๐Ÿ“– Contents Chapter 7 ยท Normal Distribution
Statistics · Chapter 7

The Normal Distribution

Measure enough of almost anything that varies for lots of small, independent reasons — heights, birth weights, test scores, the error in a laboratory instrument — and the histogram tends to pile up in the middle and thin out symmetrically on both sides. That shape has a name: the normal distribution, or the bell curve. It matters far beyond its own good looks, because once you know a variable is roughly normal, two numbers — the mean and the standard deviation — let you answer questions about any value in the distribution. This chapter teaches you to read the curve, to reason with the 68–95–99.7 rule, to convert any score into a z-score, to turn that z-score into a probability, and to run the whole machine backward when you are handed a percentile instead of a score. There is arithmetic here, but it is all addition, subtraction, multiplication, and division — nothing harder.

By the end of this chapter you can

  • List the defining properties of a normal curve and explain what μ and σ each control.
  • Explain why probability for a continuous variable is an area, not a height, and why total area equals 1.
  • Apply the empirical rule (68–95–99.7) to estimate proportions above, below, and between values.
  • Compute a z-score with z = (x − μ) ÷ σ and interpret its sign and size correctly.
  • Use z-scores to compare values measured on different scales.
  • Read a standard normal table to find left-tail, right-tail, and between-two-values probabilities.
  • Work backward from a percentile to a raw score using x = μ + zσ.
  • Recognize when the normal model does not apply, and avoid the classic misreadings.

Section 1The normal curve

Several normal probability density curves with different means and standard deviations plotted on the same axes
Reference. Normal curves with different means and standard deviations — the mean slides the curve left or right, the standard deviation makes it narrow or wide. Source and licence: see the figure credits

What the shape is telling you

A normal distribution is a smooth, symmetric, single-peaked curve — the bell curve. It is not a picture of data you collected; it is a model, an idealized mathematical shape that real data can resemble more or less closely. When we say "adult female heights are approximately normal," we mean the histogram of real heights looks a lot like that curve, close enough that we can use the curve's mathematics to make good predictions.

Every normal curve is completely pinned down by exactly two numbers, called parameters. The mean μ (mu) sits at the center and locates the peak. The standard deviation σ (sigma) controls the spread: a small σ gives a tall, narrow curve where values huddle near the middle; a large σ gives a short, wide curve where values sprawl. Change μ and the whole curve slides sideways without changing shape. Change σ and the curve stretches or squeezes without moving its center. Statisticians write this compactly as X ~ N(μ, σ2), read "X follows a normal distribution with mean μ and variance σ2."

The equation behind the picture is f(x) = [ 1 ÷ (σ √(2π) ) ] · e−(x − μ)² / (2σ²). You will never have to evaluate it by hand, and nothing in this chapter depends on it. It is worth seeing once only so you can confirm the claim above: the only symbols in it besides x are μ and σ. Two numbers, one curve.

The properties, one at a time

Six features define the shape, and every technique in the rest of this chapter leans on one of them.

PropertyWhat it meansWhy it matters
SymmetricThe left half is a mirror image of the right half about μLets you convert a right-tail question into a left-tail one, and vice versa
UnimodalExactly one peak, located at μA two-humped histogram is not normal — do not use these tools on it
Mean = median = modeAll three centers land on the same pointThe center is unambiguous; 50% of the area lies on each side
Total area = 1The area under the whole curve is exactly 1 (100%)Area is probability — this is what makes the curve usable
Asymptotic tailsThe tails approach the axis forever but never touch itNo value is strictly impossible, but extremes are vanishingly rare
Inflection at μ ± σThe curve switches from bending down to bending out one σ from centerYou can literally see σ on the picture — it is where the curve changes its bend

The fourth property deserves a slow read, because it is where most students first get uneasy. For a continuous variable — one that can take any value in a range, like height in inches carried out to as many decimals as you like — the probability of landing on exactly one specific value is zero. Nobody is exactly 64.000000… inches tall. So we never ask "what is the probability height equals 64?" We ask "what is the probability height falls between 63 and 65?", and the answer is the area under the curve over that stretch of the axis. Height of the curve is density, not probability. Area is probability.

