TC3 → Stan Brown → TI-83/84/89 → Normal Calculations
Updated 23 Oct 2012

Normal Calculations on TI-83/84 or TI-89
for Individuals and Samples

Copyright © 2002–2015 by Stan Brown, Oak Road Systems

Summary: Earlier, we used the empirical rule (68–95–99.7 rule) to find probabilities between certain values in a ND. Now we extend that to calculate probabilities between any values. There are really only a few calculations, but the variations can be hard to manage. This page summarizes all the normal calculations, along with some important related ideas.

1. Calculations for Individuals (Use with Chapter 7)

Have boundary(ies), need area Have area, need boundary
working with z scores normalcdf(left boundary, right boundary) invNorm(area to left)
working with raw (x) scores normalcdf(left boundary, right boundary, mean, standard deviation) invNorm(area to left, mean, standard deviation)
TI-83/84 keystrokes [2nd VARS makes DISTR] [2] [2nd VARS makes DISTR] [3]
TI-89 keystrokes [CATALOG] [F3] [plain 6 makes N] [ENTER] [CATALOG] [F3] [plain 9 makes I] [ 3 times] [ENTER]

When you have only one boundary, and thus you’re computing area in a tail, use ∞ or −∞ as the other boundary. You get ∞ on your TI-83 as 1 [2nd , makes EE] 99, or as 10^99.

Caution: invNorm works from area to left. If the problem actually gives you an area to the right or an area in the middle, you must convert it to an area to left by 1−area.


These are already in terms of area to left. For instance, P40 is the x value such that 40% (0.40) of the population scored lower. Therefore, invNorm(0.40, mean, standard deviation) will give you the 40th percentile.


Suppose the mean score on the math SAT is 500 and the standard deviation is 100.

Example 1: What proportion of test takers earn a score between 650 and 700?

Solution: This is a calculation for individuals: μ = 500, σ = 100. normalcdf(650, 700, 500, 100) = 0.0441. About 4.4% of test takers earn a score between 650 and 700.

Example 2: What percentile is represented by a score of 735?

Solution: This, too, is a calculation for individuals. The percentile is the percent of area to the left of x = 735. normalcdf(−10^99, 735, 500, 100) = 0.9906, so a score of 735 is at the 99th percentile: 99% of individual scores are lower than or equal to 735.

Example 3: What score is earned by the best 15% of test takers?

Solution: The area of the right-hand tail is 0.15, so the area to left is 1−0.15. invNorm(1−.15, 500, 100) = 603.64. A score of 604 or above puts you in the top 15%.

2. Calculations for Sample Means (Use with Chapter 8)

As you know, the mean of sample means is the same as the population mean μ, but the standard deviation of sample means (symbol σ, also known as the standard error of the mean, or SEM) is different, equal to σ/√n, the population standard deviation divided by the square root of sample size.

With just that one change, SEM instead of s.d., the calculations for a distribution of sample means are quite similar to the calculations for individuals.

Have boundary(ies), need area Have area, need boundary
working with raw (x) scores normalcdf(left boundary, right boundary, mean,÷n) invNorm(area to left, mean,÷n)


Suppose the mean score on the math SAT is 500 and the standard deviation is 100.

Example 4: You randomly select 48 test takers, and compute their average score as 550. Does this surprise you?

Solution: To ask whether that sample mean is surprising, compute how likely it is to that random chance would give a sample mean of 550 or further away from the supposed population mean. Since 550>500, the question is the probability of finding ≥550 for n=48. You’re concerned with a sample mean, not an individual measurement, so the relevant standard deviation is the SEM, σ/√n. normalcdf(550, 10^99, 500, 100/√48) = 2.66×10-4, or about 0.000 266.

Yes, this would be a surprising result. If μ = 500 and σ = 100, there’s less than a 3 in 10,000 chance that the mean of a random sample of size 48 will be ≥550.

This page is used in instruction at Tompkins Cortland Community College in Dryden, New York; it’s not an official statement of the College. Please visit to report errors or ask to copy it.

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