The Normal Distribution: Study Guide (Statistics 2.11)

Continuous random variables

In Topics 2.8 to 2.10 you met discrete random variables, such as a count of successes. You could list their values and give each one a probability.

A continuous random variable can take any value in an interval: a mass, a time, a height. You cannot list every value. So probability works differently:

• Probability belongs to intervals, not single values. “The loaf weighs between 790 g and 815 g” has a probability. “The loaf weighs exactly 800 g” has probability 0, because exactly 800.000… g is one point among infinitely many.
• The probability that the variable falls in an interval is the area under the density curve over that interval.
• The total area under the curve is 1, just as the probabilities of a discrete distribution add to 1.

Because a single value has probability 0, P(X < a) and P(X ≤ a) are the same for a continuous variable.

The normal curve

Many continuous variables are well modelled by a normal distribution: lengths and masses of manufactured items, measurement errors, many biological measurements. A normal curve is:

• continuous: a smooth curve, with no gaps;
• unimodal: one peak;
• symmetric about its centre;
• bell-shaped: it falls away on both sides and gets very close to the axis in the tails, without ever touching it.

A normal distribution is completely fixed by two parameters:

• the mean μ, which is the centre (and, by symmetry, also the median and the mode);
• the standard deviation σ, which sets the spread.

We write X ~ N(μ, σ). For example, N(800, 12) means a normal distribution with mean 800 and standard deviation 12. (Some books put the variance σ² second. In this course, the second number is the standard deviation; say so if there could be doubt.)

A normal curve can model two things: a distribution of data (the masses of all the loaves a bakery made last month) or a random variable (the mass of one loaf chosen at random). The same areas answer both kinds of question: “what proportion of loaves…?” and “what is the probability that a randomly chosen loaf…?”.

How σ changes the curve

The area under every normal curve is 1. So if the values are more spread out, the curve has to be lower:

• smaller σ: the curve is taller and more concentrated around μ;
• larger σ: the curve is shorter and more spread out.

Changing μ just slides the curve left or right without changing its shape.

The standard normal distribution

The standard normal distribution is the normal distribution with mean 0 and standard deviation 1, written N(0, 1). Its variable is usually called Z.

You already know z-scores from Topic 1.9: z = (x − μ) / σ counts how many standard deviations a value lies above or below the mean. If X ~ N(μ, σ), then the z-score of X follows N(0, 1). This is why a single table of areas for Z works for every normal distribution. To go back from a z-score to a value, use x = μ + zσ.

A standard normal table gives the area to the left of z, that is P(Z < z), with z rounded to 2 decimal places. For example, P(Z < 1.25) = 0.8944.

The empirical rule (68–95–99.7)

For any normal distribution, approximately:

• 68% of values lie within 1 standard deviation of the mean;
• 95% lie within 2 standard deviations;
• 99.7% lie within 3 standard deviations.

Using symmetry, you can split these into strips: 34% between μ and μ + σ, 13.5% between μ + σ and μ + 2σ, 2.35% between μ + 2σ and μ + 3σ, and 0.15% beyond μ + 3σ (the same on the left). The rule gives quick estimates; technology gives more precise values (for example, 68.27% rather than 68%).

The fictional Corran Bakery makes sourdough loaves. Their masses are approximately normal with mean 800 g and standard deviation 12 g: X ~ N(800, 12).

Worked example 1: estimating with the empirical rule

Question. Use the empirical rule to estimate the percentage of Corran loaves that (a) weigh between 776 g and 824 g, (b) weigh more than 824 g, (c) weigh between 788 g and 836 g, (d) weigh less than 788 g.

