"

Chapter 6: The Normal Distribution and The Central Limit Theorem

6.4 The Central Limit Theorem for Sums

Learning Objectives

By the end of this section, the student should be able to:

  • Recognize characteristics of the Central Limit Theorem for the Sums
  • Apply and interpret the Central Limit Theorem for the Sums and use it to solve real-world applications.

Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:

  1. [latex]\mu_X = \text{the mean of } X[/latex]
  2. [latex]\sigma_X = \text{the standard deviation of } X[/latex]

If you draw random samples of size n, then as n increases, the random variable ΣX consisting of sums tends to be normally distributed and [latex]\Sigma{X} \sim N \left((n)(\mu_x), (\sqrt{n})(\sigma_X) \right)[/latex].

The central limit theorem for sums says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution), which approaches a normal distribution as the sample size increases. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size.

The random variable [latex]\Sigma{X}[/latex] has the following z-score associated with it:

  1. [latex]\Sigma{X}[/latex] is one sum.
  2. [latex]z = \frac{\Sigma x–(n)({\mu }_{X})}{(\sqrt{n})({\sigma }_{X})}[/latex]
    1. [latex](n)(\mu_X) = \text{the mean of } \Sigma{X}[/latex]
    2. [latex](\sqrt{n})({\sigma }_{X}) = \text{standard deviation of }\Sigma X[/latex]

Note

It is important for you to understand when to use the central limit theorem and when to differentiate between the CLT for the Mean versus the CLT for the Sum (or Total). If you are being asked to find the probability of the mean, use the clt for the mean, but if you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums.

Example

An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population.

  1. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500.
  2. Find the sum that is 1.5 standard deviations above the mean of the sums.
Solution

Let [latex]X =[/latex] one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values.

[latex]\Sigma{X} = \text{the sum or total of 80 values}[/latex]. Since [latex]\mu_X = 90[/latex], [latex]\sigma_X = 15[/latex], and [latex]n = 80[/latex], [latex]\Sigma X \sim N \left((80)(90), (\sqrt{80})(15) \right)[/latex]

  • [latex]\text{mean of the sums} = (n)(\mu_X) = (80)(90) = 7,200[/latex]
  • [latex]\text{standard deviation of the sums} = (\sqrt{n})(\sigma _{X}) = (\sqrt{80})(15)[/latex]
  • [latex]\text{sum of 80 values} = \Sigma{X} = 7,500[/latex]
  1. Find [latex]P(\Sigma{x} > 7,500)[/latex]. [latex]P(\Sigma{x} > 7,500) = 0.0127[/latex].
  2. Find [latex]\Sigma{x}[/latex] where [latex]z = 1.5[/latex]. [latex]\Sigma{x} = (n)(\mu_x)+(z)(\sqrt{n})(\sigma_x) = (80)(90) + (1.5)(\sqrt{80})(15) = 7,401.2[/latex]
A normal distribution curve with the at 7200 and 7500 to the right. The area to the right of 7500 below the curve is shaded.
Figure 1. The shaded area represents the probability of the sum being greater than 7500

Your Turn!

An unknown distribution has a mean of 45 and a standard deviation of eight. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400.

Solution

0.0040

Section 6.4 Review

The central limit theorem tells us that for a population with any distribution, the distribution of the sums for the sample means approaches a normal distribution as the sample size increases. In other words, if the sample size is large enough, the distribution of the sums can be approximated by a normal distribution even if the original population is not normally distributed. Additionally, if the original population has a mean of [latex]\mu_x[/latex] and a standard deviation of [latex]\sigma_x[/latex], the mean of the sums is [latex]n \mu_x[/latex] and the standard deviation is [latex]\left(\sqrt{n}\right)(\sigma_x)[/latex] where [latex]n[/latex] is the sample size.

Formula Review

The Central Limit Theorem for Sums: [latex]\Sigma{X} \sim N[(n)(\mu_x), (\sqrt{n})(\sigma_x)][/latex]

Mean for Sums ([latex]\Sigma{X}[/latex]): [latex](n)(\mu_x)[/latex]

The Central Limit Theorem for Sums z-score and standard deviation for sums: [latex]z\text{ (for the sample mean)} = \frac{\Sigma x–\left(n\right)\left({\mu }_{X}\right)}{\left(\sqrt{n}\right)\left({\sigma }_{X}\right)}[/latex]

Standard deviation for Sums ([latex]\Sigma{X}[/latex]): [latex]\left(\sqrt{n}\right)(\sigma_x)[/latex]

Section 6.4 Practice

An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population.

  1. Find the probability that the sum of the 95 values is greater than 7,650.
  2. Find the probability that the sum of the 95 values is less than 7,400.
  3. Find the sum that is two standard deviations above the mean of the sums.
  4. Find the sum that is 1.5 standard deviations below the mean of the sums.
Solution
  1. 0.3345
  2. 0.0436
  3. 7,833.92
  4. 7,424.56

A researcher measures the amount of sugar in several cans of the same soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100.

