Welcome to Statistical Distributions
Welcome to your study notes for Statistical Distributions in CCEA AS Further Mathematics (Unit AS 2: Section C). In standard mathematics, you have encountered basic probabilities and simple data summaries. In Further Mathematics, we take this a step further by using powerful mathematical functions to model uncertainty in the real world.
Whether you are calculating how many attempts a basketball player takes before scoring their first basket, predicting how many customer queries an IT helpdesk receives in an hour, or analyzing continuous measurements like waiting times, probability distributions provide the exact blueprint. Don't worry if this seems abstract at first—we will break down each distribution step-by-step with intuitive analogies and clear methods.
1. General Discrete Random Variables
A discrete random variable (usually written as an uppercase letter like \(X\)) takes distinct, separate numerical values (like \(0, 1, 2, 3\dots\)). The function that assigns a probability to each possible value is called the Probability Mass Function (PMF), written as \(P(X = x)\).
Two Golden Rules of Any Discrete Distribution
• Non-negative probabilities: For every value \(x\), \(0 \le P(X = x) \le 1\).
• Total probability equals 1: The sum of all probabilities across all possible outcomes must equal 1: \(\sum P(X = x) = 1\).
Expectation (The Mean) and Variance (The Spread)
Expectation \(E(X)\) or \(\mu\): This is the theoretical mean (average) value of \(X\) if we repeated the experiment infinitely many times. It is calculated as the sum of each outcome multiplied by its probability:
\(E(X) = \mu = \sum x P(X = x)\)
Variance \(\operatorname{Var}(X)\) or \(\sigma^2\): This measures how spread out the values of \(X\) are around the mean. To calculate variance, we first find \(E(X^2) = \sum x^2 P(X = x)\), and then use the fundamental formula:
\(\operatorname{Var}(X) = \sigma^2 = E(X^2) - [E(X)]^2\)
Memory Aid: Calculating Variance
Always remember the phrase: "The mean of the squares minus the square of the mean."
That is: \(\operatorname{Var}(X) = E(X^2) - (\text{Mean})^2\). Watch out for the minus sign—never add them!
Worked Example: Discrete Distribution
A discrete random variable \(X\) has the probability distribution given by \(P(X = 1) = 0.2\), \(P(X = 2) = 0.5\), and \(P(X = 3) = 0.3\).
Step 1: Find the mean \(E(X)\):
\(E(X) = (1 \times 0.2) + (2 \times 0.5) + (3 \times 0.3) = 0.2 + 1.0 + 0.9 = 2.1\)
Step 2: Find \(E(X^2)\):
\(E(X^2) = (1^2 \times 0.2) + (2^2 \times 0.5) + (3^2 \times 0.3) = (1 \times 0.2) + (4 \times 0.5) + (9 \times 0.3) = 0.2 + 2.0 + 2.7 = 4.9\)
Step 3: Calculate \(\operatorname{Var}(X)\):
\(\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = 4.9 - (2.1)^2 = 4.9 - 4.41 = 0.49\)
Key Takeaway: For any discrete random variable, check that all probabilities sum to 1, find \(E(X)\) by multiplying outcomes by probabilities, and find \(\operatorname{Var}(X)\) using \(E(X^2) - [E(X)]^2\).
2. The Geometric Distribution: \(X \sim \operatorname{Geo}(p)\)
Imagine rolling a standard six-sided die until you roll a 6. How many rolls will it take? You might get it on your 1st roll, 4th roll, or 10th roll. This scenario is modelled by the Geometric Distribution.
The random variable \(X\) represents the number of trials up to and including the first success.
Modelling Assumptions / Conditions
To use the Geometric distribution \(X \sim \operatorname{Geo}(p)\), four conditions must hold:
1. The experiment consists of a sequence of repeated, independent trials.
2. Each trial has only two outcomes: Success (with probability \(p\)) or Failure (with probability \(q = 1 - p\)).
3. The probability of success \(p\) remains constant on every single trial.
4. Trials are repeated until the first success occurs.
Probability Formulae
To succeed on trial \(r\), you must fail the first \(r-1\) trials and then succeed on the \(r\text{-th}\) trial:
\(P(X = r) = (1 - p)^{r-1} p \quad \text{for } r = 1, 2, 3, \dots\)
Cumulative and Tail Probabilities:
• Probability of needing more than \(r\) trials: To need more than \(r\) trials, the first \(r\) trials must all be failures!
