D. Statistical Analysis · One-Variable Data Analysis

Modelling Data with the Normal Distribution

D1.2About 75 minutes

Learning goals

I can describe the key properties of a normal distribution and use the 68–95–99.7 rule.
I can decide whether a data set is approximately normal by checking its shape and the percent of values within 1, 2 and 3 standard deviations.
I can use z-scores and a standard normal table (or technology) to estimate the percent of data below, above or between values.
I can find the value that corresponds to a given percentile in a normal model.

Words to know

normal distribution
A continuous, symmetric, bell-shaped distribution completely described by its mean μ and standard deviation σ. Written N(μ, σ²).

68–95–99.7 rule (empirical rule)
In a normal distribution, about 68% of values are within 1σ of the mean, about 95% within 2σ, and about 99.7% within 3σ.

standard normal distribution
The normal distribution of z-scores: mean 0 and standard deviation 1. Its areas are listed in z-tables.

model
A mathematical description that approximates real data. A good model is simple and fits the data well enough to make useful predictions.

area under the curve
For a normal model, the proportion of values in an interval equals the area under the curve over that interval; the total area is 1.

A shape that keeps showing up

Measure the lengths of brook trout, the masses of bags of flour from one machine, the heights of adult women in Canada, or the errors in a careful scientist's repeated measurements, and you often get the same shape: a single symmetric mound that tails off smoothly on both sides. Mathematicians describe that shape with one formula, the normal distribution.

The curriculum asks you to "use the normal distribution to model suitable one-variable data sets". Modelling means: check whether the normal shape fits the data, and if it does, use the model to answer questions about position and percent quickly. The probability side of the normal distribution (normal curves for continuous random variables, and normal approximations to the binomial) is developed fully in the unit Continuous Distributions and the Normal Distribution. Here the focus is on describing data; if you have already studied that unit, the properties, the 68–95–99.7 rule and the z-table below are a quick review, and the new part is checking whether a real data set suits the model.

Properties of a normal distribution

It is symmetric and bell-shaped, with a single peak.
mean = median = mode, all at the centre μ.
It is completely determined by μ (where the centre is) and σ (how wide it is). The curve changes from bending down to bending up exactly one σ on each side of μ.
The total area under the curve is 1 (100%), and the area over an interval is the proportion of values in that interval.
The tails go on forever but get very thin very quickly.

The 68–95–99.7 rule: about 68% of values lie within 1σ of μ, about 95% within 2σ, and about 99.7% within 3σ.

The 68–95–99.7 rule. By symmetry, about 34% of values are between μ and μ + σ, about 13.5% between μ + σ and μ + 2σ, and about 2.35% between μ + 2σ and μ + 3σ.

Worked example

Example 1: Using the 68–95–99.7 rule

The masses of bags of flour filled by one machine are approximately normal with μ = 1002 g and σ = 4 g. Use the 68–95–99.7 rule to estimate (a) the percent of bags between 994 g and 1010 g, (b) the percent above 1006 g, (c) the percent below 990 g.

(a) 994 = μ − 2σ and 1010 = μ + 2σ: about 95%.
(b) 1006 = μ + σ. About 68% are within 1σ, so 32% are outside, half on each side: about 16%.
(c) 990 = μ − 3σ. About 99.7% are within 3σ, so 0.3% are outside, half below: about 0.15%.

(a) about 95%, (b) about 16%, (c) about 0.15%.

Is the normal model suitable?

The curriculum says "suitable" data sets, so always check before you use the model:

Shape: draw a histogram. Is it single-peaked and roughly symmetric? Strong skew, two peaks or a hard boundary near the centre are warning signs.
Centre: the mean and median should be close.
Spread: count the percent of data within x̄ ± s, x̄ ± 2s and x̄ ± 3s and compare with 68%, 95% and 99.7%.
With technology, a normal probability plot (also called a normal quantile plot) should be close to a straight line. You will see these in first-year university statistics.

Worked example

Example 2: Checking the trout data

For the 40 brook trout from lesson 3, x̄ = 30.3 cm, median = 30.2 cm and s ≈ 5.12 cm. The histogram is single-peaked and roughly symmetric. Counting values: 27 fish are within x̄ ± s, 38 within x̄ ± 2s and all 40 within x̄ ± 3s. Is a normal model suitable?

