🎯What you need to be able to do
- Standardise a normal variable and use the tables of \( \Phi(z) \).
- Find probabilities for ranges, including those below the mean.
- Work backwards from a probability to a value of \(x\).
- Find an unknown mean or standard deviation from given probabilities.
- Use the normal approximation to the binomial, with a continuity correction.
- State when that approximation is valid.
📚The mathematics
Standardising
If \( X \sim \mathrm{N}(\mu, \sigma^{2}) \), then
\[ Z = \frac{X - \mu}{\sigma} \sim \mathrm{N}(0, 1) \]
Note the notation: the second parameter is the variance, not the standard
deviation. \( \mathrm{N}(70, 64) \) has \( \sigma = 8 \). Reading it as \( \sigma = 64 \) wrecks
every subsequent line, and it is one of the most common slips on this paper.
The tables give \( \Phi(z) = \mathrm{P}(Z < z) \) — always the area to the
left. Everything else is built from that by symmetry.
\( \mathrm{P}(Z > z) = 1 - \Phi(z) \)
\( \Phi(-z) = 1 - \Phi(z) \)
\( \mathrm{P}(a < Z < b) = \Phi(b) - \Phi(a) \)
Sketch the curve every time. Mark the mean, mark the value, shade the region you
want. It takes ten seconds and it settles whether you need \( \Phi(z) \) or \( 1 - \Phi(z) \) —
which is the decision that actually costs marks. The normal distribution is symmetric about
\( \mu \), so any shaded area to the left of the mean must come out below 0.5.
Working backwards
Given a probability, find the \(z\) that produces it from the “critical values” table,
then reverse the standardisation:
\[ x = \mu + z\sigma \]
Worth memorising: \( \Phi(1.645) = 0.95 \), \( \Phi(1.282) = 0.90 \) and
\( \Phi(1.96) = 0.975 \). If the required probability is below 0.5 the \(z\) is
negative — look up the complement and attach the minus sign.
Unknown \( \mu \) or \( \sigma \)
Two given probabilities produce two equations in \( \mu \) and \( \sigma \). Standardise each,
read off its \(z\), and solve simultaneously:
\( \dfrac{x_1 - \mu}{\sigma} = z_1 \)
\( \dfrac{x_2 - \mu}{\sigma} = z_2 \)
Rewrite each as \( x = \mu + z\sigma \) and subtract to eliminate \( \mu \). If only one unknown
appears, one equation is enough.
Normal approximation to the binomial
When \(n\) is large, \( \mathrm{B}(n, p) \) is awkward — a hundred terms is not a
calculation. If
\[ np > 5 \quad\text{and}\quad n(1-p) > 5 \]
then \( X \) is approximately \( \mathrm{N}\!\left(np,\; np(1-p)\right) \). Both conditions must be
checked and stated; they ensure the binomial is not so skewed that a symmetric curve misrepresents
it.
The continuity correction is not optional. A discrete variable is being replaced by
a continuous one, so each whole number is stretched to an interval of width 1. Move the boundary
outwards when you want to include a value and inwards when you do
not:
\( \mathrm{P}(X \ge 45) \to \mathrm{P}(Y > 44.5) \),
\( \mathrm{P}(X > 45) \to \mathrm{P}(Y > 45.5) \),
\( \mathrm{P}(X \le 60) \to \mathrm{P}(Y < 60.5) \),
\( \mathrm{P}(X = 45) \to \mathrm{P}(44.5 < Y < 45.5) \).
Omitting it costs marks even when the answer is close; shifting it the wrong way costs more.
✏️Worked example
(a) The masses of apples are normally distributed with mean 70 g and standard deviation 8 g.
Find the probability that an apple has mass (i) more than 78 g, (ii) between 62 g and 80 g.
(b) Find the mass exceeded by the heaviest 5% of apples.
(c) In a large batch, 40% of apples are of the top grade. A random sample of 100 apples is taken.
Use a suitable approximation to find the probability that at least 45 are of the top grade.
(a) \( X \sim \mathrm{N}(70, 8^{2}) \).
(i) Standardise: \( z = \dfrac{78 - 70}{8} = 1 \). We want the area to the right, so
\[ \mathrm{P}(X > 78) = 1 - \Phi(1) = 1 - 0.8413 = 0.159 \]
(ii) The two \(z\)-values are \( \dfrac{62 - 70}{8} = -1 \) and \( \dfrac{80 - 70}{8} = 1.25 \).
Then
\[ \mathrm{P}(62 < X < 80) = \Phi(1.25) - \Phi(-1)
= 0.8944 - 0.1587 = 0.736 \]
using \( \Phi(-1) = 1 - \Phi(1) = 1 - 0.8413 \).
