Recognize the properties of a normal distribution; use the empirical (68-95-99.7) rule; compute a z-score; and use z-scores (with a calculator or table) to find the proportion of data in an interval.

What this topic is asking

The Regents Algebra II exam (the Interpreting Data, S-ID, and Making Inferences, S-IC, clusters) wants you to know the properties of a normal distribution, apply the empirical (68-95-99.7) rule, compute a z-score, and use z-scores to find the proportion of data in an interval. The normal model underlies much of the statistics on the exam.

Properties of the normal distribution

A normal distribution is the familiar symmetric bell curve. It is centered at its mean $\mu$, which is also its median and mode, and its width is governed by the standard deviation $\sigma$: a larger $\sigma$ spreads the curve out, a smaller one makes it tall and narrow. The total area under the curve is 1 (or 100%), and the curve is symmetric about the mean, so half the data lies on each side.

The empirical rule

For data that is approximately normal, the 68-95-99.7 rule gives quick proportions without a calculator.

So for a mean of 100 and standard deviation 15, about 95% of values fall between 70 and 130. Because the curve is symmetric, you can split these: about 34% lies between the mean and one standard deviation above it (half of 68%).

The z-score

A z-score standardizes a value by measuring its distance from the mean in standard deviations.

From z-scores to proportions

The power of the z-score is that it lets you find the proportion of data in any interval using a calculator's normal distribution function (or a standard normal table), which works for any mean and standard deviation. A clarifying point worth stressing is the difference between the empirical rule and a z-score calculation: the empirical rule gives the round 68, 95, and 99.7 percentages for whole-number standard deviations, while a z-score with a calculator handles any value, such as $z = 1.5$ or a score 137 units from the mean. Use the empirical rule for the clean cases and the z-score method for everything else. Computing the z-score by dividing by the mean instead of the standard deviation is the most common error, so always divide the difference $x - \mu$ by $\sigma$.

Try this

Q1. Data is normal with mean 50, standard deviation 5. About what percent lies between 40 and 60? [1 credit]

• Cue. That is $\mu \pm 2\sigma$, so about 95%.

Q2. Find the z-score of $x = 88$ when $\mu = 80$ and $\sigma = 4$. [1 credit]

• Cue. $z = \frac{88 - 80}{4} = 2$.