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6.1 The central limit theorem for sample means (averages) -- rrc  (Page 2/17)

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The 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< x ¯ <3) = 1

In 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.

  1. What are the mean and standard deviation for the sample mean ages of tablet users?
  2. What does the distribution look like?
  3. Find the probability that the sample mean age is more than 30 years (the reported mean age of tablet users in this particular study).
  4. Find the 95 th percentile for the sample mean age (to one decimal place).
  1. Since the sample mean tends to target the population mean, we have μ χ = μ = 34. The sample standard deviation is given by σ χ = σ n = 15 100 = 15 10 = 1.5
  2. The central limit theorem states that for large sample sizes( n ), the sampling distribution will be approximately normal.
  3. The probability that the sample mean age is more than 30 is given by P ( Χ >30) = normalcdf (30,E99,34,1.5) = 0.9962
  4. Let k = the 95 th percentile.
    k = invNorm ( 0. 95,34, 15 100 ) = 36.5

Try it

In 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.

P (29< x ¯ <35) = 0.0186

You can conclude there is approximately a 1.9% chance that your game will be played by men whose mean age is between 29 and 35.

The 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.

  1. What are the mean and standard deviation for the sample mean number of app engagement by a tablet user?
  2. What is the standard error of the mean?
  3. Find the 90 th percentile for the sample mean time for app engagement for a tablet user. Interpret this value in a complete sentence.
  4. Find the probability that the sample mean is between eight minutes and 8.5 minutes.
  1. μ x ¯ = μ = 8.2   σ x ¯ = σ n = 1 60 = 0.13
  2. This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60.
  3. Let k = the 90 th percentile
    k = 8.37. This values indicates that 90 percent of the average app engagement time for table users is less than 8.37 minutes.
  4. P (8< x ¯ <8.5) = 0.9293

Try it

Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are n = 34, x ¯ = 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 (( x ¯ >16.01) = 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.

References

Baran, 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.

Chapter review

In 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 review

The Central Limit Theorem for Sample Means: X ¯ ~ N ( μ x ,  σ x n )

The Mean X ¯ : μ x

Central Limit Theorem for Sample Means z-score and standard error of the mean: z = x ¯ − μ x ( σ x n )

Standard Error of the Mean (Standard Deviation ( X ¯ )): σ x n

Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the random variable representing the time it takes her to complete one review. Assume Χ is normally distributed. Let X ¯ be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.

What is the mean, standard deviation, and sample size?

mean = 4 hours; standard deviation = 1.2 hours; sample size = 16

Complete the distributions.

  1. X ~ _____(_____,_____)
  2. X ¯ ~ _____(_____,_____)

Find the probability that one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.

  1. This is a frequency curve for a normal distribution. It shows a single peak in the center with the curve tapering down to the horizontal axis on each side. The distribution is symmetrical. The horizontal axis represents the random variable X.
  2. P (________< x <________) = _______

a. Check student's solution.
b. 3.5, 4.25, 0.2441

Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.

  1. This is a frequency curve for a normal distribution. It shows a single peak in the center with the curve tapering down to the horizontal axis on each side. The distribution is symmetrical. The horizontal axis represents the random variable X.
  2. P (________________) = _______

What causes the probabilities in [link] and [link] to be different?

The fact that the two distributions are different accounts for the different probabilities.

Find the 95 th percentile for the mean time to complete one month's reviews. Sketch the graph.

  1. This is a frequency curve for a normal distribution. It shows a single peak in the center with the curve tapering down to the horizontal axis on each side. The distribution is symmetrical. The horizontal axis represents the random variable X.
  2. The 95 th Percentile =____________
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Source:  OpenStax, Statistics i - math1020 - red river college - version 2015 revision a - draft 2015-10-24. OpenStax CNX. Oct 24, 2015 Download for free at http://legacy.cnx.org/content/col11891/1.8
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