The Normal Distribution

The Bell Curve

Many real quantities — heights, test scores, measurement errors — pile up around a central value and thin out symmetrically on both sides. That shape is the normal distribution, the famous bell curve. On the GRE you will not be asked to integrate it or run a statistical test; instead you read areas under it, convert raw values into standard units, and translate between those units and percentiles.

A normal distribution is fixed by two numbers: its mean $\mu$, which sits at the exact center (and equals the median and mode), and its standard deviation $\sigma$, which sets the width. A larger $\sigma$ makes a wider, flatter curve; a smaller $\sigma$ makes a narrow, tall one. The curve is perfectly symmetric about $\mu$, so exactly half the area lies on each side of the mean.

The 68-95-99.7 Empirical Rule

The single most useful GRE fact about the normal curve is the empirical rule, which gives the area (proportion of data) within a few standard deviations of the mean:

The GRE's own reference figure breaks the region between $-2\sigma$ and $+2\sigma$ into six slices with these rounded areas: 2%, 14%, 34%, 34%, 14%, 2%, reading left to right across the ticks at $-2, -1, 0, +1, +2$ standard deviations. These add to $100\%$ and are worth memorizing:

• Between the mean and $+1\sigma$: about 34%.
• Between $+1\sigma$ and $+2\sigma$: about 14%.
• Beyond $+2\sigma$ (either tail): about 2%.

Because the curve is symmetric, the left side mirrors the right. So the area below $-1\sigma$ is $2\% + 14\% = 16\%$, and the area below $+1\sigma$ is $2\% + 14\% + 34\% + 34\% = 84\%$.

Standardizing: the z-score

To use the empirical rule on real numbers, convert a raw value $x$ into a z-score — the number of standard deviations it sits from the mean:

$z = \frac{x - \mu}{\sigma}$

A z-score of $+1.5$ means "one and a half standard deviations above the mean"; a z-score of $-2$ means "two standard deviations below the mean." The sign tells you the side, and the magnitude tells you how far. To go the other way — from a z-score back to a raw value — rearrange to $x = \mu + z\sigma$.

For example, if scores are normal with $\mu = 500$ and $\sigma = 100$, a score of $650$ has $z = \dfrac{650 - 500}{100} = 1.5$, and the value one standard deviation below the mean is $x = 500 + (-1)(100) = 400$.

Reading Percentages Between Intervals

Most GRE normal-curve questions ask for the share of data in some interval. The reliable method:

1. Convert both endpoints of the interval to z-scores.
2. Split the interval at the mean if it straddles $\mu$.
3. Add the empirical-rule slices between the z-scores.

Suppose $\mu = 500$, $\sigma = 100$, and you want the percent of scores between $400$ and $700$. The endpoints are $z = -1$ and $z = +2$. From $-1\sigma$ to the mean is $34\%$; from the mean to $+1\sigma$ is $34\%$; from $+1\sigma$ to $+2\sigma$ is $14\%$. The total is $34 + 34 + 14 = 82\%$. Do not try to halve $95\%$ here — the interval is asymmetric, so you must add the individual slices.

Percentiles

A percentile is the percent of the data at or below a value, which equals the area to the left of it. Combining the slices:

So a value one standard deviation above the mean sits at roughly the 84th percentile — it exceeds about 84% of the data. This is why "one SD above the mean" and "84th percentile" describe the same place.

Staying On-Syllabus

The GRE's use of the normal distribution stops at the empirical rule, z-scores, and percentile reading. It never requires hypothesis testing, confidence intervals, or precise tail areas beyond the $68/95/99.7$ landmarks — treat those as out of scope. Master the six-slice picture and the z-score conversion, and you can handle every normal-curve item the test offers.