A practical consequence: for continuous distributions, P(X < 64) and P(X ≤ 64) are the same number, because the single point 64 contributes zero area. That is not true for the discrete distributions of Chapter 6, and it trips people up on exams.

How close to normal does real data have to be?

Close enough, and no closer. Real heights are bounded below by zero and above by about nine feet; the normal curve technically extends to infinity in both directions. Real test scores are integers; the curve is smooth. None of that matters in practice, because the curve assigns so little area to the impossible regions that the error is negligible. What does matter is gross violations: strong skew, two peaks, or a hard boundary sitting close to the mean. Income is the classic bad fit — it is right-skewed, with a long tail of high earners, so its mean sits well above its median and the empirical rule you are about to learn will give you badly wrong answers.

Two numbers, and you know everything. That is the whole payoff of the normal model. For most distributions, knowing the mean and standard deviation tells you the center and the spread and little else. For a normal distribution, μ and σ determine the entire curve — so they let you compute the proportion of the population above any value, below any value, or between any two values, without ever looking at the original data again.
Interactive Drag the mean and the standard deviation โ€” with sliders or by grabbing the peak and the inflection point directly โ€” and watch the bell slide and stretch while the area inside ยฑ1, ยฑ2, and ยฑ3 SDs is recomputed live and a real sample piles up underneath it. Open full ↑
Key points
  • A normal distribution is a symmetric, unimodal, bell-shaped model defined completely by μ (center) and σ (spread).
  • Mean, median, and mode coincide at μ; changing μ slides the curve, changing σ widens or narrows it.
  • Area under the curve is probability, and the total area is 1; the height of the curve is density, not probability.
  • The model is an approximation — do not use it on strongly skewed or two-peaked data.

Section 2The empirical rule in action

Normal curve divided into standard deviation bands showing 68 percent, 95 percent, and 99.7 percent of the area
Reference. The empirical rule — about 68%, 95%, and 99.7% of a normal distribution falls within one, two, and three standard deviations of the mean. Source and licence: see the figure credits

The rule, and the finer breakdown

Because every normal curve has the same shape, the proportion of area inside a given number of standard deviations is always the same, no matter what the units are. That constancy is the empirical rule, also called the 68–95–99.7 rule:

Split those bands in half using symmetry and you get a map of the curve precise enough for most quick reasoning. Each half-band below is one side of the center.

RegionShare of the areaRunning total from the left
Below μ − 3σ0.15%0.15%
μ − 3σ to μ − 2σ2.35%2.5%
μ − 2σ to μ − 1σ13.5%16%
μ − 1σ to μ34%50%
μ to μ + 1σ34%84%
μ + 1σ to μ + 2σ13.5%97.5%
μ + 2σ to μ + 3σ2.35%99.85%
Above μ + 3σ0.15%100%

Memorize the sequence 0.15, 2.35, 13.5, 34 — then 34, 13.5, 2.35, 0.15 going back out. Those eight numbers sum to 100% and answer a surprising share of exam questions with no table at all.

A worked example

Suppose adult female heights in a large population are approximately normal with μ = 64.0 inches and σ = 2.5 inches. First, build the ruler by marking off standard deviations from the center:

Now answer four questions using only that ruler and the table above.

(a) What percent are taller than 69.0 inches? 69.0 is exactly μ + 2σ. The middle 95% lies inside ±2σ, so 100% − 95% = 5% lies outside. Symmetry splits that 5% evenly between the two tails: 5% ÷ 2 = 2.5%.

(b) What percent are shorter than 61.5 inches? 61.5 is μ − 1σ. Outside ±1σ lies 100% − 68% = 32%, split evenly: 32% ÷ 2 = 16%.

(c) What percent are between 66.5 and 69.0 inches? That is the band from μ + 1σ to μ + 2σ. Take the difference of the two symmetric bands and halve it: (95% − 68%) ÷ 2 = 27% ÷ 2 = 13.5%.

(d) What percent are between 59.0 and 66.5 inches? Split at the mean. From 59.0 (μ − 2σ) up to 64.0 is half of the 95% band = 47.5%. From 64.0 up to 66.5 (μ + 1σ) is half of the 68% band = 34%. Add them: 47.5% + 34% = 81.5%.

