Main Index > Central Limit Theorem Examples

A

**Central Limit Theorem**word problem will most likely contain the phrase “assume the variable is

**normally distributed**”, or one like it. You will be given:

- A population (i.e. 29-year-old males, seniors between 72 and 76, all registered vehicles, all cat owners)
- An average (i.e. 125 pounds, 24 hours, 15 years, $15.74)
- A standard deviation (i.e. 14.4lbs, 3 hours, 120 months, $196.42)
- A sample size (i.e. 15 males, 10 seniors, 79 cars, 100 households)

**Click the link to skip down to one of three central limit theorem examples: **

- I want to find the probability that the mean is
**greater**than a certain number - I want to find the probability that the mean is
**less**than a certain number - I want to find the probability that the mean is
**between**a certain set of numbers either side of the mean

### Central Limit Theorem Examples: Greater than

For Central Limit Theorem word problems that contain the phrase “greater than” (or a similar phrase such as “above”).

**Step 1:***Identify the parts of the problem*. Your question should state:

- the mean (average or μ)
- the standard deviation (σ)
- population size
- sample size (n)
- a number associated with “greater than” ( )

**Step 2:** *Draw a graph. *Label the center with the mean. Shade the area roughly above (i.e. the “greater than” area). This step is optional, but it may help you see what you are looking for.

**Step 3:** *Use the following formula to find the z-score.* Plug in the numbers from step 1.

Click here if you want easy, step-by-step instructions for solving this formula.

- Subtract the mean (μ in step 1) from the greater than’ value (in step 1). Set this number aside for a moment.
- Divide the standard deviation (σ in step 1) by the square root of your sample (n in step 1). For example, if thirty six children are in your sample and your standard deviation is 3, then 3/√36=0.5
- Divide your result from step 1 by your result from step 2 (i.e. step 1/step2)

**Step 4:** *Look up the z-value you calculated in step 3 in the z-table. *If you don’t remember how to look up z-scores, you can find an explanation in step 1 of this article: Area to the right of a z-score on a normal distribution curve.

**Step 5: ***Subtract your z-score from 0.5.* For example, if your z-score is 0.1554, then 0.5 – 0.1554 = 0.3446.

**Step 6:** *Convert the decimal in Step 5 to a percentage*. In our example, 0.3446 = 34.46%.

That’s it!

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### Central Limit Theorem Examples: Less than

Solving **Central Limit Theorem** word problems that contain the phrase “less than” (or a similar phrase such as “lower”).

**Step 1:** *Identify the parts of the problem*. Your question should state:

- the mean (average or μ)
- the standard deviation (σ)
- population size
- sample size (n)
- a number associated with “less than” ( )

**Step 2:** *Draw a graph. *Label the center with the mean. Shade the area roughly below (i.e. the “less than” area). This step is optional, but it may help you see what you are looking for.

**Step 3:** *Use the following formula to find the z-value. *Plug in the numbers from step 1.

Click here if you want simple, step-by-step instructions for using this formula.

If formulas confuse you, all this formula is asking you to do is:

- Subtract the mean (μ in step 1) from the less than’ value ( in step 1). Set this number aside for a moment.
- Divide the standard deviation (σ in step 1) by the square root of your sample (n in step 1). For example, if thirty six children are in your sample and your standard deviation is 2, then 3/√36=0.5
- Divide your result from step 1 by your result from step 2 (i.e. step 1/step2)

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**Step 4:** *Look up the z-value you calculated in step 4 in the z-table. *If you don’t remember how to look up z-scores, you can find an explanation in step 1 of this article on area to the right of a z-score in a normal distribution curve.

**Step 5:** *Add your z-score to 0.5.* For example, if your z-score is 0.1554, then 0.5 + 0.1554 is 0.6554.

**Step 6:***Convert the decimal in Step 6 to a percentage*. In our example, 0.6554=65.54%.

That’s it!

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### Central Limit Theorem Examples: Between

Sample problem: There are 250 dogs at a dog show who weigh an average of 12 pounds, with a standard deviation of 8 pounds. If 4 dogs are chosen at random, what is the probability they have an average weight of greater than 8 pounds and less than 25 pounds?

Step 1:Identify the parts of the problem. Your question should state:

- mean (average or μ) standard deviation (σ) population size
- sample size (n)
- number associated with “less than” 1
- number associated with “greater than” 2

Step 2: Draw a graph. Label the center with the mean. Shade the area between 1 and 2. This step is optional, but it may help you see what you are looking for.

Step 3: Use the following formula to find the z-values.

All this formula is asking you to do is:

a) Subtract the mean (μ in Step 1) from the greater than value (Xbar in Step 1): 25-12=13.

b) Divide the standard deviation (σ in Step 1) by the square root of your sample (n in Step 1): 8/sqrt4=4

c) Divide your result from *a* by your result from *b*: 13/4=3.25

Step 4 Use the formula from Step 3 to find the z-values. This time, use Xbar2 from Step 1 (8).

a) Subtract the mean (μ in Step 1) from the greater than value (Xbar in Step 1): 8-12=-4.

b) Divide the standard deviation (σ in Step 1) by the square root of your sample (n in Step 1): 8/sqrt4=4

c) Divide your result from *a* by your result from *b:* -4/4= -1

Step 5: Look up the z-value you calculated in Step 3 in the z-table.

Z value of 3.25 corresponds to .4994

Step 6: Look up the z-value you calculated in Step 4 in the z-table.

Z value of 1 corresponds to .3413

Note that the z-table is symmetrical, so if you want to look up a negative value like -1, then just look up the positive counterpart. The area will be the same.

Step 7: Add Step 5 and 6 together:

.4994 + .3413 = .8407

Step 8: Convert the decimal in Step 7 to a percentage:

.8407 = 84.07%

That’s it!

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I did not find the internet link to the explanation for finding the mean between a certain set of numbers as helpful as the instructions you post here on the blog. I thought that it was more difficult to follow.

This one is pretty easy. I think I understand these types of problems.