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This chapter is from the book

Excel Functions for the Normal Distribution

Excel names the functions that pertain to the normal distribution so that you can tell whether you're dealing with any normal distribution, or the unit normal distribution with a mean of 0 and a standard deviation of 1.

Excel refers to the unit normal distribution as the "standard" normal, and therefore uses the letter s in the function's name. So the NORM.DIST() function refers to any normal distribution, whereas the NORMSDIST() compatibility function and the NORM.S.DIST() consistency function refer specifically to the unit normal distribution.

The NORM.DIST() Function

Suppose you're interested in the distribution in the population of high-density lipoprotein (HDL) levels in adults over 20 years of age. That variable is normally measured in milligrams per deciliter of blood (mg/dl). Assuming HDL levels are normally distributed (and they are), you can learn more about the distribution of HDL in the population by applying your knowledge of the normal curve. One way to do so is by using Excel's NORM.DIST() function.

NORM.DIST() Syntax

The NORM.DIST() function takes the following data as its arguments:

  • x—This is a value in the distribution you're evaluating. If you're evaluating high-density lipoprotein (HDL) levels, you might be interested in one specific level—say, 60. That specific value is the one you would provide as the first argument to NORM.DIST().
  • Mean—The second argument is the mean of the distribution you're evaluating. Suppose that the mean HDL among humans over 20 years of age is 54.3.
  • Standard Deviation—The third argument is the standard deviation of the distribution you're evaluating. Suppose that the standard deviation of HDL levels is 15.
  • Cumulative—The fourth argument indicates whether you want the cumulative probability of HDL levels from 0 to x (which we're taking to be 56 in this example), or the probability of having an HDL level of specifically x (that is, 56). If you want the cumulative probability, use TRUE as the fourth argument. If you want the specific probability, use FALSE.

Requesting the Cumulative Probability

The formula

  • =NORM.DIST(60, 54.3, 15, TRUE)

returns .648, or 64.8%. This means that 64.8% of the area under the distribution of HDL levels is between 0 and 60 mg/dl. Figure 7.4 shows this result.

Figure 7.4

Figure 7.4 You can adjust the number of gridlines by formatting the vertical axis to show more or fewer major units.

If you hover your mouse pointer over the line that shows the cumulative probability, you'll see a small pop-up window that tells you which data point you are pointing at, as well as its location on both the horizontal and vertical axes. Once created, the chart can tell you the probability associated with any of the charted data points, not just the 60 mg/dl this section has discussed. As shown in Figure 7.4, you can use either the chart's gridlines or your mouse pointer to determine that a measurement of, for example, 60.3 mg/dl or below accounts for about 66% of the population.

Requesting the Point Estimate

Things are different if you choose FALSE as the fourth, cumulative argument to NORM.DIST(). In that case, the function returns the probability associated with the specific point you specify in the first argument. Use the value FALSE for the cumulative argument if you want to know the height of the normal curve at a specific value of the distribution you're evaluating. Figure 7.5 shows one way to use NORM.DIST() with the cumulative argument set to FALSE.

Figure 7.5

Figure 7.5 The height of the curve at any point is the probability that the point appears in a random sample from the full distribution.

It doesn't often happen that you need a point estimate of the probability of a specific value in a normal curve, but if you do—for example, to draw a curve that helps you or someone else visualize an outcome—then setting the cumulative argument to FALSE is a good way to get it. (You might also see this value—the probability of a specific point, the height of the curve at that point—referred to as the probability density function or probability mass function. The terminology has not been standardized.)

If you're using a version of Excel prior to 2010, you can use the NORMDIST() compatibility function. It is the same as NORM.DIST() as to both arguments and returned values.

The NORM.INV() Function

As a practical matter, you'll find that you usually have need for the NORM.DIST() function after the fact. That is, you have collected data and know the mean and standard deviation of a sample or population. A question then arises: Where does a given value fall in a normal distribution? That value might be a sample mean that you want to compare to a population, or it might be an individual observation that you want to assess in the context of a larger group.

In that case, you would pass the information along to NORM.DIST(), which would tell you the probability of observing up to a particular value (cumulative = TRUE) or that specific value (cumulative = FALSE). You could then compare that probability to the alpha rate that you already adopted for your experiment.

