² = 10.89 / 234) Xbar ~ Normal (μ = 4.2, σ ² = 0.04653846) Xbar ~ Normal (μ = 4.2, σ = 3.3 / sqrt (234)) Xbar ~ Normal (μ = 4.2, σ = 0.2157277)

The Average American Male on CNBC

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