(b) The heaviest 5% lie above a value \(k\) with
\( \mathrm{P}(X > k) = 0.05 \), so \( \Phi(z) = 0.95 \) and \( z = 1.645 \). Then
\[ k = \mu + z\sigma = 70 + 1.645(8) = 83.2\ \text{g} \]
(c) Here \( X \sim \mathrm{B}(100, 0.4) \), and
\( np = 40 > 5 \) with \( n(1-p) = 60 > 5 \), so the normal approximation is valid:
\[ X \approx \mathrm{N}(40,\ 24), \qquad \sigma = \sqrt{24} = 4.899 \]
“At least 45” includes 45, so the boundary moves out to 44.5:
\[ \mathrm{P}(Y > 44.5),\qquad z = \frac{44.5 - 40}{4.899} = 0.919 \]
giving \( 1 - \Phi(0.919) = 1 - 0.8208 = 0.179 \).
Check it. The exact binomial value in (c) is 0.1789, so the approximation is
right to three decimal places — that is what a well-conditioned \( np \) and \( n(1-p) \)
buys you. In (a)(i), 78 g is exactly one standard deviation above the mean, and about 16% of a
normal distribution lies beyond one standard deviation, matching 0.159. In (b), 83.2 g is roughly
1.6 standard deviations above 70, which is where a 5% tail belongs.
\( \mathrm{N}(70, 8^{2}) \) means \( \sigma = 8 \), not 64. Written as
\( \mathrm{N}(70, 64) \) — which is how a question may present it — the temptation to
divide by 64 is real. Every standardisation divides by the standard deviation, so take the square
root first and write \( \sigma = 8 \) at the top of your working before touching a \(z\).
📝Practise
Work through these, then reveal the answer.
1. \( X \sim \mathrm{N}(50, 36) \). Find \( \mathrm{P}(X < 56) \) and \( \mathrm{P}(41 < X < 59) \).
Here \( \sigma = \sqrt{36} = 6 \). For the first, \( z = \dfrac{56-50}{6} = 1 \), so \( \mathrm{P}(X < 56) = \Phi(1) = 0.841 \). For the second, \( z = \pm 1.5 \), so the probability is \( \Phi(1.5) - \Phi(-1.5) = 0.9332 - 0.0668 = 0.866 \). The interval is symmetric about the mean, so its probability must exceed a half ✓.
2. \( X \sim \mathrm{N}(20, 5^{2}) \). Find the value of \(k\) such that \( \mathrm{P}(X < k) = 0.9 \).
From the critical-value table, \( \Phi(1.282) = 0.9 \), so \( z = 1.282 \). Then \( k = 20 + 1.282(5) = 26.4 \). Since the probability exceeds 0.5, \(k\) must lie above the mean ✓.
3. \( X \sim \mathrm{N}(\mu, \sigma^{2}) \) with \( \mathrm{P}(X < 40) = 0.1 \) and \( \mathrm{P}(X > 55) = 0.2 \). Find \( \mu \) and \( \sigma \).
The first gives \( \dfrac{40-\mu}{\sigma} = -1.282 \) — negative, since 0.1 is below a half. The second gives \( \Phi(z) = 0.8 \), so \( \dfrac{55-\mu}{\sigma} = 0.842 \). Writing both as \( x = \mu + z\sigma \): \( 40 = \mu - 1.282\sigma \) and \( 55 = \mu + 0.842\sigma \). Subtracting, \( 15 = 2.124\sigma \), so \( \sigma = 7.06 \) and \( \mu = 40 + 1.282(7.06) = 49.1 \).
4. State the conditions for the normal approximation to the binomial to be appropriate, and check them for \( \mathrm{B}(50, 0.06) \).
The conditions are \( np > 5 \) and \( n(1-p) > 5 \). Here \( np = 50(0.06) = 3 \), which fails the first condition, so the normal approximation should not be used — the distribution is too skewed. (A Poisson approximation would be the appropriate one here; see
S2 1.)
5. The times taken to complete a task are \( \mathrm{N}(24, 3^{2}) \) minutes. In a group of 200 people, how many would be expected to take longer than 30 minutes?
\( z = \dfrac{30-24}{3} = 2 \), so \( \mathrm{P}(X > 30) = 1 - \Phi(2) = 1 - 0.9772 = 0.0228 \). The expected number is \( 200 \times 0.0228 = 4.56 \), so about 5 people. Leave the expectation as 4.56 if the question asks for it — only round when reporting a count of people.
6. 35% of voters support a policy. In a random sample of 200, use a suitable approximation to find the probability that at most 60 support it.
\( X \sim \mathrm{B}(200, 0.35) \) with \( np = 70 > 5 \) and \( n(1-p) = 130 > 5 \), so \( X \approx \mathrm{N}(70,\ 45.5) \) and \( \sigma = 6.745 \). "At most 60" includes 60, so the boundary moves out to 60.5: \( z = \dfrac{60.5 - 70}{6.745} = -1.408 \), giving \( \Phi(-1.408) = 1 - 0.9205 = 0.0795 \). (The exact binomial value is 0.0783.)
🔗Go deeper — other people’s work
These are external resources, not mine. If one stops working, tell me and
everything above it on this page still stands.
- Seeing Theory (Brown University) — the normal distribution and the central limit theorem
- Khan Academy — \(z\)-scores and normal probability
- Desmos — overlay a binomial histogram on a normal curve and watch the fit improve with \(n\)