Notice the pattern in all four: draw the curve, mark μ and the three standard-deviation ticks on each side, shade what the question asks for, and add or subtract the block percentages. The sketch is not optional decoration — it is the method.

Where the rule stops working

The empirical rule is a fact about normal distributions only. Apply it to right-skewed data such as household income or hospital length of stay and it will overstate the left tail and understate the right, sometimes badly. For a distribution of unknown shape there is a weaker but universal backstop, Chebyshev's inequality: at least 1 − 1/k2 of any distribution lies within k standard deviations of the mean, for any k > 1. That guarantees at least 75% within 2σ and at least 88.9% within 3σ — true always, but much looser than 95% and 99.7%. The precision of the empirical rule is the reward for the normal assumption being justified.

±2σ is 95.45%, not 95%. The rounding is harmless in casual reasoning, but the two get confused later. When someone says "the 95% interval," they almost always mean the interval built from 1.96 standard deviations, not 2. The empirical rule's "95" is a friendly approximation; 1.96 is the exact number, and it is the one that shows up in confidence intervals.
Interactive Apply 68โ€“95โ€“99.7 to six real measurements: pick a quantity, build a question with ฯƒ-snapped cuts, and watch the slice arithmetic, the exact area, and a live simulated sample all compute themselves โ€” including one skewed case where the rule returns an impossible answer. Open full ↑
Key points
  • The empirical rule: about 68% / 95% / 99.7% of a normal distribution lies within 1, 2, and 3 standard deviations of μ.
  • Symmetry lets you halve any band: outside ±2σ is 5%, so each tail is 2.5%.
  • The block percentages 0.15 / 2.35 / 13.5 / 34 (each side) answer most questions with no table.
  • The rule applies to approximately normal data only; Chebyshev's inequality is the weaker rule that always holds.

Section 3The standard normal

The standard normal distribution with mean zero and standard deviation one, marked in z units
Reference. The standard normal distribution — mean 0, standard deviation 1, with the horizontal axis measured in z units. Source and licence: see the figure credits

One curve to rule them all

There are infinitely many normal curves — one for every pair (μ, σ). Nobody wants infinitely many tables. The fix is to rescale every normal distribution onto a single common curve, the standard normal distribution: the normal distribution with μ = 0 and σ = 1, written Z ~ N(0, 1).

The rescaling is done with the z-score:

z = ( x − μ ) ÷ σ

Read it as two steps in plain English. Subtract the mean: this recenters the value, telling you how far above or below average it sits, in the original units. Divide by the standard deviation: this converts that distance from original units into "standard deviations," which have no units at all. A z-score of 1.5 means "one and a half standard deviations above the mean," whether the original data were inches, dollars, milliseconds, or exam points.

The sign carries meaning. Positive z means above the mean, negative z means below, and z = 0 means exactly at the mean. The magnitude carries the rest: |z| under 1 is thoroughly ordinary, around 2 is somewhat unusual, and 3 or more is genuinely rare — only about 0.3% of a normal distribution sits beyond ±3.

Worked example: one score, two ways

An exam has μ = 72 points and σ = 8 points. A student scores x = 86.

z = (86 − 72) ÷ 8 = 14 ÷ 8 = 1.75

The score is 14 points above average, and 14 points is 1.75 standard deviations. Now a second student scores x = 60:

z = (60 − 72) ÷ 8 = (−12) ÷ 8 = −1.50

That student sits one and a half standard deviations below the mean. Note that z = −1.50 and z = +1.50 are the same distance from center in opposite directions — a fact we will exploit constantly in Section 4.

Why standardizing is worth the trouble: comparing apples to oranges

The real power of a z-score is that it strips away the units and the scale, so scores from completely different measurements become comparable. Suppose a student takes two exams:

ExamScore xMean μSD σz = (x − μ) ÷ σReading
Exam A86728(86 − 72) ÷ 8 = +1.751.75 SD above average
Exam B91856(91 − 85) ÷ 6 = +1.001.00 SD above average
Exam C687510(68 − 75) ÷ 10 = −0.700.70 SD below average

The raw scores say Exam B (91) was the best performance and Exam A (86) second. The z-scores say otherwise: relative to everyone else who took each test, Exam A was the strongest showing at +1.75, ahead of Exam B at +1.00. The 91 looks impressive until you notice the class averaged 85 with a tight spread, so a 91 was only one standard deviation up. This is exactly why z-scores appear on growth charts, standardized tests, and laboratory reference ranges — they answer "how unusual is this?" rather than "how big is this?"