1. Locate each boundary in standard deviations. 776 = 800 − 2(12), 788 = 800 − 12, 824 = 800 + 2(12), 836 = 800 + 3(12).
2. (a) 776 g to 824 g is μ ± 2σ, so about 95%.
3. (b) The 5% outside μ ± 2σ is split equally between the two tails by symmetry: 5% ÷ 2 = 2.5%.
4. (c) From μ − σ to μ is half of 68%, which is 34%. From μ to μ + 3σ is half of 99.7%, which is 49.85%. Total: 34% + 49.85% = 83.85%.
5. (d) 32% lies outside μ ± σ, half in each tail: 32% ÷ 2 = 16%.

Interpretation. About 83.85% of Corran sourdough loaves weigh between 788 g and 836 g.

Check. Technology gives 84.00% for (c) and 15.87% for (d). The rule is close but approximate, so say “about” or “approximately”.

Probabilities as areas: technology or a table

When a boundary is not a whole number of standard deviations from the mean, use technology or a z-table. Show your work in three parts:

1. Name the distribution and its parameters: X ~ N(800, 12).
2. Write the probability with its boundary and direction, and sketch it: P(790 < X < 815). Shade the region.
3. Give the area, with the calculator command and its inputs, or the z-scores and table values.

A bare calculator command such as “normalcdf(790, 815, 800, 12)” with no labels is not a complete answer. Label which number is the mean and which is the standard deviation.

Worked example 2: probability for an interval

Question. A Corran loaf is chosen at random. Find (a) the probability that it weighs less than 785 g, (b) the probability that it weighs between 790 g and 815 g. (c) The bakery bakes 400 loaves on Saturday. About how many would you expect to weigh between 790 g and 815 g?

(a)

1. X ~ N(800, 12). We want P(X < 785).
2. z = (785 − 800) ÷ 12 = −15 ÷ 12 = −1.25.
3. Table: P(Z < −1.25) = 0.1056. Technology: normalcdf(lower = −10⁹⁹, upper = 785, μ = 800, σ = 12) = 0.1056.

(b)

1. We want P(790 < X < 815), the hatched area in Figure 3.
2. z-scores: (790 − 800) ÷ 12 = −0.8333…, which rounds to −0.83 for a table; (815 − 800) ÷ 12 = 1.25.
3. Table: P(Z < 1.25) − P(Z < −0.83) = 0.8944 − 0.2033 = 0.6911.
4. Technology: normalcdf(lower = 790, upper = 815, μ = 800, σ = 12) = 0.6920.

(c) Expected number = 400 × 0.6920 ≈ 276.8, so about 277 loaves.

Interpretation. There is about a 0.69 probability that a randomly chosen Corran loaf weighs between 790 g and 815 g, so about 69% of loaves are in this range.

Check. The table answer (0.6911) and the technology answer (0.6920) differ because the table needs z = −0.83 instead of −0.8333…. Both are acceptable if the work is shown. The answer is close to the 68% within one standard deviation (788 g to 812 g). That makes sense: the interval 790 g to 815 g is 25 g wide, almost the same as the 24 g width of μ ± σ.

Working backwards: boundaries for a given area

Sometimes you know the area and need the boundary. Think about where the area is before you calculate. With p a percentage:

You want…	Write it as…	Area to the left of the boundary
the lowest p%	P(X < xₐ) = p/100	p/100
the highest p%	P(X > x_b) = p/100	1 − p/100
the middle p%	P(xₐ < X < x_b) = p/100	(1 − p/100)/2 for xₐ and its mirror for x_b
the most extreme p% (both tails)	P(X < xₐ) = P(X > x_b) = ½(p/100)	½(p/100) for xₐ and its mirror for x_b

Then find z with invNorm or by reading the table backwards (find the area inside the table, read off z), and convert with x = μ + zσ. Most calculators’ invNorm uses the area to the left, so a “highest p%” question needs 1 − p/100.

Worked example 3: finding boundaries

Question. For the Corran loaves, X ~ N(800, 12), find (a) the mass below which the lightest 5% of loaves fall, (b) the mass above which the heaviest 10% fall, (c) the interval containing the middle 80%, (d) the boundaries of the most extreme 1% of loaves.

