Statistical distributions

Q: What is the probability of a continuous random variable taking an exact single value, and how does this affect inequality signs?

\(\text{P}(X = c) = 0\), which means strict and non-strict inequalities are identical: \(\text{P}(X \le c) = \text{P}(X < c)\).

Q: In the standard notation \(X \sim \text{N}(\mu, \sigma^2)\), what do the two parameters represent?

\(\mu\) represents the population mean, and \(\sigma^2\) represents the population variance.

Q: What are the key symmetry and shape properties of a normal distribution curve?

It is a bell-shaped curve symmetrical about \(x = \mu\) with total area \(1\), \(\text{Mean} = \text{Median} = \text{Mode} = \mu\), and points of inflection at \(x = \mu \pm \sigma\).

Q: How do you standardise a normal random variable \(X\) into a standard normal variable \(Z\)?

Use the formula \(Z = \frac{X - \mu}{\sigma}\), where \(Z \sim \text{N}(0, 1)\).

Q: How do you determine the algebraic sign of a \(z\)-score when performing an inverse normal calculation?

If the cumulative probability to the left is greater than \(0.5\), \(z > 0\); if it is less than \(0.5\), \(z < 0\).

Q: How do you set up simultaneous equations to find unknown parameters \(\mu\) and \(\sigma\)?

Find the \(z\)-scores for both probabilities and form two equations of the type \(x = \mu + z\sigma\), then solve by elimination.

Q: What is the distribution of the sample mean \(\bar{X}\) drawn from a normal population \(X \sim \text{N}(\mu, \sigma^2)\)?

\(\bar{X} \sim \text{N}\left(\mu, \frac{\sigma^2}{n}\right)\)

Q: What is the standard error of the sample mean, and what formula standardises \(\bar{X}\)?

The standard error is \(\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\), and the standardised score is \(Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}\).

Mathematics 2210 · CCEA A Level · 7 min read · Updated: 31 Aug 2026

Statistical Distributions: The Normal Distribution (CCEA A2 Mathematics)

Welcome to Statistical Distributions for CCEA A2 Unit 2 (Applied Mathematics)! In AS Mathematics, you explored discrete variables like the Binomial distribution, where outcomes are distinct counts (such as counting the number of heads in coin tosses). In A2, we take a leap into the continuous world with the most famous distribution in all of mathematics: the Normal Distribution.

Whether you are predicting human heights, manufacturing tolerances, test scores, or scientific measurement errors, the normal distribution is the foundational tool. Don't worry if statistics has felt daunting in the past—we will break down every formula, calculator trick, and examiner trap step by step!

---

1. Continuous Random Variables & The Normal Curve

Continuous vs Discrete Variables

A continuous random variable \(X\) can take any real value within a given interval (for example, time, weight, or distance). Because there are infinitely many possible values, the probability of an exact single value is always zero:

\(\text{P}(X = c) = 0\)

Because of this, strict inequalities and non-strict inequalities mean the exact same thing in continuous distributions:
\(\text{P}(X \le c) = \text{P}(X < c)\)
\(\text{P}(X \ge c) = \text{P}(X > c)\)
Key Takeaway: You never need to apply discrete continuity corrections or worry about whether the boundary is included when working directly with a normal variable.

Notation

A normally distributed variable with population mean \(\mu\) and population variance \(\sigma^2\) is written as:

\(X \sim \text{N}(\mu, \sigma^2)\)

Examiner Alert: The second number written in the notation is the variance (\(\sigma^2\)), NOT the standard deviation (\(\sigma\)). If you are given \(X \sim \text{N}(50, 16)\), the mean is \(\mu = 50\), but the standard deviation is \(\sigma = \sqrt{16} = 4\).