The NORM.INV() function is closely related to the NORM.DIST() function and gives you a slightly different angle on things. Instead of returning a value that represents an area—that is, a probability—NORM.INV() returns a value that represents a point on the normal curve's horizontal axis. That's the point that you provide as the first argument to NORM.DIST().

For example, the prior section showed that the formula

  • =NORM.DIST(60, 54.3, 15, TRUE)

returns .648. The value 60 is at least as large as 64.8% of the observations in a normal distribution that has a mean of 54.3 and a standard deviation of 15.

The other side of the coin: the formula

  • =NORM.INV(0.648, 54.3, 15)

returns 60. If your distribution has a mean of 54.3 and a standard deviation of 15, then 64.8% of the distribution lies at or below a value of 60. That illustration is just, well, illustrative. You would not normally care that 64.8% of a distribution lies below a particular value.

But suppose that in preparation for a research project you decide that you will conclude that a treatment has a reliable effect only if the mean of the experimental group is in the top 5% of the population. (This is consistent with the traditional null hypothesis approach to experimentation, which Chapters 8 and 9 discuss in considerably more detail.) In that case, you would want to know what score would define that top 5%.

If you know the mean and standard deviation, NORM.INV() does the job for you. Still taking the population mean at 54.3 and the standard deviation at 15, the formula

  • =NORM.INV(0.95, 54.3, 15)

returns 78.97. Five percent of a normal distribution that has a mean of 54.3 and a standard deviation of 15 lies above a value of 78.97.

As you see, the formula uses 0.95 as the first argument to NORM.INV(). That's because NORM.INV assumes a cumulative probability—notice that unlike NORM.DIST(), the NORM.INV() function has no fourth, cumulative argument. So asking what value cuts off the top 5% of the distribution is equivalent to asking what value cuts off the bottom 95% of the distribution.

In this context, choosing to use NORM.DIST() or NORM.INV() is largely a matter of the sort of information you're after. If you want to know how likely it is that you will observe a number at least as large as X, hand X off to NORM.DIST() to get a probability. If you want to know the number that serves as the boundary of an area—an area that corresponds to a given probability—hand the area off to NORM.INV() to get that number.

In either case, you need to supply the mean and the standard deviation. In the case of NORM.DIST, you also need to tell the function whether you're interested in the cumulative probability or the point estimate.

The consistency function NORM.INV() is not available in versions of Excel prior to 2010, but you can use the compatibility function NORMINV() instead. The arguments and the results are as with NORM.INV().

Using NORM.S.DIST()

There's much to be said for expressing distances, weights, durations, and so on in their original unit of measure. That's what NORM.DIST() is for. But when you want to use a standard unit of measure for a variable that's distributed normally, you should think of NORM.S.DIST(). The S in the middle of the function name of course stands for standard.

It's quicker to use NORM.S.DIST() because you don't have to supply the mean or standard deviation. Because you're making reference to the unit normal distribution, the mean (0) and the standard deviation (1) are known by definition. All that NORM.S.DIST() needs is the z-score and whether you want a cumulative area (TRUE) or a point estimate (FALSE). The function uses this simple syntax:

  • =NORM.S.DIST(z, cumulative)

Thus, the formula

  • =NORM.S.DIST(1.5, TRUE)

informs you that 93.3% of the area under a normal curve is found to the left of a z-score of 1.5. (See Chapter 3, "Variability: How Values Disperse," for an introduction to the concept of z-scores.)

Using NORM.S.INV()

It's even simpler to use the inverse of NORM.S.DIST(), which is NORM.S.INV(). All the latter function needs is a probability:

  • =NORM.S.INV(.95)

This formula returns 1.64, which means that 95% of the area under the normal curve lies to the left of a z-score of 1.64. If you've taken a course in elementary inferential statistics, that number probably looks familiar: as familiar as the 1.96 that cuts off 97.5% of the distribution.

These are frequently occurring numbers because they are associated with the all-too-frequently occurring "p<.05" and "p<.025" entries at the bottom of tables in journal reports—a rut that you don't want to get caught in. Chapters 8 and 9 have much more to say about those sorts of entries, in the context of the t-distribution (which is closely related to the normal distribution).

The compatibility function NORMSINV() takes the same argument and returns the same result as does NORM.S.INV().

There is another Excel worksheet function that pertains directly to the normal distribution: CONFIDENCE.NORM(). To discuss the purpose and use of that function sensibly, it's necessary first to explore a little background.

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