Normal & Standard Normal

Squint at almost anything built from many small influences — the heights of a thousand people, the errors in a repeated measurement, the totals on a long exam — and the same shape keeps surfacing. A hump in the middle, thinning smoothly on both sides. That bell is the normal distribution, the most important continuous shape there is, and the “standardise first” the last lesson hinted at is the key to using it. You never integrate this bell by hand. Instead you reshape every normal back to one fixed reference curve, and look the answer up.

One bell, and the reference bell

A normal is fixed by just two numbers: its mean μ, where the peak sits, and its variance σ², how wide it spreads. The one special case μ = 0, σ = 1 is the standard normal, written Z ~ N(0, 1) — the reference curve every other normal is compared against.

To turn any normal into that reference, subtract the mean and divide by the standard deviation:

A standardised value Z = (X − μ)/σ just says how many standard deviations from the mean the value sits. A score of +2 is two SDs above; −1 is one below. Every question about any normal collapses to a question about this one Z.

Reading a Φ table

Φ(z) is the cumulative standard-normal function — the area to the left of z, that is Φ(z) = P(Z ≤ z). A Φ table lists these areas, so to find P(X ≤ a) you standardise a to a z-score and read Φ(z) off the table. Two values are worth memorising: Φ(1) ≈ 0.8413 and Φ(2) ≈ 0.9772.

Because the bell is symmetric about 0, the left tail mirrors the right, which is how you handle a negative z from a table that only prints positive ones:

Φ(−z) = 1 − Φ(z)        e.g.   Φ(−1) = 1 − 0.8413 = 0.1587

The same symmetry gives the 68-95-99.7 rule: about 68% of the mass lies within one SD of the mean (μ ± σ), about 95% within two, and about 99.7% within three. Open the Normal tab below, bracket an interval, and watch the shaded area — that area is exactly the Φ(z_b) − Φ(z_a) you would otherwise look up.

A worked example

Let X ~ N(50, 10²), so μ = 50 and σ = 10. Find P(X ≤ 60) and P(40 ≤ X ≤ 60). Use Φ(1) ≈ 0.8413.

Standardise the bound, then read Φ:

P(X ≤ 60):   z = (60 − 50)/10 = 1     →   Φ(1) = 0.8413

P(40 ≤ X ≤ 60):
  z_low  = (40 − 50)/10 = −1
  z_high = (60 − 50)/10 = +1
  P = Φ(1) − Φ(−1) = 0.8413 − (1 − 0.8413) = 0.8413 − 0.1587 = 0.6826

So P(X ≤ 60) ≈ 0.8413 and P(40 ≤ X ≤ 60) ≈ 0.6826. That second answer is the 68% band in disguise — 40 and 60 sit exactly one SD either side of the mean, so the 68-95-99.7 rule could have given it with no lookup at all.

A question to carry forward

The normal turned up whenever many small influences piled together — but that was just an observation, not a reason. Here is the thread onward: if you take a sample from any population, however lopsided, and average it, the averages come out bell- shaped. Why should averaging manufacture a normal out of nothing, and how tight does that bell get as the sample grows?

In one breath

• A normal N(μ, σ²) is the symmetric bell, fixed by mean μ (location) and variance σ² (width).
• Standardise Z = (X − μ)/σ (divide by σ, never σ²) to the standard normal N(0,1); every question becomes one about Z.
• Φ(z) = P(Z ≤ z) from the table; symmetry Φ(−z) = 1 − Φ(z). Memorise Φ(1)≈0.8413, Φ(2)≈0.9772. Interval: P(a≤X≤b) = Φ(z_b) − Φ(z_a).
• 68-95-99.7: ~68% within μ±σ, ~95% within μ±2σ, ~99.7% within μ±3σ.
• It underlies the CLT and z-tests — standardise, look up Φ, combine.

Practice