Key Properties of the Normal Curve

The probability density function forms the iconic symmetrical "bell curve":

• Total Area: The total area under the probability density curve is exactly \(1\).
• Symmetry: The curve is perfectly symmetrical about the vertical line \(x = \mu\).
• Central Measures: \(\text{Mean} = \text{Median} = \text{Mode} = \mu\).
• Points of Inflection: The curve changes concavity at exactly \(x = \mu - \sigma\) and \(x = \mu + \sigma\).
• The Empirical Rule (Rough Percentages):
— Approximately \(68.27\%\) of the data lies within \(1\) standard deviation of the mean: \(\mu \pm 1\sigma\)
— Approximately \(95.45\%\) lies within \(2\) standard deviations: \(\mu \pm 2\sigma\) (exactly \(95\%\) lies within \(\mu \pm 1.96\sigma\))
— Approximately \(99.73\%\) lies within \(3\) standard deviations: \(\mu \pm 3\sigma\)

---

2. The Standard Normal Distribution (\(Z\)) & Standardisation

What is the Standard Normal Distribution?

Every normal distribution has its own unique mean and spread. To compare them or use statistical tables, we convert any normal variable \(X\) into the standard normal variable \(Z\), which has a mean of \(0\) and a variance of \(1\):

\(Z \sim \text{N}(0, 1)\)

The Standardisation Formula

To convert any value \(x\) into a standard \(z\)-score (which tells you how many standard deviations \(x\) is above or below the mean), use:

\(Z = \frac{X - \mu}{\sigma}\)

Cumulative Distribution Function: \(\Phi(z)\)

The notation \(\Phi(z)\) represents the cumulative probability up to \(z\):

\(\Phi(z) = \text{P}(Z \le z)\)

Because the standard normal curve is symmetrical about \(z = 0\), we have these essential symmetry relationships:

• Upper tail: \(\text{P}(Z > z) = 1 - \Phi(z)\)
• Lower tail (negative \(z\)): \(\Phi(-z) = \text{P}(Z \le -z) = 1 - \Phi(z)\)
• Between two values: \(\text{P}(a \le X \le b) = \Phi\left(\frac{b - \mu}{\sigma}\right) - \Phi\left(\frac{a - \mu}{\sigma}\right)\)

Calculator vs Formula Booklet Tables

In the CCEA examination, you are expected to use an advanced scientific calculator (such as the Casio fx-991EX ClassWiz or fx-991CW):
• Use Normal CD (Cumulative Distribution) to find probabilities \(\text{P}(a \le X \le b)\) by entering lower limit, upper limit, \(\sigma\), and \(\mu\).
• For tail probabilities on a calculator, use a large extreme value like \(-9999\) for lower or \(9999\) for upper.
• You may also use the \(\Phi(z)\) and percentage points tables in the CCEA Mathematics Formulae and Statistical Tables booklet.

---

3. Inverse Normal Calculations & Finding Unknown Parameters

Working Backwards: The Inverse Normal

When you are given a probability \(p\) and need to find the boundary value \(x\), you use the Inverse Normal function on your calculator (or the percentage points table):

\(\text{P}(X \le x) = p \implies x = \mu + z\sigma\)

Crucial Sign Rule:
• If the area to the left is greater than 0.5, \(z\) is positive (\(x > \mu\)).
• If the area to the left is less than 0.5, \(z\) is negative (\(x < \mu\)).

Finding Two Unknown Parameters (\(\mu\) and \(\sigma\))

Exam questions frequently provide two probability statements and ask you to find both \(\mu\) and \(\sigma\). Follow this 4-step foolproof method:

Step 1: Write both probability statements in standard cumulative form: \(\text{P}(X < x_1) = p_1\) and \(\text{P}(X < x_2) = p_2\).
Step 2: Find the corresponding \(z\)-scores (\(z_1\) and \(z_2\)) using Inverse Normal (keep at least 4 decimal places to prevent rounding error).
Step 3: Set up two simultaneous linear equations using \(x = \mu + z\sigma\):
\(x_1 = \mu + z_1\sigma\)
\(x_2 = \mu + z_2\sigma\)
Step 4: Subtract the equations to eliminate \(\mu\), solve for \(\sigma\), and substitute back to find \(\mu\).

