Grade 11 · Statistics 1 · Cambridge A-Level 9709 · Age 16–17
The normal distribution is the most important continuous probability distribution in statistics. It describes how many real-world quantities — heights, weights, exam scores, measurement errors — are naturally distributed. In Cambridge A-Level Statistics 1 (9709), you learn to use the standard normal distribution, standardise values, find probabilities from tables, work backwards using inverse normal, and approximate binomial distributions. This is essential for all further statistics work.
X~N(μ,σ²): symmetric, mean=median=mode=μ, total area = 1
68% within 1σ, 95% within 2σ, 99.7% within 3σ of mean
Z = (X−μ)/σ transforms any N(μ,σ²) to N(0,1)
P(X<a) = Φ(z); use symmetry Φ(−z) = 1−Φ(z)
Given p, find x: z = Φ²(p), then x = μ + zσ
Set up simultaneous equations from two probability conditions
B(n,p) ≈ N(np, npq) when np > 5 and nq > 5
P(X=k) becomes P(k−0.5<Y<k+0.5) for approximation
The normal distribution is a continuous probability distribution described by a symmetric bell-shaped curve. It is completely determined by two parameters: the mean μ and the variance σ². We write X~N(μ,σ²).
For any X~N(μ,σ²), the following approximate probabilities hold:
Because the normal distribution is continuous, the probability of X taking any exact value is zero: P(X = a) = 0. We can only find probabilities over intervals:
The standard normal distribution is the special case with mean 0 and variance 1: Z~N(0,1). Tables of probabilities are published for Z~N(0,1), so we convert any normal distribution to this standard form before looking up probabilities.
This transforms a value x from X~N(μ,σ²) into a z-score: how many standard deviations x is above or below the mean. A positive z means x is above the mean; a negative z means x is below the mean.
If you know the z-score and want to find the original x-value, rearrange the formula:
For Z~N(0,1), Φ(z) = P(Z < z). The normal distribution tables give values of Φ(z) for z ≥ 0. We combine these with the symmetry rule to find all other probabilities.
| Probability required | z-score(s) | Formula |
|---|---|---|
| P(X < a), z > 0 | z = (a−μ)/σ | Φ(z) |
| P(X < a), z < 0 | z = (a−μ)/σ | 1 − Φ(|z|) |
| P(X > a) | z = (a−μ)/σ | 1 − Φ(z) or Φ(−z) |
| P(a < X < b) | z⊂1;, z⊂2; | Φ(z⊂2;) − Φ(z⊂1;) |
In inverse normal problems, you are given a probability and asked to find the value of x. This is the reverse of finding P(X < a).
When X~B(n,p) and n is large, the binomial becomes hard to compute directly. If n is large enough, X is approximately normal:
The binomial is discrete but the normal is continuous. We must apply a continuity correction of 0.5 to account for this gap.
X~N(45, 100). Find P(X < 58).
X~N(30, 64). Find P(X > 22).
X~N(200, 400). Find P(190 < X < 216).
X~N(70, 25). Find x such that P(X < x) = 0.8944.
X~N(μ, 36). P(X < 80) = 0.9772. Find μ.
X~N(50, σ²). P(X > 65) = 0.0668. Find σ.
X~N(μ, σ²). P(X < 20) = 0.1151 and P(X < 40) = 0.8849. Find μ and σ.
X~B(120, 0.35). Using normal approximation, find P(X ≥ 50).
The formula Z = (X−μ)/σ uses σ, not σ². Always take the square root of the variance first.
For example, Φ(−1.5) = 1 − Φ(1.5) = 1 − 0.9332 = 0.0668. The symmetry rule means the left tail equals the right tail.
Without continuity correction, the approximation is less accurate and you lose accuracy marks.
Mnemonic: If the inequality includes the endpoint (le/ge), move 0.5 towards it. If it excludes (lt/gt), move 0.5 away from it.