  1. Find the probability that the sum of the 100 values is greater than 3,910.
  2. Find the probability that the sum of the 100 values is less than 3,900.
  3. Find the probability that the sum of the 100 values falls between the numbers you found in number 1 and number 2.
  4. Find the sum with a z–score of –2.5.
  5. Find the sum with a z–score of 0.5.
  6. Find the probability that the sums will fall between the z-scores –2 and 1.
Solution
  1. 0.0359
  2. 0.4207
  3. 0.5433
  4. 3,888.5
  5. 3,903.5
  6. 0.8186

An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object of interest.

  1. What is the mean of [latex]\Sigma{X}[/latex]?
  2. What is the standard deviation of [latex]\Sigma{X}[/latex]?
  3. What is [latex]P(\Sigma{X} = 290)[/latex]?
  4. What is [latex]P(\Sigma{X} > 290)[/latex]?
Solution
  1. 300
  2. 5
  3. 0
  4. 0.9772

True or False: Only the sums of normal distributions are also normal distributions.

Solution

False, the sums of any distribution approach a normal distribution as the sample size increases.

In order for the sums of a distribution to approach a normal distribution, what must be true?

Solution

The sample size, n, gets larger.

What three things must you know about a distribution to find the probability of sums?

Solution

the mean of the distribution, the standard deviation of the distribution, and the sample size

An unknown distribution has a mean of 25 and a standard deviation of six. Let [latex]X =[/latex] one object from this distribution. What is the sample size if the standard deviation of [latex]\Sigma{X}[/latex] is 42?

Solution

49

An unknown distribution has a mean of 19 and a standard deviation of 20. Let [latex]X =[/latex] the object of interest. What is the sample size if the mean of [latex]\Sigma{X}[/latex] is 15,200?

Solution

800

A market researcher analyzes how many electronics devices customers buy in a single purchase. The distribution has a mean of three with a standard deviation of 0.7. She samples 400 customers.

  1. What is the z-score for [latex]\Sigma{X} = 840[/latex]?
  2. What is the z-score for [latex]\Sigma{X} = 1,186[/latex]?
  3. What is [latex]P(\Sigma{X} \lt 1,186)[/latex]?
Solution
  1. -25.7
  2. –1
  3. 0.1587

An unknown distribution has a mean of 100, a standard deviation of 100, and a sample size of 100. Let [latex]X =[/latex] one object of interest.

  1. What is the mean of [latex]\Sigma{X}[/latex]?
  2. What is the standard deviation of [latex]\Sigma{X}[/latex]?
  3. What is [latex]P(\Sigma{X} > 9,000)[/latex]?
Solution
  1. 10,000
  2. 1,000
  3. 0.8413

Which of the following is NOT TRUE about the theoretical distribution of sums?

  1. The mean, median and mode are equal.
  2. The area under the curve is one.
  3. The curve never touches the x-axis.
  4. The curve is skewed to the right.
Solution

The curve is skewed to the right.

Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days and a standard deviation of seven days. We randomly sample nine trials.

  1. In words, [latex]\Sigma{X} = \underline{\hspace{2cm}}[/latex]
  2. [latex]\Sigma{X} \sim \underline{\hspace{2cm}}(\underline{\hspace{2cm}}, \underline{\hspace{2cm}})[/latex]
  3. Find the probability that the total length of the nine trials is at least 225 days.
  4. Ninety percent of the total of nine of these types of trials will last at least how long?
Solution
  1. the total length of time for nine criminal trials
  2. [latex]N(189, 21)[/latex]
  3. 0.0432
  4. 162.09; ninety percent of the total nine trials of this type will last 162 days or more.

Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from two to six pounds with a mean of four pounds and standard deviation of 1.1547. We randomly survey 64 homes with children.

  1. In words, [latex]X = \underline{\hspace{2cm}}[/latex]
  2. The distribution is [latex]\underline{\hspace{2cm}}[/latex].
  3. In words, [latex]\Sigma{X} = \underline{\hspace{2cm}}[/latex]
  4. [latex]\Sigma{X} \sim \underline{\hspace{2cm}}(\underline{\hspace{2cm}}, \underline{\hspace{2cm}})[/latex]
  5. Find the probability that the total weight of open boxes is less than 250 pounds.
  6. Find the 35th percentile for the total weight of open boxes of cereal.

 

Solution
  1. [latex]X =[/latex] weight of open cereal boxes
  2. The distribution is uniform.
  3. [latex]\Sigma{X} = the sum of the weights of 64 randomly chosen cereal boxes[/latex]
  4. [latex]N(256, 9.24)[/latex]
  5. 0.2581
  6. 252.44 pounds

References

Farago, Peter. “The Truth About Cats and Dogs: Smartphone vs Tablet Usage Differences.” The Flurry Blog, October 29, 2012. Available online at https://www.flurry.com/blog (accessed May 17, 2013).

License

Icon for the Creative Commons Attribution-ShareAlike 4.0 International License

Introductory Statistics Copyright © 2024 by LOUIS: The Louisiana Library Network is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book