Why does the IQ score 115 have the same percentile as the SAT score 1250?

Both are exactly one standard deviation above their mean: (115 − 100)/15 = 1 and (1250 − 1050)/200 = 1. The z-score strips away the units and the scale, leaving only how many standard deviations a value lies from the centre. Every normal distribution has the same shape in z units, so one table of Φ(z) answers questions about all of them. Φ(1) = 0.8413, so both scores sit at the 84.13th percentile.

If 95 % of values lie within ±2σ, why does the two-tailed 95 % interval use z = 1.96 and not 2?

The ±2σ figure is a rounded rule of thumb: the exact probability inside ±2σ is 95.45 %, not 95 %. Solving Φ(z) = 0.975 with the inverse normal gives z = 1.95996…, so the central 95 % lies between −1.96σ and +1.96σ and each tail holds exactly 2.5 %. Likewise the exact 99 % interval uses 2.576, not 3; ±3σ holds 99.73 %. The 68–95–99.7 rule is for mental arithmetic, while the inverse normal gives the exact cut-offs.

Why is the probability that X equals a particular value, such as exactly 100, zero?

For a continuous variable, probability is area under the density curve, and a single point has zero width and therefore zero area. P(X = 100) = 0 even though 100 is the most likely region in the sense that the density is highest there. Meaningful questions are about intervals: P(99.5 < X < 100.5) ≈ 0.0266 for IQ scores. This is also why P(X < x) and P(X ≤ x) are the same number for a normal variable, and why the left-tail, right-tail and interval areas always add up to 1.

When I drop 200 points, why do I not get exactly 136 inside ±1σ (68.27 % of 200)?

Each point lands inside ±1σ independently with probability 0.6827, so the count is a binomial random variable with mean 136.5 and standard deviation √(200 × 0.6827 × 0.3173) ≈ 6.6. Counts anywhere from about 123 to 150 are therefore ordinary outcomes, and the observed share scatters by roughly ±3 percentage points around 68.3 %. The rule describes the limit of a very large sample, not any single sample; with n = 20 000 the scatter would shrink to about ±0.3 points. This is the law of large numbers in action.

Normal Distribution & z-Scores

Q: When I drop 200 points, why do I not get exactly 136 inside ±1σ (68.27 % of 200)?

Each point lands inside ±1σ independently with probability 0.6827, so the count is a binomial random variable with mean 136.5 and standard deviation √(200 × 0.6827 × 0.3173) ≈ 6.6. Counts anywhere from about 123 to 150 are therefore ordinary outcomes, and the observed share scatters by roughly ±3 percentage points around 68.3 %. The rule describes the limit of a very large sample, not any single sample; with n = 20 000 the scatter would shrink to about ±0.3 points. This is the law of large numbers in action.

Bell curve with μ and σ; shaded tails, the 68–95–99.7 rule, and a built-in z-score converter x → z = (x − μ)/σ.

Bell curve N(μ, σ²)

Mean μ100

Standard deviation σ15

Shaded area

Show 68–95–99.7 bands

z-score converter

x (raw value)Percentile (%)

x, the percentile and the pink marker are one and the same value: edit any of them and the others follow.

Φ(z) is computed from the error function with Cody’s rational approximation (relative error below 10⁻¹⁵), the inverse with Acklam’s algorithm plus one Halley refinement. The 200 dropped points are exact N(μ, σ²) draws (Box–Muller) and are stored as z-scores, so they move with the curve when you change μ or σ. The window always shows μ ± 4σ.

Shortcuts

Space / Enter — drop 200 points
R — reset
Drag the pink marker on the canvas to move x (and b)

Measured values

Mean μ100.0

Standard deviation σ15.0

x115.0

z = (x − μ)/σ1.000

Shaded probability P(X < x)0.84134

Percentile of x = 100·Φ(z)84.13%

About this model

The simulator draws the density of a normal (Gaussian) random variable X ~ N(μ, σ²), f(x) = exp(−(x − μ)²/(2σ²)) / (σ√(2π)), over the window μ ± 4σ. The mean μ fixes the position of the peak and the standard deviation σ its width; the area under the whole curve is 1, so any shaded part of it is a probability. Below the raw x axis a second axis shows the same positions as z-scores, z = (x − μ)/σ, which map every normal distribution onto the standard normal N(0, 1). All probabilities are computed from the standard normal cumulative distribution function Φ(z) = ½·erfc(−z/√2), evaluated with W. J. Cody's rational Chebyshev approximation of the complementary error function (relative error below 10⁻¹⁵, far inside the 10⁻⁷ needed to agree with printed z tables to every digit). Four shading modes are available: the left tail P(X < x) = Φ(z); the right tail P(X > x) = 1 − Φ(z); the two-tailed probability P(|Z| > |z|) = 2(1 − Φ(|z|)), the quantity used for two-sided tests and confidence intervals; and the interval P(a < X < b) = Φ(z_b) − Φ(z_a). The marker on the canvas, the x field and the percentile field are the same state, so the converter works in both directions: x → z → percentile 100·Φ(z), and percentile → z → x = μ + zσ. The inverse Φ⁻¹ uses P. J. Acklam's rational approximation followed by one Halley refinement step with the accurate Φ, which brings it to full double precision (Φ⁻¹(0.975) = 1.959964…). The 68–95–99.7 bands shade ±1σ, ±2σ and ±3σ and label each slice with its exact share (34.1 %, 13.6 %, 2.1 %, 0.1 %); the exact totals are 68.27 %, 95.45 % and 99.73 %. Pressing Drop 200 points draws 200 independent values from N(μ, σ²) with the Box–Muller transform on a seeded generator and shows them as a coloured dot plot above the axis; the readouts count how many fell inside ±1σ, ±2σ and ±3σ so the rule can be checked against a finite sample, which scatters around the theoretical shares by roughly ±3 percentage points for n = 200. Simplifications: the model is the idealised continuous normal with no truncation, rounding or discreteness, so real IQ or SAT scores, which are integers on a bounded scale, only approximately follow it; sample points are replaced on each drop rather than accumulated. Related simulators on this site: statistics/galton-board and statistics/binomial-distribution (how the bell curve arises from many small random steps), statistics/law-of-large-numbers, statistics/confidence-interval and statistics/p-value (which use the two-tailed and one-tailed areas drawn here), and statistics/type-i-ii-errors-power.

