For X ~ N(100, 36), the standard deviation is:

6. The second parameter is the variance, so σ = √36 = 6. Using 36 directly is the most common error in the topic.

For a continuous distribution, P(X < 7) compared with P(X ⩽ 7) is:

equal. P(X = 7) = 0 because a single point encloses no area, so including or excluding the endpoint changes nothing.

If z = −1.5, then P(Z < −1.5) equals:

1 − Φ(1.5). By symmetry the area left of −1.5 equals the area right of +1.5, which is 1 − Φ(1.5). A probability can never be negative, ruling out two options immediately.

X ~ N(50, 16) and x = 58. The z-score is:

2. σ = √16 = 4, so z = (58 − 50)/4 = 2. Dividing by 16 instead of 4 gives 0.5.

P(a < Z < b) is given by:

Φ(b) − Φ(a). Both table values are areas to the left, so subtracting leaves the strip between them.

About what proportion lies within two standard deviations of the mean?

95%. The 68-95-99.7 rule gives about 95% within two standard deviations. 68% is one, and 99.7% is three.

To find the mark exceeded by the top 5%, you need the z value:

1.645. The top 5% lies above the 95th percentile, whose z-score is 1.645. The value 1.960 corresponds to the top 2.5%.

Standardising a normal distribution changes:

only the labelling of the axis. The curve is identical; only the scale beneath it is renamed in standard deviations. That is precisely why a single table can serve every normal distribution.

The Normal Distribution — Mathematics Notes, Worked Examples and Quiz

Q: For a continuous distribution, P(X < 7) compared with P(X ⩽ 7) is:

equal. P(X = 7) = 0 because a single point encloses no area, so including or excluding the endpoint changes nothing.

Q: If z = −1.5, then P(Z < −1.5) equals:

1 − Φ(1.5). By symmetry the area left of −1.5 equals the area right of +1.5, which is 1 − Φ(1.5). A probability can never be negative, ruling out two options immediately.

Q: X ~ N(50, 16) and x = 58. The z-score is:

2. σ = √16 = 4, so z = (58 − 50)/4 = 2. Dividing by 16 instead of 4 gives 0.5.

Q: P(a < Z < b) is given by:

Φ(b) − Φ(a). Both table values are areas to the left, so subtracting leaves the strip between them.

Q: About what proportion lies within two standard deviations of the mean?

95%. The 68-95-99.7 rule gives about 95% within two standard deviations. 68% is one, and 99.7% is three.

Q: To find the mark exceeded by the top 5%, you need the z value:

1.645. The top 5% lies above the 95th percentile, whose z-score is 1.645. The value 1.960 corresponds to the top 2.5%.

Q: Standardising a normal distribution changes:

only the labelling of the axis. The curve is identical; only the scale beneath it is renamed in standard deviations. That is precisely why a single table can serve every normal distribution.

The shape that keeps appearing

Heights, exam marks, measurement errors, the diameters of manufactured parts — plot enough of them and the same bell-shaped curve keeps emerging. It is symmetric about the mean, most of the data sits near the middle, and the tails thin out quickly in both directions.

The curve is continuous, which changes what probability means. For a discrete variable a probability is a height on a bar chart; here it is an area under the curve. The total area is 1, matching the fact that some value must occur.

One consequence catches everybody: the probability of any single exact value is zero, because a single point has no width and therefore encloses no area. That is why P(X < a) and P(X ⩽ a) are the same for a normal distribution, though they differ for a discrete one.

X ~ N(μ, σ²)μ = the mean, the centre of symmetryσ² = the variance; σ = standard deviationtotal area under the curve = 1the bracket holds the variance σ², not the standard deviation

μ: the meanthe curve is symmetric about it
σ: the standard deviationcontrols the width — larger σ, flatter and wider
σ²: the variancewhat the notation N(μ, σ²) actually contains

N(μ, σ²) holds the variance

X ~ N(50, 16) means the variance is 16, so the standard deviation is 4, not 16. Using 16 as σ is one of the most frequent errors in the topic, and it corrupts every subsequent calculation. Take the square root before standardising.

Standardising: relabelling the axis

There are infinitely many normal distributions, one for every pair of μ and σ. Tabulating them all is impossible, so instead every question is converted to a single reference distribution: the standard normal, Z ~ N(0, 1).

The conversion asks a simple question: how many standard deviations is this value from the mean? That count is the z-score, and it is all the table needs. Subtracting the mean shifts the centre to zero; dividing by σ rescales the width to one.