Shape: single peak, roughly symmetric ✓.
Centre: mean 30.3 and median 30.2 are almost equal ✓.
Within 1s: 27 ÷ 40 = 67.5% (normal: 68%) ✓. Within 2s: 38 ÷ 40 = 95% (normal: 95%) ✓. Within 3s: 40 ÷ 40 = 100% (normal: 99.7%) ✓.

Yes. The trout lengths are well modelled by a normal distribution with mean about 30.3 cm and standard deviation about 5.1 cm.

The normal curve with the same mean and standard deviation as the trout data, drawn over the histogram (scaled to match the bar heights). It follows the bars closely, so the model is suitable.

Beyond 1, 2 and 3: the standard normal table

For any normal model, convert to z-scores. Then the proportion of values below z is the area to the left of z under the standard normal curve, written P(Z < z). A z-table (usually printed at the back of statistics textbooks) lists these areas. Here are some values you will use in this unit:

z	−2.00	−1.50	−1.25	−1.00	−0.75	−0.50	−0.25	0
P(Z < z)	0.0228	0.0668	0.1056	0.1587	0.2266	0.3085	0.4013	0.5000

Areas to the left of negative z-scores.

z	0.25	0.50	0.75	1.00	1.25	1.50	2.00	2.50
P(Z < z)	0.5987	0.6915	0.7734	0.8413	0.8944	0.9332	0.9772	0.9938

Areas to the left of positive z-scores. By symmetry, P(Z < −z) = 1 − P(Z < z).

Three kinds of question:

Below x: P(X < x) = P(Z < z).
Above x: P(X > x) = 1 − P(Z < z).
Between a and b: P(a < X < b) = P(Z < z_b) − P(Z < z_a).

Technology: in a spreadsheet, =NORM.DIST(x, mean, sd, TRUE) gives the area to the left of x directly, and =NORM.S.DIST(z, TRUE) gives it for a z-score. On a TI calculator, normalcdf(lower, upper, μ, σ) gives the area between two values.

Worked example

Example 3: Percents from the normal model

Flour bags: μ = 1002 g, σ = 4 g. What percent of bags (a) weigh less than the labelled 1000 g, (b) weigh more than 1007 g, (c) weigh between 998 g and 1010 g?

(a) z = (1000 − 1002) ÷ 4 = −0.5. P(Z < −0.5) = 0.3085, so about 30.9%.
(b) z = (1007 − 1002) ÷ 4 = 1.25. P(Z > 1.25) = 1 − 0.8944 = 0.1056, so about 10.6%.
(c) z-scores −1 and 2. P = 0.9772 − 0.1587 = 0.8185, so about 81.9%.

(a) about 30.9% are under the labelled mass, (b) about 10.6%, (c) about 81.9%. (A company that did not want so many bags under the labelled mass could raise μ, or buy a machine with a smaller σ.)

The shaded area to the left of 1000 g (z = −0.5) is about 0.3085 of the total area: about 31% of bags weigh less than 1000 g.

Working backwards: from a percentile to a value

To find the value at a given percentile, find the z-score with that area to its left, then use x = μ + zσ. Useful z-values: P90: z ≈ 1.28, P95: z ≈ 1.645, P97.5: z ≈ 1.96, P75: z ≈ 0.67, P25: z ≈ −0.67. In a spreadsheet, =NORM.INV(percentile, mean, sd) does the whole calculation; on a TI calculator use invNorm.

Worked example

Example 4: Setting a cut-off

The flour company wants to know the mass that only 10% of bags exceed (the 90th percentile), and the IQR predicted by the normal model.

P90 has z ≈ 1.28: x = 1002 + 1.28(4) ≈ 1007.1 g.
Q1 (P25) has z ≈ −0.67: x ≈ 1002 − 0.67(4) ≈ 999.3 g. Q3 (P75) has z ≈ 0.67: x ≈ 1004.7 g.
IQR ≈ 1004.7 − 999.3 = 5.4 g (about 1.35σ, which is true for every normal distribution).

About 10% of bags weigh more than 1007.1 g; the model's IQR is about 5.4 g.

