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Example: From the previous example, μ=20, and σ=5. Suppose we draw a sample of size n=16 from this population and want to know how likely we are to see a sample average greater than 22, that is P(> 22)? So the probability that the sample mean is greater than 22 is between 0.005 and 0.025 (or between 0.5% and 2.5%) Learning Outcomes
The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by, the sample size. The variable n is the number of values that are averaged together, not the number of times the experiment is done. Central Limit TheoremSuppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose:
If you draw random samples of size n, then as n increases, the random variable [latex]\displaystyle\overline{{X}}[/latex]. To put it more formally, if you draw random samples of size n, the distribution of the random variable [latex]\displaystyle\overline{{X}}[/latex], which consists of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases. The random variable [latex]\displaystyle\overline{{X}}[/latex] in one sample. [latex]\displaystyle\frac{{\overline{X}-{\mu}_{x}}}{{\frac{{\sigma{x}}}{{\sqrt{n}}}}}[/latex] [latex]\displaystyle{\mu}_{x}[/latex] is the average of both X and [latex]\displaystyle\overline{X}[/latex]
ExampleAn unknown distribution has a mean of 90 and a standard deviation of 15. Samples of sizen = 25 are drawn randomly from the population.
Solution:
To find the value that is two standard deviations above the expected value 90, use the formula:
The value that is two standard deviations above the expected value is 96. The standard error of the mean is [latex]\displaystyle\frac{{\sigma}}{{\sqrt{n}}}[/latex] = [latex]\displaystyle\frac{{15}}{{\sqrt{25}}}[/latex] = 3. Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n. Try itAn unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size n = 30 are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50. P(42 < [latex]\displaystyle\overline{x}[/latex] < 50) = 42, 50, 45, [latex]\displaystyle\frac{{8}}{{\sqrt{30}}}[/latex] = 0.9797 ExampleThe length of time, in hours, it takes an “over 40” group of people to play one soccer match is normally distributed with a mean of two hours and a standard deviation of 0.5 hours. A sample of size n = 50 is drawn randomly from the population. Find the probability that the sample mean is between 1.8 hours and 2.3 hours. The length of time, in hours, it takes an “over 40” group of people to play one soccer match is normally distributed with a mean of two hours and a standard deviation of 0.5 hours. A sample of size n = 50 is drawn randomly from the population. Find the probability that the sample mean is between 1.8 hours and 2.3 hours. Let [latex]\displaystyle\overline{X}[/latex] = the mean time, in hours, it takes to play one soccer match. If [latex]{\mu}_{x}[/latex]= _________, [latex]{\sigma}_{x}[/latex] = __________, and n = ___________, then X~ N(______, ______) by the central limit theorem for means. [latex]{\mu}_{x}[/latex] = 2 Find P(1.8 < [latex]\displaystyle\overline{x}[/latex] < 2.3) Solution: P(1.8 < [latex]\displaystyle\overline{x}[/latex] < 2.3) P(1.8 < [latex]\displaystyle\overline{x}[/latex] < 2.3) = 0.9977
The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977. Try itThe length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of n = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours. P(2 < [latex]\overline{x}[/latex] < 3) = normalcdf( 2, 3, 2.5, [latex]\frac{0.25}{\sqrt{60}}[/latex] = 1 To find percentiles for means on the calculator, follow these steps.
ExampleIn a recent study reported Oct. 29, 2012 on the Flurry Blog, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size n = 100.
Solution:
Try itIn an article on Flurry Blog, a gaming marketing gap for men between the ages of 30 and 40 is identified. You are researching a startup game targeted at the 35-year-old demographic. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article’s data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. You take a sample of 100 randomly selected gamers. If your target market is 29- to 35-year-olds, should you continue with your development strategy? You need to determine the probability for men whose mean age is between 29 and 35 years of age wanting to play a strategy game. Find P(29 < [latex]\displaystyle\overline{x}[/latex] < 35) = normalcdf = 0.0186 (29, 35, 28, [latex]\displaystyle\frac{{4.8}}{{\sqrt{100}}}[/latex] = 0.0186 You can conclude there is approximately a 2% chance that your game will be played by men whose mean age is between 29 and 35. ExampleThe mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60.
Solution:
Try itCans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are n = 34,[latex]\displaystyle\overline{x}[/latex] = 16.01 ounces. If the cans are filled so that μ = 16.00 ounces (as labeled) and σ= 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces? We have P([latex]\displaystyle\overline{x}[/latex] > 16.01) = normalcdf(16.01, E99, 16, [latex]\displaystyle\frac{{0.143}}{{\sqrt{34}}}[/latex]= 0.3417 Since there is a 34.17% probability that the average sample weight is greater than 16.01 ounces, we should be skeptical of the company’s claimed volume. If I am a consumer, I should be glad that I am probably receiving free cola. If I am the manufacturer, I need to determine if my bottling processes are outside of acceptable limits.
ReferencesBaran, Daya. “20 Percent of Americans Have Never Used Email.”WebGuild, 2010. Available online at http://www.webguild.org/20080519/20-percent-of-americans-have-never-used-email (accessed May 17, 2013). Data from The Flurry Blog, 2013. Available online at http://blog.flurry.com (accessed May 17, 2013). Data from the United States Department of Agriculture. Concept ReviewIn a population whose distribution may be known or unknown, if the size ( n) of samples is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size (n). Formula ReviewThe Central Limit Theorem for Sample Means:[latex]\displaystyle\overline{X}{\sim}{N}({\mu}_{x},\frac{{{\sigma}_{x}}}{{\sqrt{n}}})[/latex] When the mean of a distribution of scores is 50 and the standard deviation is 10?Normal distribution with a mean of 50 and standard deviation of 10. 68% of the area is within one standard deviation (10) of the mean (50).
When the mean of a number is 18 what is the mean of the sampling distribution?So when mean of a number is going to be equals to 18, then mean of a sampling distribution? It remains the same, which is 18 only.
What is the standard error of the mean if the sample size is 25 with standard deviation?Thus, for a sample of N = 25 and population standard deviation of s = 100, the standard error of the mean is 100/5 or 20.
What is the mean of the distribution of sample means?The mean of the distribution of sample means is called the Expected Value of M and is always equal to the population mean μ.
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