Cambridge uses N(μ, σ²) notation. The second parameter is always the variance.
Standardise both boundaries, then subtract the smaller Φ value from the larger.
Examiners award a method mark for checking conditions. Always verify both np > 5 and n(1−p) > 5.
Whenever the given probability is less than 0.5, the z-value is negative and x is below the mean.
| Formula | Description | When to use |
|---|---|---|
| Z = (X−μ)/σ | Standardise X~N(μ,σ²) to Z~N(0,1) | Before every probability calculation |
| X = μ + Zσ | Back-transform z-score to original units | Inverse normal problems |
| P(X<a) = Φ(z) | CDF of standard normal | z > 0; look up in tables |
| P(X>a) = 1−Φ(z) | Upper tail probability | Right-hand probabilities |
| Φ(−z) = 1−Φ(z) | Symmetry of N(0,1) | Negative z-scores |
| P(a<X<b) = Φ(z⊂2;)−Φ(z⊂1;) | Interval probability | Both boundaries given |
| Item | Formula / Rule |
|---|---|
| Conditions | np > 5 AND nq > 5 (where q = 1−p) |
| Approximating distribution | Y~N(np, npq) |
| Mean | μ = np |
| Variance | σ² = npq = np(1−p) |
| Binomial (discrete) | Normal approximation (continuous) |
|---|---|
| P(X = k) | P(k−0.5 < Y < k+0.5) |
| P(X ≤ k) | P(Y < k+0.5) |
| P(X < k) | P(Y < k−0.5) |
| P(X ≥ k) | P(Y > k−0.5) |
| P(X > k) | P(Y > k+0.5) |
| P(a ≤ X ≤ b) | P(a−0.5 < Y < b+0.5) |
| z | Φ(z) | 1−Φ(z) |
|---|---|---|
| 1.00 | 0.8413 | 0.1587 |
| 1.28 | 0.8997 | 0.1003 |
| 1.645 | 0.9500 | 0.0500 |
| 1.96 | 0.9750 | 0.0250 |
| 2.00 | 0.9772 | 0.0228 |
| 2.326 | 0.9900 | 0.0100 |
| 2.576 | 0.9950 | 0.0050 |
| 3.00 | 0.9987 | 0.0013 |
For Z~N(0,1), we want to show P(−1 < Z < 1) ≈ 0.6827.
By definition, P(−1 < Z < 1) = Φ(1) − Φ(−1) = Φ(1) − (1−Φ(1)) = 2Φ(1) − 1.
From standard normal tables: Φ(1.00) = 0.8413.
Therefore P(−1 < Z < 1) = 2(0.8413) − 1 = 1.6826 − 1 = 0.6826 ≈ 68.27%
Similarly: P(−2 < Z < 2) = 2Φ(2)−1 = 2(0.9772)−1 = 0.9544 ≈ 95.44%
P(−3 < Z < 3) = 2Φ(3)−1 = 2(0.9987)−1 = 0.9974 ≈ 99.74%
For a general X~N(μ,σ²), we standardise: P(μ−σ < X < μ+σ) = P(−1 < Z < 1) ≈ 0.6827, and similarly for 2σ and 3σ bounds. These results hold for any normal distribution regardless of μ and σ.
The core problem: A discrete distribution (binomial) assigns positive probability to integer values only. A continuous distribution (normal) assigns zero probability to any single point. We must bridge this gap.
Geometric justification: Imagine a bar chart of the binomial distribution. Each integer k is represented by a bar from k−0.5 to k+0.5 with height P(X=k). The area of bar k equals P(X=k).
When we approximate with a continuous normal curve, P(X ≤ k) for the binomial is the total area of bars up to and including bar k — but bar k extends from k−0.5 to k+0.5. So the continuous analogue should include up to k+0.5:
P(X ≤ k) ≈ P(Y ≤ k + 0.5)
Without this correction, we would be cutting bar k in half, systematically underestimating the probability by approximately P(X=k)/2.