The reverse conversion is just algebra. Multiply both sides of the z-score formula by σ and add μ:

x = μ + zσ

Check it on Exam A: x = 72 + (1.75)(8) = 72 + 14 = 86. It returns the original score, as it must. Section 5 is built entirely on this rearranged form.

Standardizing does not make data normal. Converting to z-scores subtracts a constant and divides by a constant — a purely linear rescaling. It moves the center to 0 and the spread to 1, and it does absolutely nothing to the shape. If the original distribution was right-skewed, the z-scores are right-skewed too, and the standard normal table will give you wrong answers. Check the shape first; standardize second.
Interactive Convert raw scores to z-scores with live arithmetic and watch any normal curve collapse onto the one standard normal โ€” with tail areas computed two independent ways, a real simulated sample that standardizes to mean 0 and SD 1, and a reverse lookup from percentile back to a raw cutoff. Open full ↑
Key points
  • The standard normal distribution has μ = 0 and σ = 1, written Z ~ N(0, 1).
  • z = (x − μ) ÷ σ converts any value into "number of standard deviations from the mean," a unitless quantity.
  • Sign gives direction (above or below μ); magnitude gives unusualness, so z-scores let you compare different scales.
  • x = μ + zσ converts back; standardizing never changes the shape of a distribution.

Section 4Finding probabilities

A printed table of standard normal cumulative probabilities with z values down the left column
Reference. A standard normal probability table — each entry is the area to the left of the listed z value. Source and licence: see the figure credits

What the table actually gives you

A standard normal table (or z-table) is a lookup of areas. The standard version used almost everywhere reports cumulative left-tail area: the entry for a given z is the proportion of the distribution lying below that z, written Φ(z) or P(Z < z). So the entry for z = 0.00 is 0.5000 — half the curve lies below center — and the entries climb toward 1 as z increases.

Reading it takes two coordinates. The row gives z to one decimal place; the column gives the hundredths digit. To find Φ(1.75), go down to the row 1.7 and across to the column .05:

z.00.01.02.03.04.05.06.07
1.5.9332.9345.9357.9370.9382.9394.9406.9418
1.6.9452.9463.9474.9484.9495.9505.9515.9525
1.7.9554.9564.9573.9582.9591.9599.9608.9616
1.8.9641.9649.9656.9664.9671.9678.9686.9693
1.9.9713.9719.9726.9732.9738.9744.9750.9756

Φ(1.75) = 0.9599. That single number means: 95.99% of a standard normal distribution lies below z = 1.75. Everything else in this section is arithmetic performed on numbers like that one.

The three question types

Every normal probability problem is one of three shapes, and each has a one-line rule. In all of them, first convert your raw value to a z-score, then look it up.

QuestionRuleWhy
P(X < a) — left tailΦ(za)Read the table directly; it already reports left-tail area
P(X > a) — right tail1 − Φ(za)Total area is 1, so what is not below must be above
P(a < X < b) — betweenΦ(zb) − Φ(za)Take everything below the top, remove everything below the bottom

Two habits keep you out of trouble. Sketch first: draw the bell, mark the mean, mark your value, shade the region the question asks about. If your final answer is more than 0.5 but you shaded a small sliver, you have made an error. Sanity-check against the empirical rule: an area beyond z = 2 should be near 2.5%, an area below z = 0 should be exactly 0.5000.

Worked example: the same exam, three questions

Exam scores are approximately normal with μ = 72 and σ = 8.

(a) What proportion of students score below 86?
Step 1 — standardize: z = (86 − 72) ÷ 8 = 14 ÷ 8 = 1.75.
Step 2 — look up: Φ(1.75) = 0.9599.
Step 3 — answer: P(X < 86) = 0.9599, about 96.0% of students score below 86.

