Normal & Standard Normal

Heights of people. Errors in a measurement. Scores on a long test. Stand back and squint at any of them and you’ll see the same shape: a hump in the middle, thinning smoothly on both sides. That hump is the normal distribution — the bell curve — and it shows up so reliably that it underpins z-tests, the Central Limit Theorem, and most of the machine learning you’ll meet later.

You will never integrate the bell curve by hand on the exam. The whole trick is to standardize any normal back to one fixed reference curve, then look up the answer in a Phi table. Master that single move and this entire topic becomes plug-and-play.

N(mu, sigma squared) and the standard normal

A normal is fixed by two numbers: its mean mu (where the peak sits) and its variance sigma² (how wide it spreads). The special case mu = 0, sigma = 1 is the standard normal Z ~ N(0, 1).

To convert any normal to the standard one, subtract the mean and divide by the standard deviation:

Z = (X − mu) / sigma

Now Z is a standard normal, and a Z value (a z-score) just says “how many standard deviations from the mean.” Everything reduces to questions about Z.

Reading a Phi table

Phi(z) is the cumulative standard-normal function: the area to the left of z, i.e. Phi(z) = P(Z ≤ z). A Phi table lists these areas. To find P(X ≤ a): standardize a to a z-score, then read Phi(z) off the table.

Two values worth memorizing: Phi(1) ≈ 0.8413 and Phi(2) ≈ 0.9772.

Because the bell curve is symmetric about 0, the left tail mirrors the right:

Phi(-z) = 1 - Phi(z)

So Phi(−1) = 1 − 0.8413 = 0.1587. Tables usually only print positive z, so this identity is how you handle negatives.

Open the Normal tab below. Drag the two handles to bracket an interval and the shaded area is the probability — that area is exactly the Phi(z_b) − Phi(z_a) you would otherwise look up in a table. Slide the mean and standard deviation to see how the same bell shape sits at any location with any width: that is why one standardised reference curve is enough for all of them.

How GATE asks this

Reliably a NAT or MCQ: a normal with given mu and sigma, a target value or interval, and Phi values supplied (or expected from memory). You standardize, look up Phi, and combine. Interval problems use P(a ≤ X ≤ b) = Phi(z_b) − Phi(z_a). This appears in essentially every paper, sometimes wrapped inside a CLT question.

Worked example

Let X ~ N(50, 10²), so mu = 50 and sigma = 10. Find P(X ≤ 60) and P(40 ≤ X ≤ 60). Use Phi(1) ≈ 0.8413.

Step 1 — standardize the upper bound.

z = (60 - 50) / 10 = 1
P(X <= 60) = Phi(1) = 0.8413

Step 2 — the interval. Standardize both ends, then subtract:

z_low  = (40 - 50)/10 = -1
z_high = (60 - 50)/10 = +1

P(40 <= X <= 60) = Phi(1) - Phi(-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 exactly the 68% band — 40 and 60 sit one standard deviation either side of the mean, so the 68-95-99.7 rule predicts it without a single lookup.

Quick check