How accurate is the approximation? The error in the normal approximation (after continuity correction) is of order 1/√n, which is small when n is large. This is why we require np > 5 and nq > 5 to ensure n is large enough for the error to be acceptably small.
Formal statement (Berry–Esseen theorem, simplified): If X~B(n,p) and Y~N(np, npq), then for all x:
|P(X ≤ k) − P(Y ≤ k+0.5)| ≤ C/(n√(pq))
for a constant C, so the error shrinks as n grows.
The standard normal distribution has probability density function f(z) = (1/√(2π)) e^(−z²/2), which is symmetric about z = 0, i.e. f(−z) = f(z).
By definition, Φ(z) = P(Z ≤ z) = area under curve to the left of z.
By symmetry: area to left of −z = area to right of z (mirror image).
Area to right of z = 1 − Φ(z).
Therefore: Φ(−z) = 1 − Φ(z). □
Enter the distribution parameters and boundaries to compute and visualise a probability. Uses JavaScript approximation of Φ(z).
Self-marking. Enter numeric answers where asked; enter 1 for Yes and 0 for No for yes/no questions.
Self-marking. Give z-scores to 2 decimal places where needed.
Self-marking. Give answers to 4 decimal places. Use standard normal table values.
Self-marking. Give answers to 2 decimal places.
Self-marking. Always apply continuity correction. Give answers to 4 decimal places.
Mixed normal distribution questions. Confetti on 100%!
Harder and multi-step problems requiring deeper understanding.
Cambridge-style structured questions. Show full working. Marks shown in brackets.
The heights of students in a school are modelled by X~N(165, 81).
(a) Find P(X < 174). [2]
(b) Find P(153 < X < 174). [3]
X~N(μ, 49). Given that P(X > 85) = 0.0401, find μ.
X~N(40, σ²). Given that P(X < 33) = 0.1587, find σ and hence find P(X > 46).
X~N(μ, σ²). Given P(X < 28) = 0.2266 and P(X < 42) = 0.9332. Find μ and σ.
The mass of apples is normally distributed with mean 140 g and standard deviation 18 g.
(a) Find the probability that a randomly chosen apple has mass greater than 158 g. [2]
(b) Find the mass m such that 90% of apples have mass less than m. [3]
X~B(200, 0.45). Using a suitable approximation, find:
(a) P(X ≤ 80), (b) P(85 ≤ X ≤ 95). [3, 3]
X~N(50, σ²) and P(45 < X < 55) = 0.6826. Find σ, and hence P(X > 58).
The lifetimes of batteries are normally distributed. 10% last less than 8 hours and 5% last more than 20 hours. Find the mean and standard deviation of battery lifetime.
Representative Cambridge 9709 S1 past paper questions on Normal Distribution. Attempt fully before revealing.
The random variable X~N(μ, σ²). It is given that P(X > 23.5) = 0.2 and P(X < 14.5) = 0.1.
(a) Find μ and σ. [5]
(b) Find P(17 < X < 22). [2]
The masses of oranges are normally distributed with mean 190 g and standard deviation 15 g. An orange is selected at random.
(a) Find the probability the orange has mass between 175 g and 208 g. [3]
(b) Find the value of m such that P(X > m) = 0.25. [2]
X~B(180, 0.3). Using a normal approximation, find P(X ≤ 60) and P(50 < X < 65).
The times taken by students to complete a test follow X~N(45, σ²). Given that 20% of students take more than 55 minutes, find σ. Hence find the probability a student completes the test in less than 35 minutes.
The lengths of steel rods produced are normally distributed with mean μ and standard deviation 2.5 cm. A rod is acceptable if its length is between 95 cm and 103 cm.
(a) Given μ = 99, find the probability that a randomly chosen rod is acceptable. [3]
(b) The manufacturer wants at least 98% of rods to be acceptable. If σ = 2.5, find the range of values of μ. [5]