Just for fun

The normal distribution is not called "normal" because other distributions are weird. Skewed and bimodal distributions are perfectly respectable; the name stuck in the late 1800s and statisticians have been apologizing for it ever since.

Hands-on activity

Is it normal? Rice, beans or pasta

You need: a bag of long-grain rice or dried beans or short pasta, a ruler with millimetres (or, for beans, a kitchen scale that reads to 0.1 g if you have one), paper and a pencil.

Choose 40 pieces at random (stir the bag, then scoop without looking).
Measure each piece's length to the nearest millimetre (or mass to the nearest 0.1 g).
Make a histogram, then find x̄, the median and s (a spreadsheet is fine).
Count how many values are within x̄ ± s, x̄ ± 2s and x̄ ± 3s and compare with 68%, 95% and 99.7%.
Decide whether a normal model is suitable, and use it to predict the percent of pieces longer than x̄ + 1.5s. Check the prediction against your data.

What to look for: natural or manufactured items of one kind often give a roughly normal shape. Broken pieces can create a tail on the left (left skew), which is a good example of why you must check before you model. With 40 values, expect about 27 within 1s, about 38 within 2s, and the percent above x̄ + 1.5s to be near 7% (1 − 0.9332 = 0.0668), which is about 3 pieces.

Your turn

Heights of a population are normal with μ = 165 cm and σ = 7 cm. Between which two heights do about 95% of people fall?

For the same population, about what percent are taller than 172 cm?

Use the table to find P(Z < 1.5) and P(Z > 1.5).

Test marks are normal with μ = 68 and σ = 10. What percent are below 63?

For the same test, what percent are between 58 and 88?

For the same test, what mark is at the 95th percentile?

A data set has mean 52 and median 41 and a long right tail. Should you use a normal model? Explain.

More practice: warm-up

In a normal distribution, how do the mean, median and mode compare?

About what percent of values are within 1 standard deviation of the mean?

What is the total area under a normal curve?

P(Z < 0) = ?

About what percent of values are more than 2σ above the mean?

More practice: on your own

A population of maple trees has trunk diameters normal with μ = 42 cm and σ = 8 cm. What percent are between 36 cm and 52 cm?

Reaction times are normal with μ = 0.25 s and σ = 0.04 s. What percent are slower (longer) than 0.33 s?

A sample of 50 values has x̄ = 80 and s = 6. Exactly 25 values are within 74 to 86, 47 within 68 to 92, and all 50 within 62 to 98. Is a normal model suitable?

Lifetimes of a type of LED bulb are normal with μ = 25 000 h and σ = 2 000 h. The company wants to guarantee its bulbs so that only 2.5% fail before the guarantee ends. What guarantee should it offer?

Explain why the normal model gives a small but non-zero percent of negative heights, and why that does not ruin the model.

Challenge

Applicants' scores are normal with μ = 72 and σ = 9. A program admits the top 25%. Using z ≈ 0.67 for P75, find the cut-off score. Then find the percent of admitted students who scored above 90.

In a normal distribution, 15.87% of values are above 120 and 2.28% are above 135. Find μ and σ.

Show that for any normal distribution, the 1.5 × IQR upper fence is about 2.7σ above the mean, and estimate the percent of normal data that the IQR rule flags as outliers.

Check your understanding

Quiz

About what percent of a normal distribution lies within 2 standard deviations of the mean?

X is normal with μ = 50 and σ = 5. What is P(X > 55)?

Which data set is least suitable for a normal model?

In a normal model with μ = 200 and σ = 20, which value is at about the 90th percentile?

A data set has 52% of its values within one standard deviation of the mean. What does that suggest?

Using the table, what is P(−0.5 < Z < 1)?

For parents and teachers

Big idea (D1.2): the normal distribution is a model. First check that the data is single-peaked and roughly symmetric and that about 68%, 95% and 99.7% of values fall within 1, 2 and 3 standard deviations; then use z-scores and areas to answer questions about percents and percentiles.

How to help: Ask your teen to sketch a bell curve and shade the area for every normal question before calculating. The sketch catches most "above or below?" mistakes. The rice or bean activity works well as a family project.

Common difficulties:

Using the normal model for skewed data without checking.
Forgetting that the table gives the area to the left: for "above", subtract from 1.
Mixing up percentile (an area) with z-score (a position).
Rounding z too early.