Worked Example: Simultaneous Equations

The masses of apples, \(X\) grams, follow \(X \sim \text{N}(\mu, \sigma^2)\). Given that \(10\%\) of apples weigh less than \(120\text{ g}\) and \(5\%\) weigh more than \(190\text{ g}\), find \(\mu\) and \(\sigma\).

Solution:
1. Lower tail: \(\text{P}(X < 120) = 0.10 \implies z_1 = -1.2816\)
2. Upper tail: \(\text{P}(X > 190) = 0.05 \implies \text{P}(X < 190) = 0.95 \implies z_2 = +1.6449\)
3. Form equations:
\(120 = \mu - 1.2816\sigma\) (Equation 1)
\(190 = \mu + 1.6449\sigma\) (Equation 2)
4. Subtract Equation 1 from Equation 2:
\(190 - 120 = (1.6449 - (-1.2816))\sigma\)
\(70 = 2.9265\sigma \implies \sigma = \frac{70}{2.9265} \approx 23.9\text{ g}\)
5. Substitute \(\sigma\) back into Equation 2:
\(\mu = 190 - 1.6449(23.919) \approx 150.7\text{ g}\)

---

4. Distribution of the Sample Mean (\(\bar{X}\))

The Sampling Distribution

Suppose we take an independent random sample of size \(n\) from a normal parent population \(X \sim \text{N}(\mu, \sigma^2)\), and calculate the sample mean \(\bar{X}\):

\(\bar{X} = \frac{X_1 + X_2 + \dots + X_n}{n}\)

The sample mean \(\bar{X}\) is also normally distributed, with the same mean \(\mu\), but with a reduced variance:

\(\bar{X} \sim \text{N}\left(\mu, \frac{\sigma^2}{n}\right)\)

Standard Error of the Mean

The standard deviation of the sample mean is known as the standard error:

\(\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\)

Why does this make sense? A single individual can easily be very tall or very short, but the average height of a sample of \(100\) people will almost always be very close to the true population mean. As sample size \(n\) increases, the spread of \(\bar{X}\) shrinks!

Standardising the Sample Mean

To calculate probabilities regarding a sample mean \(\bar{X}\), use the standardised test statistic:

\(Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{(\bar{X} - \mu)\sqrt{n}}{\sigma}\)

Worked Example: Individual vs Sample Mean

Cereal boxes are filled with a mean mass \(\mu = 500\text{ g}\) and standard deviation \(\sigma = 15\text{ g}\). A random sample of \(9\) boxes is selected. Find the probability that the sample mean mass is less than \(490\text{ g}\).

Solution:
1. Identify the distribution of \(\bar{X}\):
\(n = 9\), so \(\bar{X} \sim \text{N}\left(500, \frac{15^2}{9}\right) \implies \bar{X} \sim \text{N}\left(500, \frac{225}{9}\right) \implies \bar{X} \sim \text{N}(500, 25)\)
The standard error is \(\sigma_{\bar{X}} = \frac{15}{\sqrt{9}} = \frac{15}{3} = 5\).
2. Standardise \(\bar{X} = 490\):
\(Z = \frac{490 - 500}{5} = \frac{-10}{5} = -2\)
3. Compute the probability:
\(\text{P}(\bar{X} < 490) = \text{P}(Z < -2) = 1 - \Phi(2) = 1 - 0.9772 = 0.0228\) (or \(2.28\%\)).

---

5. Top Examiner Pitfalls & Revision Checklist

Avoid These 5 Common Mistakes in CCEA Exams:

1. Squaring instead of Square-Rooting:
When a question states \(X \sim \text{N}(100, 25)\), remember that \(\sigma = 5\), NOT \(25\). Never put \(\sigma = 25\) into your standardisation formula!