(a) P(X < xₐ) = 0.05. invNorm(area = 0.05, μ = 0, σ = 1) gives z = −1.6449. xₐ = 800 + (−1.6449)(12) = 780.26 g.

(b) P(X > x_b) = 0.10, so the area to the left is 0.90. z = 1.2816 and x_b = 800 + 1.2816(12) = 815.38 g.

(c) The middle 80% leaves 20% in the tails, 10% each side. The lower boundary has area 0.10 to its left, so z = −1.2816; the upper has 0.90, so z = 1.2816. Interval: 800 ± 1.2816(12) = 784.62 g to 815.38 g.

(d) The most extreme 1% is split into 0.5% in each tail. Lower boundary: area to the left 0.005, z = −2.5758. Upper boundary: area to the left 0.995, z = 2.5758. Boundaries: 800 ± 2.5758(12) = 769.09 g and 830.91 g. Loaves lighter than 769.09 g or heavier than 830.91 g are the most extreme 1%.

Check. With a table, the nearest values are z = 1.28 (area 0.8997) and z = 2.58 (area 0.9951), which give boundaries within about 0.05 g of these. Each answer should match the sketch: (a) is below the mean, (b) above it, and (c) and (d) are symmetric about 800 g.

Comparing relative positions with percentiles

A z-score tells you how many standard deviations a value is from its mean. When a distribution is normal, you can turn that into a percentile: the percentage of the distribution at or below the value. Percentiles and proportions let you compare positions within one normal distribution, or between two different ones.

Worked example 4: which loaf is more unusually heavy?

Question. Corran Bakery’s rye loaves have masses that are approximately N(650 g, 20 g). One sourdough loaf weighs 815 g and one rye loaf weighs 680 g. Which loaf is heavier relative to its own type? Use percentiles.

1. Sourdough: z = (815 − 800) ÷ 12 = 1.25. P(Z < 1.25) = 0.8944, so the loaf is at about the 89th percentile. About 10.56% of sourdough loaves are heavier.
2. Rye: z = (680 − 650) ÷ 20 = 1.5. P(Z < 1.5) = 0.9332, so the loaf is at about the 93rd percentile. About 6.68% of rye loaves are heavier.
3. Compare. 93rd percentile > 89th percentile.

Conclusion. The rye loaf is heavier relative to its type. A smaller proportion of rye loaves (about 6.68%) are heavier than it, compared with about 10.56% of sourdough loaves heavier than the sourdough loaf. The sourdough loaf is heavier in grams, but that comparison mixes two different distributions.

Common misconceptions

• “The second number in N(800, 12) is the variance.” In this course it is the standard deviation. Dividing by 12² = 144 in a z-score gives a wrong answer.
• “A taller curve has more area.” Every normal curve has total area 1. A taller curve is narrower: its σ is smaller.
• “P(X = 800) is the height of the curve at 800.” For a continuous variable the probability of one exact value is 0. Only intervals have probability.
• “Use the empirical rule for any boundary.” It only works for boundaries a whole number of standard deviations from the mean, and it gives estimates. Otherwise use technology or a table.
• Wrong direction. A table and invNorm work with the area to the left. For “more than” or “the highest p%”, use 1 minus the area.
• Putting all of the “most extreme p%” in one tail. Split it: p/2 % in each tail.
• “Every distribution is normal.” Check that the data are roughly unimodal and symmetric first. A skewed variable, or one that cannot go below 0 with a mean close to 0, is not well modelled by a normal curve.
• A bare calculator command. Name the distribution, the parameters, the boundaries and the direction, as well as the answer.

Where this leads

The normal distribution is the model behind most of the rest of the course. Next, in Topic 2.12, you will see why: the sample mean from a random sample has a sampling distribution that is approximately normal, and the approximation improves as the sample gets larger. Try the practice questions now, then use the revision notes and the checklist to consolidate. Coming from the binomial distribution? The Topic 2.10 study guide is the step before this one.