\(P(X > r) = (1 - p)^r\)
• Probability of needing at most \(r\) trials (within the first \(r\) trials):
\(P(X \le r) = 1 - P(X > r) = 1 - (1 - p)^r\)
Expectation and Variance
For \(X \sim \operatorname{Geo}(p)\):
\(E(X) = \frac{1}{p}\)
\(\operatorname{Var}(X) = \frac{1 - p}{p^2}\)
Worked Example: Geometric Distribution
A computer algorithm succeeds on any given run with probability \(p = 0.2\). Let \(X\) be the number of runs needed to achieve the first success.
(a) Find the probability that the first success occurs on the 4th run:
\(P(X = 4) = (1 - 0.2)^{4-1}(0.2) = (0.8)^3(0.2) = 0.512 \times 0.2 = 0.1024\)
(b) Find the probability that it takes more than 3 runs to get a success:
\(P(X > 3) = (1 - 0.2)^3 = (0.8)^3 = 0.512\)
(c) Find the expected number of runs and the variance:
\(E(X) = \frac{1}{0.2} = 5\)
\(\operatorname{Var}(X) = \frac{1 - 0.2}{(0.2)^2} = \frac{0.8}{0.04} = 20\)
Key Takeaway: The geometric distribution counts the trials until the first success. Use \(P(X > r) = (1-p)^r\) as a direct shortcut instead of summing individual terms.
3. The Poisson Distribution: \(X \sim \operatorname{Po}(\lambda)\)
While the Geometric distribution counts trials, the Poisson Distribution counts how many times an event occurs within a fixed continuous interval of time or space (for example: the number of meteor sightings per hour, or the number of typos per page in a book).
Modelling Assumptions / Conditions
A situation can be modelled by \(X \sim \operatorname{Po}(\lambda)\) if:
1. Independence: The occurrence of one event does not affect the occurrence of another.
2. Constant Rate: Events occur at a constant average rate (\(\lambda\)) per unit time or space.
3. Singly: Events occur one at a time (simultaneous occurrences are impossible).
4. Randomness: Events occur completely at random.
Probability Mass Function
If \(X \sim \operatorname{Po}(\lambda)\), where \(\lambda > 0\) is the average rate of occurrence:
\(P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \quad \text{for } x = 0, 1, 2, 3, \dots\)
The Hallmark of the Poisson Distribution
For a Poisson distribution, the theoretical mean and variance are equal:
\(E(X) = \lambda \quad \text{and} \quad \operatorname{Var}(X) = \lambda\)
Examiner Tip: If an exam question presents sample data where the sample mean and sample variance are roughly equal, this is strong evidence that a Poisson model is appropriate!
Scaling and Additivity Properties
• Scaling intervals: If events occur at an average rate of \(\lambda\) per hour, then in a period of \(k\) hours, the new parameter is \(\lambda_{\text{new}} = k\lambda\).
• Adding independent Poisson variables: If \(X \sim \operatorname{Po}(\lambda_1)\) and \(Y \sim \operatorname{Po}(\lambda_2)\) are independent, then the combined count is also Poisson:
\(X + Y \sim \operatorname{Po}(\lambda_1 + \lambda_2)\)
Worked Example: Poisson Distribution
Calls arrive at a switchboard at an average rate of \(3\) calls per minute. Let \(X\) be the number of calls in a 1-minute period.
(a) Find the probability of receiving exactly 2 calls in a 1-minute period:
Here, \(\lambda = 3\).
