The normal distribution and z-scores: the sketch, the table, and the three questions every normal-curve problem turns out to be

A normal distribution is the bell-shaped curve described by two numbers, its mean μ and its standard deviation σ. A z-score, z = (x − μ)/σ, says how many standard deviations a value sits from the mean, and it turns any normal question into a question about the one standard normal table. Every such question is one of three: the area below a value, the area between two values, or the value at a given percentile. Here are all three, worked on one mean and standard deviation, with the table misread named.

The curve and its two numbers

The normal distribution is symmetric about its mean μ (the Greek letter mu), and its width is set by its standard deviation σ (sigma). The total area under the curve is 1, and an area under part of the curve is the fraction of values that fall in that range, which is also the probability that one value drawn at random falls there.

The empirical rule, also called the 68 95 99.7 rule, is the shape in three numbers: about 68 % of values lie within one standard deviation of the mean, about 95 % within two, and about 99.7 % within three. The exact table values are 0.6827, 0.9545 and 0.9973; the rounded rule is what you quote and what you check against.

The standard normal distribution is the one with μ = 0 and σ = 1. Every other normal curve is this one, shifted and stretched, which is why one table serves them all. The table gives, for each z, the area under the standard normal curve to the left of z.

In the MDM4U course this is idea 12, the normal distribution and areas under the curve, and idea 13, z-scores and standardizing. Before them is idea 11, the hypergeometric distribution, the last of the discrete distributions; after them is idea 14, the normal approximation to the binomial. In the first-year statistics course it is idea 12, the normal distribution and z-scores. The MDM4U course map places the two ideas in the course, and AP Statistics asks the same three questions in its unit on the normal distribution.

The z-score

How to calculate z score: subtract the mean, divide by the standard deviation.

z = (x − μ)/σ.

A positive z is above the mean, a negative z below it, and the size is the distance in standard deviations. What does a z score mean in a sentence: z = 1.5 means one and a half standard deviations above the mean, whatever the units of x were. That is the point of standardising. A mark, a height and a reaction time can all be compared once each is a z.

Every normal-curve question is answered by the same three moves:

1. Sketch and shade. Draw the bell, mark the mean, mark the value or values, and shade the piece the question asks about. Is the shaded piece more or less than half? Write down which.
2. Standardise. Convert each boundary value to a z-score.
3. Read the table. The table gives the area to the left of z. If the shading is to the right, subtract from 1. If it is between two values, subtract two readings. If the area is given and the value is wanted, read the table backwards and un-standardise.

The sketch in move 1 is not decoration. It is how you know, before you read any number, whether the answer should be big or small, and it is what catches the misread in every example below.

The one set of numbers

A class's test marks are normal with mean μ = 70 and standard deviation σ = 8. These numbers are ours, chosen so that the z-scores come out clean. Every question on this page uses them, so you can see that the three questions differ only in the shading.

Question 1: the area below a value, line by line

What fraction of the class scored below 82?

Line 1, sketch: the bell with 70 in the middle, 82 to the right of it, everything to the left of 82 shaded. More than half is shaded.

Line 2, standardise: z = (82 − 70)/8 = 12/8 = 1.5.

Line 3, read the table: the area to the left of z = 1.5 is 0.9332.

Line 4, answer: about 93.3 % of the class scored below 82. The fraction above 82 is 1 − 0.9332 = 0.0668, about 6.7 %.

The sketch said more than half, and 0.9332 is more than half. Agreed.

The common wrong lines

Reporting 0.9332 as the fraction above 82. The table is the area to the left. The sketch shows that the right tail past 82 is the small piece, so the fraction above must be small, and 0.9332 cannot be it. Students who skip the sketch make this mistake; students who draw it do not.

Forgetting to divide by σ. A student writes z = 82 − 70 = 12 and then cannot find 12 on any table, because the table stops at about 3.5. A z-score that large is a sign the division was skipped, not a sign the table is incomplete.

Question 2: the area between two values, line by line

What fraction of the class scored between 60 and 82?

Line 1, sketch: shade between 60 and 82. The piece straddles the mean, so it is a large piece, more than half.

Line 2, standardise both boundaries: z for 60 = (60 − 70)/8 = −10/8 = −1.25. z for 82 = 1.5, from Question 1.

Line 3, read the table twice: the area to the left of z = −1.25 is 0.1056; the area to the left of z = 1.5 is 0.9332.

Line 4, subtract: 0.9332 − 0.1056 = 0.8276, about 82.8 %.

(The unrounded difference from the distribution function is 0.8275; the four-place table values give 0.8276. Either is right to three places, 0.828, and a marker accepts both.)