(b) What proportion score above 86?
Same z = 1.75. This is a right tail, so subtract from 1:
P(X > 86) = 1 − 0.9599 = 0.0401, about 4.0%. Sensible: (a) and (b) must add to 1, and 0.9599 + 0.0401 = 1.0000.

(c) What proportion score between 64 and 86?
Step 1 — standardize both ends.
Lower: z = (64 − 72) ÷ 8 = (−8) ÷ 8 = −1.00.
Upper: z = (86 − 72) ÷ 8 = 1.75.
Step 2 — look up both. Φ(1.75) = 0.9599. Φ(−1.00) = 0.1587.
Step 3 — subtract: 0.9599 − 0.1587 = 0.8012, about 80.1%.
Sanity check: the interval runs from 1σ below the mean to 1.75σ above, so it should comfortably exceed the 68% that sits inside ±1σ but stay under 95%. It does.

If your table lists only positive z values, get Φ(−1.00) from symmetry: Φ(−z) = 1 − Φ(z), so Φ(−1.00) = 1 − 0.8413 = 0.1587. Same number.

Worth committing to memory, because they recur throughout the rest of statistics:

zΦ(z) = area to the leftArea to the rightNote
0.000.50000.5000The mean splits the curve in half
1.000.84130.1587Gives the 68% rule: 0.8413 − 0.1587 = 0.6826
1.6450.95000.0500Top 5% cutoff
1.960.97500.0250The real "95%" number: 2.5% in each tail
2.000.97720.0228Gives the 95.45% rule
2.5760.99500.0050Middle 99%
3.000.99870.0013Gives the 99.73% rule
A probability is a long-run proportion, not a verdict about one person. "P(X > 86) = 0.04" means that in this population, about 4% of scores exceed 86 — or equivalently, a randomly drawn student has a 4% chance of scoring above 86 before you look. Once you know a particular student scored 90, there is no longer any probability involved: that student is simply in the top 4%. Probability describes the process of drawing, not the value already drawn.
Interactive Shade any region under a normal curve โ€” between, below, above, or outside two boundaries โ€” and read its probability off a live numerical integration, then check that answer against a real simulated sample counted one observation at a time. Open full ↑
Key points
  • The z-table gives Φ(z) = P(Z < z), the area to the left; row = tenths, column = hundredths.
  • Left tail: read directly. Right tail: 1 − Φ(z). Between: Φ(zupper) − Φ(zlower).
  • Symmetry gives negative values: Φ(−z) = 1 − Φ(z); and P(X < a) equals P(X ≤ a) because the variable is continuous.
  • Always sketch and shade before computing, and check the answer against the empirical rule.

Section 5Working backward

A bell curve populated with one hundred figures illustrating how percentile rank maps onto the normal distribution
Reference. Percentile rank on a bell curve — a value's percentile is the share of the distribution lying below it. Source and licence: see the figure credits

The question runs the other way

So far you have been handed a value and asked for a proportion. Many real questions arrive in the opposite order: What score puts a student in the top 10%? What birth weight is the 5th percentile? Where do we set the cutoff so that 95% of parts pass inspection? These are inverse problems, and they run the same machinery backward.

Begin with the vocabulary. The kth percentile is the value with k percent of the distribution below it. The 90th percentile is the value that 90% of the data falls under, and 10% exceeds. Percentiles are always stated as left-tail areas, which is convenient, because that is exactly what the z-table reports.

The three-step recipe:

  1. Convert the wording into a left-tail area. "90th percentile" is already 0.9000. "Top 10%" is not 0.10 — the cutoff for the top 10% has 90% below it, so the area is 0.9000 as well. "Bottom 5%" is 0.0500.
  2. Search the body of the table for that area and read off the z. This is the reverse lookup: you scan the interior probabilities for the one closest to your target, then read the z from its row and column.
  3. Un-standardize with x = μ + zσ. That converts the position back into the original units.

Worked example: the 90th percentile

Same exam: μ = 72, σ = 8. What score is the 90th percentile?

Step 1 — target left-tail area = 0.9000.

