How to use normalcdf and invNorm on the TI-84

normalcdf returns an area, invNorm returns a boundary

These two commands answer opposite questions, and swapping them is the fastest way to lose a normal distribution problem.

normalcdf(lower, upper, mean, SD) takes two boundary values and returns the area between them under the normal curve. That area is a probability, so the answer always lands between 0 and 1.

invNorm(area, mean, SD) runs the process backwards. It takes an area measured from the left tail and returns the boundary value that cuts off that area, reported in the original units of the variable.

Choose between them by looking at what the question hands you and what it wants back. Given two values and asked for a chance, a percent, or a proportion: normalcdf. Given a percent or a percentile and asked for a score, a height, or a time: invNorm. The word percentile is the strongest signal for invNorm, because a percentile is defined by the area to its left, which is exactly what invNorm consumes.

Where the two commands live

Both sit in the distribution menu, which you reach with 2nd VARS. The menu is labeled DISTR, and normalpdf, normalcdf, and invNorm sit near the top of the list.

Item numbers differ across TI-84 models, so scroll to the name instead of memorizing a number. The binomial commands live further down the same menu, covered in binompdf vs binomcdf.

normalpdf is the one to avoid. It returns the height of the normal curve at a single point, not an area, so its output is not a probability. You almost never need it in a statistics course.

Entry style varies too. Some calculators open a fill-in screen with a separate line for each value, and others expect the values typed inside parentheses and separated by commas. The order of the values is the same either way, so learn the order rather than the screen.

normalcdf, argument by argument

The four values go in a fixed order: normalcdf(lower bound, upper bound, mean, standard deviation).

• Lower bound: the smaller boundary, in the original units of the variable.
• Upper bound: the larger boundary, in the same units.
• Mean: the population mean $\mu$ (mu) of the distribution.
• Standard deviation: the population standard deviation $\sigma$ (sigma).

For a variable that is normal with $\mu = 100$ and $\sigma = 15$, the probability of landing between 85 and 115 is normalcdf(85, 115, 100, 15), which returns 0.6827. That number is the 68 in the empirical rule, carried to four decimals.

Leave off the mean and the standard deviation and a TI-84 assumes the standard normal distribution, $\mu = 0$ and $\sigma = 1$. So normalcdf(-1, 1) returns the same 0.6827, with the bounds read as z-scores (standardized values) rather than raw values. Both routes work as long as you do not mix them by feeding raw bounds to a command that is reading them as z-scores.

Two checks catch most typing errors. A probability has to sit between 0 and 1, and a negative result means the bounds went in backwards.

Open-ended bounds and the 1E99 convention

Plenty of questions have only one boundary. $P(X < 85)$ has no lower limit and $P(X > 130)$ has no upper limit, but normalcdf still wants two numbers.

The convention is to supply a bound so far out in the tail that it sweeps up all the remaining area. Type -1E99 for a missing lower bound and 1E99 for a missing upper bound. The E is scientific notation, so 1E99 means $1 \times 10^{99}$, and you enter it with the calculator's EE key rather than by typing the letter E.

For the distribution with $\mu = 100$ and $\sigma = 15$:

• $P(X < 85)$ is normalcdf(-1E99, 85, 100, 15) = 0.1587.
• $P(X > 130)$ is normalcdf(130, 1E99, 100, 15) = 0.0228.

Any bound far enough out gives the same answer. Ten standard deviations past the mean, so -50 and 250 here, agrees to four decimal places, which is a useful fallback if E notation is giving you trouble. What fails is guessing a bound that sits near the data, such as 0 or the mean, because that quietly deletes area the question wanted included.

invNorm and the left-tail rule

invNorm(area, mean, SD) takes three values, and the first one is where students lose points. That area is always measured from the left, meaning it is the area below the value you are trying to find.

For the same distribution with $\mu = 100$ and $\sigma = 15$:

• The 90th percentile is invNorm(0.90, 100, 15) = 119.223.
• The value with 5% of the distribution above it is invNorm(0.95, 100, 15) = 124.673, since 5% above means 95% below.
• The value with 5% below it is invNorm(0.05, 100, 15) = 75.327.

