The Normal Distribution

Study Sheet

The Normal Distribution

Z-scores, the empirical rule and the standard normal table

Z-Scores

Concept
Standardizing

A z-score says how many standard deviations a value sits from the mean: z=x−μσz = \dfrac{x - \mu}{\sigma}. It lets you compare values from different scales — an SAT score and an ACT score — on one ruler.

Subtracting μ\mu moves the centre to 00; dividing by σ\sigma measures distance in standard deviations instead of points. (These are the shifting and scaling rules from Center and Spread.) A z-score of 22 is unusual in any roughly bell-shaped data set, whatever the units, because the standard deviation has already absorbed the scale. Negative z means below the mean. Converting back is the same formula solved for xx: x=μ+zσx = \mu + z\sigma.

Example
Compare two scores
4506508501050125014501650SAT1350: z = 1.56111621263136ACT29: z = 1.6z
Each scale laid out in standard deviations from its own mean: 1350 sits at z = 1.5, 29 at z = 1.6.

SAT: mean 10501050, SD 200200, score 13501350: z=(1350−1050)/200=1.5z = (1350 - 1050)/200 = 1.5. ACT: mean 2121, SD 55, score 2929: z=(29−21)/5=1.6z = (29 - 21)/5 = 1.6. On a common ruler the ACT score sits slightly farther above its mean, so it is the (slightly) stronger performance — even though 300300 points sounds like more than 88.

Example
A value far out

Adult heights have mean 170170 cm and SD 88 cm. How unusual is 186186 cm? z=(186−170)/8=2z = (186 - 170)/8 = 2: two standard deviations above the mean. In a bell-shaped distribution only about 2.5%2.5\% of people are taller than that (next section).

Example
Solving for the value

A test has mean 7272 and SD 66. What score is 1.51.5 standard deviations above the mean? Use x=μ+zσ=72+1.5⋅6=81x = \mu + z\sigma = 72 + 1.5\cdot6 = 81. Rearranging the definition, rather than guessing, also handles negative zz: z=−2z = -2 gives 72−12=6072 - 12 = 60.

Side note
Where it goes wrong

Dividing by the variance instead of the standard deviation, or subtracting the wrong way round. For the SAT score, (1350−1050)/2002=0.0075(1350 - 1050)/200^2 = 0.0075 is meaningless, and (1050−1350)/200=−1.5(1050 - 1350)/200 = -1.5 puts an above-average score below the mean.

Try it
Try it: back to raw

Heights have mean 170170 cm and SD 88 cm. What height has z=−1.25z = -1.25?

Show the answer

x=μ+zσ=170+(−1.25)(8)=170−10=160x = \mu + z\sigma = 170 + (-1.25)(8) = 170 - 10 = \mathbf{160} cm.

The Empirical Rule and the Table

Concept
68–95–99.7
55708510011513014568%IQ: mean 100, SD 15
The shaded band from 85 to 115 is within one SD of the mean: about 68% of the area.

For a normal (bell-shaped) distribution, about 68%68\% of values lie within 11 SD of the mean, 95%95\% within 22, and 99.7%99.7\% within 33. For other cutoffs, the standard normal table (or a calculator) gives the area to the left of any zz.

The rule is a rounding of table values: P(−1<Z<1)=0.6827P(-1 < Z < 1) = 0.6827, P(−2<Z<2)=0.9545P(-2 < Z < 2) = 0.9545, P(−3<Z<3)=0.9973P(-3 < Z < 3) = 0.9973. Because the curve is symmetric and has total area 11, the pieces can be cut and reassembled: half of 68%68\% lies on each side of the mean, and the 5%5\% outside two SDs splits into 2.5%2.5\% in each tail.

Example
Above a cutoff

IQ scores: mean 100100, SD 1515. Fraction above 130130: z=(130−100)/15=2z = (130 - 100)/15 = 2. Within 22 SDs lies 95%95\%, so 5%5\% lies outside, split equally between the two tails: P(Z>2)≈5%/2=2.5%P(Z > 2) \approx 5\%/2 = 2.5\%.

Example
Between two values

Fraction between 8585 and 115115: these are z=−1z = -1 and z=1z = 1, so about 68%68\%. Between 7070 and 115115: z=−2z = -2 to 11. Split at the mean: from −2-2 to 00 is half of 95%95\%, 47.5%47.5\%; from 00 to 11 is half of 68%68\%, 34%34\%. Total 81.5%81.5\%.