2. Forgetting the Negative Sign on Lower Tails:
If \(\text{P}(X < x) = 0.05\), the \(z\)-score MUST be negative (\(z = -1.6449\)). If you forget the minus sign, your equations will produce an impossible negative standard deviation (\(\sigma < 0\)).

3. Forgetting \(\sqrt{n}\) in Sample Mean Questions:
Always check: is the question asking about one item (\(X\)) or a sample mean of \(n\) items (\(\bar{X}\))? If it mentions a sample of size \(n\), divide \(\sigma\) by \(\sqrt{n}\).

4. Premature Rounding of \(z\)-values:
Do not round \(z\) to \(2\) decimal places (like \(1.64\)) during intermediate steps of simultaneous equations. Use at least \(4\) decimal places (\(1.6449\)) or store the exact value in your calculator memory.

5. Discrete Confusion:
Do not subtract or add \(0.5\) (continuity corrections) for standard continuous normal problems. Remember: \(\text{P}(X \le 10) = \text{P}(X < 10)\).

---

Summary Quick-Sheet

• Single Variable: \(X \sim \text{N}(\mu, \sigma^2) \implies Z = \frac{X - \mu}{\sigma}\)
• Sample Mean: \(\bar{X} \sim \text{N}\left(\mu, \frac{\sigma^2}{n}\right) \implies Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}\)
• Symmetry: \(\Phi(-z) = 1 - \Phi(z)\)
• Linear Equation: \(x = \mu + z\sigma\)

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the probability of a continuous random variable taking an exact single value, and how does this affect inequality signs?

\(\text{P}(X = c) = 0\), which means strict and non-strict inequalities are identical: \(\text{P}(X \le c) = \text{P}(X < c)\).

In the standard notation \(X \sim \text{N}(\mu, \sigma^2)\), what do the two parameters represent?

\(\mu\) represents the population mean, and \(\sigma^2\) represents the population variance.

What are the key symmetry and shape properties of a normal distribution curve?

It is a bell-shaped curve symmetrical about \(x = \mu\) with total area \(1\), \(\text{Mean} = \text{Median} = \text{Mode} = \mu\), and points of inflection at \(x = \mu \pm \sigma\).

How do you standardise a normal random variable \(X\) into a standard normal variable \(Z\)?

Use the formula \(Z = \frac{X - \mu}{\sigma}\), where \(Z \sim \text{N}(0, 1)\).

How can you express lower tail probabilities for negative values of \(z\) using the cumulative distribution function \(\Phi(z)\)?

\(\Phi(-z) = \text{P}(Z \le -z) = 1 - \Phi(z)\)

How do you determine the algebraic sign of a \(z\)-score when performing an inverse normal calculation?

If the cumulative probability to the left is greater than \(0.5\), \(z > 0\); if it is less than \(0.5\), \(z < 0\).

How do you set up simultaneous equations to find unknown parameters \(\mu\) and \(\sigma\)?

Find the \(z\)-scores for both probabilities and form two equations of the type \(x = \mu + z\sigma\), then solve by elimination.

What is the distribution of the sample mean \(\bar{X}\) drawn from a normal population \(X \sim \text{N}(\mu, \sigma^2)\)?

\(\bar{X} \sim \text{N}\left(\mu, \frac{\sigma^2}{n}\right)\)

What is the standard error of the sample mean, and what formula standardises \(\bar{X}\)?

The standard error is \(\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\), and the standardised score is \(Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}\).

Ready to test yourself?

Turn these notes into exam-style practice. Get unlimited AI questions on this topic with instant marking and explanations.

Practise This Topic

More Mathematics 2210 chapters

* The content provided by thinka is generated by AI and may not always be accurate or up-to-date. Please use it as a supplementary resource and verify with official materials.

Put These Notes into Practice

Reading the notes is a great start. Now practise with unlimited AI-generated questions and get instant feedback. 100,000+ students are already improving their grades.

Start Practising Now View Pricing

Done reading? Test yourself with AI practice questions

Practise This Topic Now