Who it's for: Upper secondary school and first-year university statistics (AP Statistics, A-level, introductory probability).

Key terms

normal distribution
mean μ and standard deviation σ
z-score
standard normal Φ(z)
percentile
68–95–99.7 rule
tail probability
inverse normal (quantile)

How it works

The normal distribution N(μ, σ²) is the bell-shaped density f(x) = exp(−(x − μ)²/2σ²)/(σ√(2π)); μ sets its centre and σ its width. Every normal variable becomes the standard normal through the z-score z = (x − μ)/σ, which is why one table of Φ(z) serves IQ scores, heights and SAT results alike. The shaded area under the curve is a probability: the left tail P(X < x) = Φ(z), the right tail 1 − Φ(z), the two-tailed P(|Z| > z) = 2(1 − Φ(|z|)), and P(a < X < b) = Φ(z_b) − Φ(z_a). The 68–95–99.7 rule says 68.27 %, 95.45 % and 99.73 % of the probability lies within ±1σ, ±2σ and ±3σ; dropping 200 random points shows how closely a real sample follows it. The curve itself never ends, so no interval has probability exactly 1 — only the sample is finite.

Key equations

f(x) = 1/(σ√(2π)) · exp(−(x − μ)² / (2σ²))

z = (x − μ) / σ ⇔ x = μ + zσ

P(X < x) = Φ(z) = ½ · erfc(−z/√2) · P(a < X < b) = Φ(z_b) − Φ(z_a)

P(|Z| ≤ 1) = 0.6827 P(|Z| ≤ 2) = 0.9545 P(|Z| ≤ 3) = 0.9973

percentile p → z = Φ⁻¹(p/100), e.g. Φ⁻¹(0.975) = 1.95996

Frequently asked questions

Why does the IQ score 115 have the same percentile as the SAT score 1250?: Both are exactly one standard deviation above their mean: (115 − 100)/15 = 1 and (1250 − 1050)/200 = 1. The z-score strips away the units and the scale, leaving only how many standard deviations a value lies from the centre. Every normal distribution has the same shape in z units, so one table of Φ(z) answers questions about all of them. Φ(1) = 0.8413, so both scores sit at the 84.13th percentile.
If 95 % of values lie within ±2σ, why does the two-tailed 95 % interval use z = 1.96 and not 2?: The ±2σ figure is a rounded rule of thumb: the exact probability inside ±2σ is 95.45 %, not 95 %. Solving Φ(z) = 0.975 with the inverse normal gives z = 1.95996…, so the central 95 % lies between −1.96σ and +1.96σ and each tail holds exactly 2.5 %. Likewise the exact 99 % interval uses 2.576, not 3; ±3σ holds 99.73 %. The 68–95–99.7 rule is for mental arithmetic, while the inverse normal gives the exact cut-offs.
Why is the probability that X equals a particular value, such as exactly 100, zero?: For a continuous variable, probability is area under the density curve, and a single point has zero width and therefore zero area. P(X = 100) = 0 even though 100 is the most likely region in the sense that the density is highest there. Meaningful questions are about intervals: P(99.5 < X < 100.5) ≈ 0.0266 for IQ scores. This is also why P(X < x) and P(X ≤ x) are the same number for a normal variable, and why the left-tail, right-tail and interval areas always add up to 1.
When I drop 200 points, why do I not get exactly 136 inside ±1σ (68.27 % of 200)?: Each point lands inside ±1σ independently with probability 0.6827, so the count is a binomial random variable with mean 136.5 and standard deviation √(200 × 0.6827 × 0.3173) ≈ 6.6. Counts anywhere from about 123 to 150 are therefore ordinary outcomes, and the observed share scatters by roughly ±3 percentage points around 68.3 %. The rule describes the limit of a very large sample, not any single sample; with n = 20 000 the scatter would shrink to about ±0.3 points. This is the law of large numbers in action.

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