Step 2 — hunt in the body of the table. The nearest entries are Φ(1.28) = 0.8997 and Φ(1.29) = 0.9015. The first is off by 0.0003, the second by 0.0015, so take z = 1.28.

Step 3 — un-standardize:
x = μ + zσ = 72 + (1.28)(8) = 72 + 10.24 = 82.24

A score of about 82.2 marks the 90th percentile: roughly 90% of students score below it, 10% above. Check the answer for plausibility — 82.24 is above the mean, as any percentile above 50 must be, and it is a bit more than one standard deviation up, which fits an area of 0.90 sitting between Φ(1.00) = 0.8413 and Φ(1.50) = 0.9332.

Two more, including a negative z

(a) What score separates the top 5%? The cutoff has 95% below it, so the target area is 0.9500 and z = 1.645 (the table straddles it: Φ(1.64) = 0.9495, Φ(1.65) = 0.9505, so 1.645 is the standard interpolated value).
x = 72 + (1.645)(8) = 72 + 13.16 = 85.16. Scores at or above about 85.2 are in the top 5%.

(b) What score is the 25th percentile (the first quartile)? Target area 0.2500, which is below the mean, so the z must be negative. By symmetry, Φ(−0.67) = 1 − Φ(0.67) = 1 − 0.7486 = 0.2514, and Φ(−0.68) = 1 − 0.7517 = 0.2483. The value 0.2500 sits between them; the conventional value is z = −0.67 (more precisely −0.6745).
x = 72 + (−0.67)(8) = 72 − 5.36 = 66.64. About a quarter of students score below 66.6.

The single most common error in inverse problems is dropping the minus sign. If the percentile is below 50, the z must be negative and the answer must land below the mean. If your arithmetic hands you a value above μ for the 25th percentile, you lost a sign somewhere.

These z values recur so often that they are worth keeping close:

PercentileLeft-tail areazScore when μ = 72, σ = 8  (x = 72 + 8z)
1st0.0100−2.32672 − 18.61 = 53.39
5th0.0500−1.64572 − 13.16 = 58.84
10th0.1000−1.28272 − 10.26 = 61.74
25th0.2500−0.67472 − 5.39 = 66.61
50th0.50000.00072.00
75th0.7500+0.67472 + 5.39 = 77.39
90th0.9000+1.28272 + 10.26 = 82.26
95th0.9500+1.64572 + 13.16 = 85.16
99th0.9900+2.32672 + 18.61 = 90.61

Finding a symmetric middle interval

A close cousin: which two scores enclose the middle 95%? Symmetry does the work. If 95% is in the middle, 5% is left over, split into 2.5% in each tail. So the lower bound has left-tail area 0.0250 (z = −1.96) and the upper bound has left-tail area 0.9750 (z = +1.96). Then:

Lower: x = 72 + (−1.96)(8) = 72 − 15.68 = 56.32
Upper: x = 72 + (1.96)(8) = 72 + 15.68 = 87.68

The middle 95% of scores runs from about 56.3 to 87.7. This calculation is the direct ancestor of the confidence interval, and it is where the number 1.96 first earns its keep.

"Middle 95%" is not "95% confidence." The interval 56.3 to 87.7 above describes where 95% of individual scores fall in a population whose μ and σ we treated as known. A confidence interval, which you will meet later, is a different object: it estimates an unknown parameter, and its "95%" refers to the long-run success rate of the procedure — if you repeated the whole study many times, about 95% of the intervals produced would contain the true parameter. It is not a 95% chance that the parameter lies inside the one interval you happen to have. Keep the two ideas in separate drawers.
Interactive Name an area instead of a value โ€” the bottom p%, the top p%, or the middle p% โ€” and run the normal curve in reverse: convert the ask into a left area, invert the standard normal to get z*, un-standardize with x = ฮผ + zฯƒ, then check the cutoff against a sample you draw yourself. Open full ↑
Key points
  • A percentile is stated as a left-tail area: the kth percentile has k% of the distribution below it.
  • Inverse problems go area → z → x, using x = μ + zσ.
  • "Top 10%" means a left-tail area of 0.90, not 0.10, and percentiles below 50 give a negative z — losing that sign is the classic error.
  • A symmetric middle interval splits the leftover area evenly between the two tails: middle 95% → z = ±1.96.

