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## TI 83 Central Limit Theorem: Overview

The Central Limit Theorem (CLT) is a way to show characteristics for the “sampling distribution of the means,” taken from a “parent population” which is created from the means of an infinite number of random population samples of size (N). It tells us that the distribution of means will be approximately normally distributed as N gets larger. In addition, the mean of the sampling distribution of the means and the standard deviation of the population means will equal the mean and standard deviation of the parent population.

The **TI 83 calculator** has a built in function that can help you calculate probabilities of central theorem word problems, which usually contain the phrase “assume the distribution is normal” (or a variation of that phrase).

**The function, normalcdf, requires you to enter a lower bound, upper bound, mean, and standard deviation.**

## TI 83 Central Limit Theorem: Steps

**Sample problem**: A fertilizer company manufactures organic fertilizer in 10 pound bags with a standard deviation of 1.25 pounds per bag. What is the probability that a random sample of 15 bags will have a mean between 9 and 9.5 pounds?

**Step 1**: 2nd VARS 2.

**Step 2**: Enter your variables (lower bound, upper bound, mean, and standard deviation). Separate each variable by a comma: 9,9.5, 10,(1.25/√15)).

**Step 3**: Press ENTER. This returns the probability of .05969, or **.05969%**.

**Tip:**If you have a question that asks for “greater than” or “less than” a certain number, enter 999999999 for the lower or upper bound. For example, if you wanted to know the probability of greater than 8 pounds you would enter:

NORMALCDF(8,999999999,10,1./√(15))

Less than 8 pounds you would enter:

NORMALCDF(999999999,8,10,1./√(15))

**Tip**: Sampling distributions require that the standard deviation of the mean is σ / √(n), so make sure you enter that as the standard deviation.

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Why is it when I enter the numbers into my Ti-83 exactly as shown, I get 4.8236763E-7

I have used this same calculation for nomalcdf(60,64,62.4, (3.1/sqrt12))= .9594453925

which according to my study materials is correct.

I have no idea how your calculation comes up with .159, is this actually incorrect or a typo ???

Please reply to email listed…I’m confused.

Hi, Dan,

You caught a typo — thank you. I took off a zero in the sample size now, so it makes sense. Basically, the larger the sample size, the higher the probability of the sample size being the ACTUAL mean. So, for a sample size of 150 bags, there’s practically zero chance that the mean will be something other than 10 (in this example). So the answer 4.8236763E-7 is technically correct for such a large sample, but it doesn’t illustrate the point too well. I lowered the sample size and redid the calculations so it should be crystal clear now.

Best,

Stephanie