Normal distribution properties, standardising, inverse normal, normal approximation to binomial.
Answer at least 3 of 3 correctly to complete this section.
The normal distribution is the most important probability distribution in statistics. It models heights, weights, exam scores, measurement errors, and countless other real-world quantities. Its bell-shaped curve appears so frequently in nature that it is sometimes called the “bell curve.”
At A-Level, you need to be fluent with normal distribution calculations — finding probabilities, working backwards from probabilities to find values, and solving for unknown parameters. These skills underpin hypothesis testing and confidence intervals later in the course.
A continuous random variable X follows a normal distribution with mean μ and variance σ2 if its probability density function forms a symmetric bell-shaped curve centred at μ.
We write:
X∼N(μ,σ2)
Critical notation point: The second parameter is the variance σ2, not the standard deviation. So X∼N(50,16) means μ=50 and σ2=16, giving σ=4.
This is known as the 68–95–99.7 rule (or the empirical rule).
Try it yourself — adjust the mean, standard deviation, and bounds to see how the probability changes. Toggle the 68-95-99.7 bands to see the empirical rule in action.
That is close to 68% — roughly one standard deviation each side of the mean!
The standard normal distribution has mean 0 and variance 1:
Z∼N(0,1)
We use Φ(z) to denote P(Z≤z), the cumulative probability up to z. Tables and calculators give you Φ(z) values.
Given Z∼N(0,1), find P(Z<1.5).
From the standard normal table (or calculator):
P(Z<1.5)=Φ(1.5)=0.9332
This means 93.32% of a standard normal distribution lies below z=1.5.
Answer at least 3 of 3 correctly to complete this section.
To find probabilities for any normal distribution X∼N(μ,σ2), we convert to the standard normal using:
Z=σX−μ
This gives us:
P(X<a)=P(Z<σa−μ)=Φ(σa−μ)
The heights of adult women in a town are normally distributed: X∼N(165,64). Find the probability that a randomly selected woman is shorter than 170 cm.
First identify the parameters: μ=165, σ2=64, so σ=8.
Standardise:
z=8170−165=85=0.625
P(X<170)=P(Z<0.625)=Φ(0.625)=0.7340
Using the same distribution, find the probability that a randomly selected woman is taller than 175 cm.
z=8175−165=810=1.25
P(X>175)=1−P(X<175)=1−Φ(1.25)=1−0.8944=0.1056
Always sketch the curve. Draw the bell curve, mark the mean, shade the region you want. For “greater than” questions, shade the right tail — this reminds you to subtract from 1.
Find the probability that a randomly selected woman has height between 155 cm and 175 cm.
z1=8155−165=−1.25z2=8175−165=1.25
P(155<X<175)=Φ(1.25)−Φ(−1.25)
By symmetry, Φ(−1.25)=1−Φ(1.25)=1−0.8944=0.1056.
P(155<X<175)=0.8944−0.1056=0.7888
Symmetry shortcut: Φ(−z)=1−Φ(z). For symmetric intervals around the mean, P(μ−a<X<μ+a)=2Φ(σa)−1.
Answer at least 3 of 3 correctly to complete this section.
Sometimes you know the probability and need to find the value of x. This is the inverse normal problem.
Method:
The length of a component is normally distributed: X∼N(30,4). The longest 10% of components are rejected. Find the minimum length that leads to rejection.
We need a such that P(X>a)=0.10, which means P(X<a)=0.90.
Find the z-value: z=Φ−1(0.90)=1.2816.
Convert back:
a=μ+zσ=30+1.2816×2=30+2.5632=32.56 (2 d.p.)
Components longer than 32.56 units are rejected.
X∼N(μ,25) and P(X>60)=0.2. Find μ.
Since σ=5 and P(X>60)=0.2, we have P(X<60)=0.8.
Φ−1(0.8)=0.8416
Using z=σx−μ:
0.8416=560−μ
60−μ=4.208
μ=55.79 (2 d.p.)
X∼N(100,σ2). Given that P(X<115)=0.95, find σ.
Φ−1(0.95)=1.6449
1.6449=σ115−100=σ15
σ=1.644915=9.12 (2 d.p.)
Exam technique: When solving for an unknown parameter, always start by writing z=σx−μ with the known z-value substituted in. This sets up a clean equation to solve.
Answer at least 3 of 3 correctly to complete this section.
X∼N(μ,σ2). Given that P(X<20)=0.15 and P(X<50)=0.90, find μ and σ.
From the first condition: Φ−1(0.15)=−1.0364.
σ20−μ=−1.0364⋯(1)
From the second condition: Φ−1(0.90)=1.2816.
σ50−μ=1.2816⋯(2)
From equation (1): 20−μ=−1.0364σ, so μ=20+1.0364σ.
Substitute into (2):
σ50−20−1.0364σ=1.2816
σ30−1.0364=1.2816
σ30=2.318
σ=2.31830=12.94 (2 d.p.)
μ=20+1.0364×12.94=33.41 (2 d.p.)
The time taken to complete a task is modelled as T∼N(45,100) minutes. A worker is given 60 minutes. Find the probability that the worker finishes on time.
σ=100=10.
z=1060−45=1.5
P(T<60)=Φ(1.5)=0.9332
There is a 93.3% probability the worker finishes on time.
| Question type | Method |
|---|---|
| Find P(X<a) | Standardise, look up Φ(z) |
| Find P(X>a) | 1−Φ(z) |
| Find P(a<X<b) | Φ(z2)−Φ(z1) |
| Find a given P(X<a)=p | a=μ+Φ−1(p)⋅σ |
| Find μ or σ | Set up z=σx−μ and solve |
| Find both μ and σ | Two conditions → simultaneous equations |
Remember: The normal distribution is continuous, so P(X≤a)=P(X<a). There is no need to worry about boundary values as you do with discrete distributions.
Answer at least 3 of 3 correctly to complete this section.
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