The chapter in one breath

  • The normal distribution is a symmetric, unimodal, bell-shaped model fixed entirely by two parameters: μ (center) and σ (spread).
  • For a continuous variable, area under the curve is probability; the total area is 1, and the probability of any single exact value is 0.
  • The empirical rule puts about 68%, 95%, and 99.7% of the data within 1, 2, and 3 standard deviations of the mean — and symmetry lets you halve any band.
  • The z-score, z = (x − μ) ÷ σ, rewrites any value as a unitless number of standard deviations from the mean, making different scales comparable.
  • The standard normal Z ~ N(0, 1) is the one curve every normal distribution is converted to, so one table serves them all.
  • The z-table reports left-tail area Φ(z); right tails are 1 − Φ(z) and between-values are Φ(zupper) − Φ(zlower).
  • Running the machine backward — area → z → x = μ + zσ — converts a percentile into a raw score.
  • All of it depends on the data being approximately normal; standardizing rescales a distribution but never repairs its shape.

ReferenceWords worth knowing

A quick reference for the terms in this chapter — skim it now, or circle back whenever a word trips you up.

Area under the curve
The region between a density curve and the horizontal axis over some interval; for a probability model, this area is the probability of landing in that interval.
Bell curve
Informal name for the normal curve, from its symmetric, single-humped shape.
Chebyshev's inequality
The guarantee that at least 1 − 1/k2 of any distribution lies within k standard deviations of the mean (k > 1); weaker than the empirical rule but always true.
Continuous random variable
A variable that can take any value in an interval; the probability of any single exact value is zero, so probabilities are computed over ranges.
Cumulative probability
The total probability at or below a given value — the left-tail area, written Φ(z) for the standard normal.
Density curve
A smooth curve whose total enclosed area equals 1, used to model the distribution of a continuous variable.
Empirical rule
The 68–95–99.7 rule: in a normal distribution, roughly 68%, 95%, and 99.7% of values lie within 1, 2, and 3 standard deviations of the mean.
Inflection point
The point where a curve changes its direction of bending; on a normal curve these sit exactly at μ − σ and μ + σ.
Lower-tail area
The proportion of the distribution below a stated value; what a standard normal table reports.
Mean (μ)
The center of a normal distribution; for this model it equals the median and the mode.
Normal distribution
A symmetric, unimodal, bell-shaped continuous probability model determined completely by μ and σ; written X ~ N(μ, σ2).
Parameter
A numerical characteristic of a population or model, such as μ or σ, as opposed to a statistic computed from a sample.
Percentile
The value below which a stated percentage of the distribution falls; the 90th percentile has 90% of values beneath it.
Probability density function
The equation that produces a density curve; its height is density, not probability.
Raw score
A value expressed in its original measurement units, before standardizing.
Standard deviation (σ)
The spread of a normal distribution; the distance from the mean to either inflection point.
Standard normal distribution
The normal distribution with μ = 0 and σ = 1, written Z ~ N(0, 1); the reference curve every other normal distribution is converted to.
Standardizing
Converting values to z-scores by subtracting the mean and dividing by the standard deviation; a linear rescaling that changes center and spread but not shape.
Symmetric distribution
One whose left and right halves are mirror images about the center.
Unimodal
Having exactly one peak.
z-score
The unitless quantity z = (x − μ) ÷ σ, giving how many standard deviations a value sits above (positive) or below (negative) the mean.
z-table
A table of standard normal cumulative probabilities, used forward to turn a z into an area and backward to turn an area into a z.