So translate before you type. A right-tail area of $a$ becomes a left-tail area of $1 - a$. Phrases like "the top 10%" and "exceeded by only 2% of values" describe right-tail areas and need converting first.

Dropping the mean and standard deviation again gives the standard normal, so invNorm(0.975) returns 1.959964, the 1.96 behind a 95% confidence interval. Some calculator versions also offer an optional tail setting on invNorm. If you are not certain which version is in front of you, convert the area to a left-tail area yourself and leave that setting alone.

The two mistakes that cost the most points

Mistake 1: reaching for invNorm when the question wants an area. If the wording is "what is the probability", "what percent", or "what proportion", the answer is a number between 0 and 1, so you want normalcdf. The tell is an answer in the original units, like 119.223, sitting where a probability belongs.

Mistake 2: feeding invNorm a right-tail area. Asking for the 90th percentile but typing invNorm(0.10, 100, 15) returns 80.777, which sits below the mean instead of above it.

One check catches the second mistake every time. If the area you typed is greater than 0.5, the answer must be above the mean; if it is less than 0.5, the answer must be below the mean. Compare your result to the mean before writing it down.

A matching check works for normalcdf. Sketch the curve, shade the region you want, and ask whether the shaded piece is more or less than half. An area of 0.9 for a sliver in the tail is wrong no matter how confident the typing felt. You can confirm any of these with the normal distribution calculator.

Getting the same answers from the z-table

The AP exam provides a normal table, so both problems work without any calculator command. Standardize first with $z = \frac{x - \mu}{\sigma}$, then read areas from the table.

For the area between 85 and 115: $z_1 = \frac{85 - 100}{15} = -1$ and $z_2 = \frac{115 - 100}{15} = 1$. The z-table gives 0.1587 to the left of $z = -1$ and 0.8413 to the left of $z = 1$, so the area between them is $0.8413 - 0.1587 = 0.6826$.

The calculator returned 0.6827 and the table returns 0.6826. Neither is wrong. Here the z-scores are exactly $-1.00$ and $1.00$, so nothing was lost rounding them; the gap comes from rounding the areas. The exact areas are 0.8413447 to the left of $z = 1$ and 0.1586553 to the left of $z = -1$. The table rounds the first down to 0.8413 and the second up to 0.1587, each by 0.0000447, and the two shifts push the subtraction the same way, so the difference loses about 0.00009.

The percentile problem runs the table backwards. Search the body of the table for 0.9000 rather than the edges. The closest entries are 0.8997 at $z = 1.28$ and 0.9015 at $z = 1.29$, so take $z = 1.28$ and unstandardize with $x = \mu + z\sigma = 100 + 1.28(15) = 119.2$, close to the 119.223 that invNorm gives.

Knowing both routes is worth the effort. It keeps you moving if a calculator fails, and it lets you recognize a calculator answer that cannot be right. For more on standardizing, see how to find a z-score.

What this looks like on the AP exam

The Course and Exam Description states that a graphing calculator with statistical capabilities is expected for both sections of the exam, and that formulas and tables are provided for both sections. The commands are allowed, and the printed table is a genuine backup rather than a penalty.

Normal distribution work is Topic 2.11 in the Fall 2026 course, inside Unit 2, Probability, Random Variables, and Probability Distributions, which carries 15 to 25 percent of the multiple-choice section. The framework says the boundaries of an interval associated with a given area can be determined using technology or using z-scores and a standard normal table. That sentence is the normalcdf and invNorm pair described in words.

On free response, a bare command communicates very little. Writing normalcdf(85, 115, 100, 15) by itself leaves out the distribution, the probability statement, and the interpretation a reader needs. Write the probability statement in symbols, then the value, then a sentence in context.

For the full response structure, see the AP Statistics FRQ guide and reading calculator output. The official course page carries the framework itself.