The curve itself does not change shape when you standardise — only the numbers written along the axis do. That is why one table serves every normal distribution that has ever been used.

z = (x − μ) / σΦ(z) = P(Z < z) = area to the left of zP(Z > z) = 1 − Φ(z)P(a < Z < b) = Φ(b) − Φ(a)Φ(−z) = 1 − Φ(z)by symmetrytables give the area to the LEFT; everything else is built from that

z: the standardised valuehow many standard deviations from the mean
Φ(z): the cumulative probabilityalways the area to the left
Φ(−z): the left tailequals 1 − Φ(z) because the curve is symmetric

On P(Z < z), drag z and watch the shaded area — that area is the probability. Then look at Standardising: the curve is identical, only the axis labels differ, which is exactly why one table is enough.

Reading the table in every direction

Tables give Φ(z), the area to the left. Every other probability is assembled from that one fact, and a sketch makes the assembly obvious rather than something to memorise.

For negative z the table usually stops at zero, so symmetry is used: the area left of −z equals the area right of +z, which is 1 − Φ(z).

You want	Use	Why
P(Z < z)	Φ(z)	straight from the table
P(Z > z)	1 − Φ(z)	the total area is 1
P(Z < −z)	1 − Φ(z)	symmetry: the left tail matches the right
P(a < Z < b)	Φ(b) − Φ(a)	the strip between two cuts
P(\|Z\| < z)	2Φ(z) − 1	both tails removed at once

Worked example

The masses of apples are normally distributed with mean 150 g and standard deviation 20 g. Find the probability that an apple weighs (a) less than 165 g, (b) between 130 g and 165 g.

(a) Standardise: z = (165 − 150)/20 = 0.75.165 g lies 0.75 standard deviations above the mean.
P(X < 165) = Φ(0.75) = 0.7734.The table gives the area to the left directly.
(b) Standardise the lower value: z = (130 − 150)/20 = −1.0.Below the mean, so the z-score is negative — as it should be.
Φ(−1) = 1 − Φ(1) = 1 − 0.8413 = 0.1587.Symmetry converts a negative z into a table lookup.
P(130 < X < 165) = 0.7734 − 0.1587 = 0.6147.The strip between the two cuts is the difference of the two left-areas. Sketching the region makes the subtraction obvious.

(a) 0.7734; (b) 0.6147

Always sketch it

A quick bell with the region shaded takes five seconds and prevents nearly every error in this topic. It shows immediately whether the answer should be bigger or smaller than a half, which catches a forgotten 1 − before it costs anything.

Working backwards from a probability

The reverse question gives a probability and asks for the value — the mark needed for the top 10%, the diameter exceeded by only 5% of parts. Here the table is used in reverse, or the inverse normal function on a calculator.

The procedure is the same one run backwards: find the z-score matching that area, then unstandardise with x = μ + zσ. The commonest slip is looking up the wrong tail, which a sketch prevents.

forwards:z = (x − μ)/σbackwards: x = μ + zσvalues worth knowing:top 5%z = 1.645top 2.5% z = 1.960top 1%z = 2.326a positive z lies above the mean, a negative z below it

x = μ + zσ: unstandardisingthe rearrangement of the z formula
1.645: the 95th percentileappears constantly in hypothesis testing too
1.960: the 97.5th percentilethe basis of a 95% confidence interval

The three shapes of question

Given x, find a probability — standardise, then read the table.
Given a probability, find x — read the table backwards, then x = μ + zσ.
Given a probability and one parameter, find the other — standardise symbolically and solve the resulting equation.
For two unknown parameters you need two pieces of information, giving simultaneous equations in μ and σ.
Sketch the region first, every time.

Approximating a binomial with a normal

Computing a binomial probability for n = 200 by hand is impractical, and when n is large and p is not too extreme the distribution is close enough to a bell that the normal can stand in for it.

The usual conditions are np > 5 and n(1 − p) > 5. Both matter: they ensure the distribution is far enough from either end to be roughly symmetric rather than crushed against zero.

The parameters carry straight across — mean np, variance np(1 − p) — but one adjustment is required. A binomial is discrete and a normal is continuous, so a whole number has to be treated as the interval half a unit either side of it. That is the continuity correction, and omitting it is the standard lost mark.

B(n, p) ≈ N(np, np(1 − p))when np > 5 and n(1 − p) > 5continuity correction:P(X ⩾ 20) → P(Y > 19.5)P(X > 20) → P(Y > 20.5)P(X ⩽ 20) → P(Y < 20.5)P(X = 20) → P(19.5 < Y < 20.5)the half unit always widens the region you actually want

np: the meancarried straight over from the binomial
np(1 − p): the variancenot np — the (1 − p) is easy to drop
± 0.5: the continuity correctiona whole number becomes an interval half a unit wide

Which way does the half go?

Sketch the bars. P(X ⩾ 20) includes the whole bar at 20, and that bar runs from 19.5 to 20.5 — so the continuous version starts at 19.5. P(X > 20) excludes that bar, so it starts at 20.5. The rule is not worth memorising as four cases; it is worth deriving each time from which bars are included.

The Normal Distribution