Check yourself

1. Name the two parameters that completely determine a normal curve, and state what each one does to the picture.
The mean μ and the standard deviation σ. Changing μ slides the entire curve left or right without altering its shape; changing σ makes the curve narrower and taller (small σ) or wider and flatter (large σ) without moving its center.
2. A classmate says "the height of the normal curve at x = 70 is the probability of getting a 70." What is wrong with that, and what is the correct idea?
Height is density, not probability; for a continuous variable the probability of any single exact value is zero. Probability is the area under the curve over an interval — you ask for P(69.5 < X < 70.5), not P(X = 70). A related consequence: P(X < 70) and P(X ≤ 70) are equal.
3. Systolic blood pressure in a population is approximately normal with μ = 120 mmHg and σ = 10 mmHg. Give the intervals covering about 68%, 95%, and 99.7% of the population.
68%: 110 to 130. 95%: 100 to 140. 99.7%: 90 to 150. Each is μ ± kσ: 120 ± 10, 120 ± 20, and 120 ± 30.
4. Using the same distribution (μ = 120, σ = 10), what percent of the population has a systolic pressure above 140? Show the reasoning without a table.
About 2.5%. 140 is μ + 2σ. The empirical rule puts about 95% inside ±2σ, leaving 100% − 95% = 5% outside; symmetry splits that evenly, so 5% ÷ 2 = 2.5% lies above 140.
5. Still with μ = 120 and σ = 10: what percent falls between 130 and 140?
About 13.5%. That band runs from μ + 1σ to μ + 2σ. Take the difference of the symmetric bands and halve it: (95% − 68%) ÷ 2 = 27% ÷ 2 = 13.5%.
6. A dataset of hospital lengths of stay is strongly right-skewed. Why is the empirical rule the wrong tool, and what can you use instead?
The 68–95–99.7 percentages are properties of the normal shape, so they do not hold for skewed data. Applying them would misstate both tails. Chebyshev's inequality still applies to any distribution: at least 1 − 1/k2 of values lie within k SD of the mean — at least 75% within 2 SD and 88.9% within 3 SD — but it is far less precise.
7. A lab value has μ = 96 and σ = 12. Compute the z-score for a reading of 78 and interpret it.
z = (78 − 96) ÷ 12 = (−18) ÷ 12 = −1.50. The reading is 18 units below the mean, which is one and a half standard deviations below average. The negative sign means "below the mean"; a magnitude of 1.5 is somewhat low but not extreme.
8. Student A scored 88 on a test with μ = 80, σ = 5. Student B scored 140 on a different test with μ = 125, σ = 12. Who performed better relative to their group?
Student A. zA = (88 − 80) ÷ 5 = 8 ÷ 5 = +1.60; zB = (140 − 125) ÷ 12 = 15 ÷ 12 = +1.25. The raw 140 is the larger number, but relative to its own distribution it is only 1.25 SD above average, while A's 88 is 1.60 SD above. z-scores remove the units and the scale so the comparison is fair.
9. Using μ = 72 and σ = 8, find the proportion of scores below 78. Use Φ(0.75) = 0.7734.
About 0.7734, or 77.3%. Standardize first: z = (78 − 72) ÷ 8 = 6 ÷ 8 = 0.75. The table gives left-tail area directly, so P(X < 78) = Φ(0.75) = 0.7734. As a check, the answer should exceed 0.50 because 78 is above the mean, and it does.
10. Same distribution (μ = 72, σ = 8). Find the proportion of scores between 60 and 78, given Φ(0.75) = 0.7734 and Φ(1.50) = 0.9332.
About 0.7066, or 70.7%. Lower: z = (60 − 72) ÷ 8 = −1.50, and by symmetry Φ(−1.50) = 1 − 0.9332 = 0.0668. Upper: z = 0.75, Φ(0.75) = 0.7734. Subtract: 0.7734 − 0.0668 = 0.7066.
11. With μ = 500 and σ = 100, find the score at the 95th percentile. Use z = 1.645.
x = 500 + (1.645)(100) = 500 + 164.5 = 664.5. The 95th percentile means a left-tail area of 0.9500, whose z is 1.645; un-standardize with x = μ + zσ. Note the same value answers "what score separates the top 5%?" — the top 5% cutoff has 95% below it.
12. With μ = 500 and σ = 100, find the 10th percentile, and explain the sign of the z. Use z = −1.28.
x = 500 + (−1.28)(100) = 500 − 128 = 372. The 10th percentile has a left-tail area of 0.1000, which is less than 0.50, so the value must sit below the mean and the z must be negative. Dropping that minus sign is the most common error in inverse problems — it would give 628, which is above the mean and therefore obviously wrong for a 10th percentile.