The common wrong line

Adding the two areas, or subtracting the z-scores. A student adds 0.1056 and 0.9332 to get 1.0388, which is more than the whole curve and so impossible. Another subtracts the z-scores, 1.5 − (−1.25) = 2.75, looks up 2.75 and reports 0.9970. Areas subtract; z-scores are positions on the axis, not areas, and the table is read once for each boundary. The sketch catches the first error at once, because no shaded piece can be more than 1.

Question 3: the value at a percentile, line by line

What mark is the 90th percentile?

Line 1, sketch: shade the left 90 % of the curve. The boundary of the shading, somewhere to the right of the mean, is the mark wanted.

Line 2, read the table backwards: find 0.90 in the body of the table and read off its z. The area 0.8997 sits at z = 1.28, and the exact z for 0.9000 is 1.2816. Use 1.28 from a table, 1.2816 from a calculator's inverse function.

Line 3, un-standardise, which is the z formula solved for x: x = μ + zσ = 70 + 1.2816 × 8 = 70 + 10.25 = 80.25.

Line 4, answer: the 90th percentile is a mark of about 80. With z = 1.28 the line gives 70 + 10.24 = 80.24, the same mark to the nearest whole number.

How to find a percentile from a z score, then, is Question 1 run in reverse: the area is given, the z is looked up from the inside of the table, and the formula is turned around.

The common wrong line

Looking up 0.90 as if it were a z-score. A student finds the row for z = 0.90, reads 0.8159 and reports it, or un-standardises it to get 70 + 0.90 × 8 = 77.2. Here the area is given and the z is wanted, so the table is read from the inside out, not from the edge in. The sketch helps again: the 90th percentile must be well above the mean, and 77.2 is less than one standard deviation up, which is only about the 82nd percentile. Booth and colleagues (2013) found that students learn more from a worked example when it is paired with an incorrect example and an explanation of why the wrong step is wrong, which is why this page shows each misread beside its correction.

Siblings to try cold

Scores on a test are normal with μ = 500 and σ = 100. With the page closed, find the fraction below 650, the fraction above 620, and the 95th percentile. Then check. Attempting first is chapter 1 of the free playbook, and testing yourself before you reread is why the attempt is where the learning happens.

Check

Below 650: z = 1.5, area 0.9332. Above 620: z = 1.2, area to the left 0.8849, so above is 1 − 0.8849 = 0.1151. The 95th percentile: z = 1.645 (the table's 1.64 and 1.65 straddle 0.9500), so x = 500 + 1.645 × 100 = 664.5, about 664 or 665 depending on the rounding your course uses. If your second answer was 0.8849, you reported the area below when the question asked for above.

Reading the z table without misreading it

A standard normal table lists z to one decimal place down the side and the second decimal place across the top. To read the area to the left of z = 1.25, find the row for 1.2 and the column for 0.05; the entry there, 0.8944, is the area. If your table has no negative half, use the symmetry of the curve: the area to the left of −z equals 1 minus the area to the left of z, so the area to the left of −1.25 is 1 − 0.8944 = 0.1056. Always compare the number you read with the sketch: a shaded piece that is most of the curve must be a number above 0.5, and a small tail must be a number below it. Your calculator's normal function returns the same areas without the table, and the sketch is still the check.

Where this sits in the course, and where it goes next

In MDM4U the normal distribution is ideas 12 and 13 of 24, the first continuous distribution after the discrete ones. In the statistics course it is idea 12 of 25. The μ and σ you are given come from the computations on the descriptive statistics page, where the mean and standard deviation are worked by hand.

What follows uses the same three moves. The normal approximation to the binomial (MDM4U idea 14) replaces a count with a normal curve and asks Question 1 or 2 of it. Sampling distributions and the central limit theorem (statistics idea 14) say that the mean of a sample is itself normally distributed, which is why the 1.96 in a 95 % confidence interval is a z-score, and confidence intervals and p-values is where that goes. For practice, work the three questions on one μ and σ as this page did, then a mixed set where you must decide which question you are being asked; worked examples, then practice problems has the order.

The normal curve on one screen

• The z-score: z = (x − μ)/σ; un-standardise with x = μ + zσ.
• The three moves: sketch and shade; standardise; read the table.
• Question 1, area below: read the table at z. Above: 1 minus that.
• Question 2, area between: two readings, subtract the smaller from the larger. Never add; never subtract z-scores.
• Question 3, a percentile: find the area inside the table, read its z, then x = μ + zσ.
• The empirical rule: 68 % within 1σ, 95 % within 2σ, 99.7 % within 3σ.
• The check: a big shaded piece is a number above 0.5; a tail is below it.