Side note
Where it goes wrong

Using the table area (which is always the area to the LEFT) as if it were the area to the right, or applying the empirical rule to data that is clearly skewed. Waiting times with mean 44 minutes and SD 44 minutes cannot have 16%16\% of values below 00.

Try it
Try it: percentile from zz

A value has z=1z = 1. Roughly what percentile is it?

Show the answer

Left of the mean is 50%50\%; from 00 to 11 is half of 68%68\%, 34%34\%. Total 84%84\%: about the 84\mathbf{84}th percentile.

Using the Standard Normal Table

Concept
Areas from Φ(z)\Phi(z)
1461541601701801861940.7887heights: mean 170 cm, SD 8 cm
Between 160 and 180 cm: z from −1.25 to 1.25, area Φ(1.25) − Φ(−1.25) ≈ 0.7887.

The table gives Φ(z)=P(Z<z)\Phi(z) = P(Z < z) for the standard normal ZZ. Every question reduces to it: P(Z>z)=1−Φ(z)P(Z > z) = 1 - \Phi(z); P(a<Z<b)=Φ(b)−Φ(a)P(a < Z < b) = \Phi(b) - \Phi(a); and Φ(−z)=1−Φ(z)\Phi(-z) = 1 - \Phi(z) by symmetry. For a normal XX with mean μ\mu and SD σ\sigma, first standardize. To go backwards from an area to a value, find zz in the body of the table, then x=μ+zσx = \mu + z\sigma.

Standardizing works because every normal curve is the standard one, shifted and stretched; areas are unchanged by relabelling the axis, so P(X<x)=P(Z<(x−μ)/σ)P(X < x) = P(Z < (x - \mu)/\sigma). One table therefore serves every normal distribution. Sketch the curve and shade the region before reading anything — the picture tells you whether you need Φ\Phi, 1−Φ1 - \Phi, or a difference.

Example
One table lookup

Find P(Z<1.25)P(Z < 1.25). Row 1.21.2, column 0.050.05 of the table: Φ(1.25)=0.8944\Phi(1.25) = 0.8944. So about 89.4%89.4\% of any normal population lies below 1.251.25 SDs above its mean.

Example
Between two heights

Heights: mean 170170, SD 88. Find P(160<X<180)P(160 < X < 180). Standardize both ends: z=−1.25z = -1.25 and z=1.25z = 1.25. Then Φ(1.25)−Φ(−1.25)=0.8944−0.1056=0.7887\Phi(1.25) - \Phi(-1.25) = 0.8944 - 0.1056 = 0.7887 (using Φ(−1.25)=1−0.8944\Phi(-1.25) = 1 - 0.8944; the table rounds, so the last digit may differ by one). About 79%79\% of adults are between 160160 and 180180 cm.

Example
The top 10%10\%

What height marks the tallest 10%10\%? We need Φ(z)=0.90\Phi(z) = 0.90; the table gives z≈1.28z \approx 1.28. Then x=170+1.28⋅8≈180.2x = 170 + 1.28\cdot8 \approx 180.2 cm. Going backwards means reading the table from the inside out: area first, then zz, then xx.

Side note
Where it goes wrong

Forgetting to subtract from 11. "P(X>182)P(X > 182): z=1.5z = 1.5, table 0.93320.9332, answer 93%93\%" gives the area BELOW 182182. A sketch shows at once that the tail above 1.51.5 SDs is small.

Try it
Try it: an upper tail

Heights have mean 170170 cm and SD 88 cm. What fraction of people are taller than 182182 cm? (The table gives Φ(1.5)=0.9332\Phi(1.5) = 0.9332.)

Show the answer

z=(182−170)/8=1.5z = (182 - 170)/8 = 1.5. The area to the right is 1−Φ(1.5)=1−0.9332=0.06681 - \Phi(1.5) = 1 - 0.9332 = \mathbf{0.0668}, about 6.7%6.7\%.

Why It Works

Add up many small independent effects — measurement errors, genetic contributions, rounding — and the total tends toward a normal distribution no matter what the individual pieces look like. This is the central limit theorem, and it is why heights, test scores and averages of anything are bell-shaped while single coin flips are not. The density 1σ2πe−(x−μ)2/2σ2\dfrac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2} is the unique shape that is preserved under this adding-up, which is why the same table serves every normal distribution once you standardize.