\(P(X = 2) = \frac{e^{-3} \times 3^2}{2!} = \frac{e^{-3} \times 9}{2} \approx 0.224\)
(b) Find the probability of receiving at least 1 call in a 2-minute period:
For 2 minutes, scale the parameter: \(\lambda_{\text{new}} = 3 \times 2 = 6\). Let \(Y \sim \operatorname{Po}(6)\).
\(P(Y \ge 1) = 1 - P(Y = 0) = 1 - \frac{e^{-6} \times 6^0}{0!} = 1 - e^{-6} \approx 1 - 0.00248 = 0.9975\)
Key Takeaway: Always check the size of the interval first and scale \(\lambda\) proportionally before calculating probabilities.
4. Continuous Random Variables
A continuous random variable \(X\) can take any real value within an interval (such as height, mass, or time). Because there are infinitely many possible values, the probability of obtaining any exact single point is zero: \(P(X = c) = 0\).
Instead of summing probabilities, we use calculus (integration) to find the probability over an interval.
Probability Density Function (PDF): \(f(x)\)
A function \(f(x)\) is a valid PDF if:
1. \(f(x) \ge 0\) for all \(x\) (the curve never drops below the \(x\)-axis).
2. The total area under the entire curve is equal to 1:
\(\int_{-\infty}^{\infty} f(x) \, dx = 1\)
To find the probability that \(X\) falls between \(a\) and \(b\), integrate the PDF:
\(P(a \le X \le b) = \int_a^b f(x) \, dx\)
Note: Because point probabilities are zero, \(P(a < X < b) = P(a \le X \le b)\).
Cumulative Distribution Function (CDF): \(F(x)\)
The CDF, denoted by \(F(x)\), gives the running total probability up to \(x\):
\(F(x) = P(X \le x) = \int_{-\infty}^x f(t) \, dt\)
To move backwards from the CDF to the PDF, simply differentiate:
\(f(x) = \frac{d}{dx}[F(x)]\)
Mean, Variance, and Percentiles
• Expectation: \(E(X) = \int_{-\infty}^{\infty} x f(x) \, dx\)
• Expected Square: \(E(X^2) = \int_{-\infty}^{\infty} x^2 f(x) \, dx\)
• Variance: \(\operatorname{Var}(X) = E(X^2) - [E(X)]^2\)
• Median (\(m\)): The value of \(m\) such that \(F(m) = 0.5\) (or \(\int_{-\infty}^m f(x) \, dx = 0.5\)).
• Quartiles: Lower quartile \(Q_1\) satisfies \(F(Q_1) = 0.25\); Upper quartile \(Q_3\) satisfies \(F(Q_3) = 0.75\).
• Mode: The value of \(x\) at which \(f(x)\) reaches its maximum value within its valid domain.
Worked Example: Continuous Variable
A continuous random variable \(X\) has the PDF given by:
\(f(x) = \frac{3}{8}x^2 \quad \text{for } 0 \le x \le 2\), and \(f(x) = 0\) otherwise.
(a) Verify that the total probability is 1:
\(\int_0^2 \frac{3}{8}x^2 \, dx = \left[ \frac{3}{8} \times \frac{x^3}{3} \right]_0^2 = \left[ \frac{x^3}{8} \right]_0^2 = \frac{8}{8} - 0 = 1\)
(b) Find \(E(X)\):
\(E(X) = \int_0^2 x \left(\frac{3}{8}x^2\right) \, dx = \int_0^2 \frac{3}{8}x^3 \, dx = \left[ \frac{3x^4}{32} \right]_0^2 = \frac{3(16)}{32} - 0 = \frac{48}{32} = 1.5\)
(c) Find the median \(m\):
Set \(F(m) = 0.5 \implies \int_0^m \frac{3}{8}x^2 \, dx = 0.5\)
\(\left[ \frac{x^3}{8} \right]_0^m = 0.5 \implies \frac{m^3}{8} = 0.5 \implies m^3 = 4 \implies m = \sqrt[3]{4} \approx 1.587\)
Key Takeaway: Integration finds probabilities, expectations, and CDFs. Differentiation recovers the PDF from the CDF.
5. Linear Coding / Transformations of Random Variables
When you transform a random variable by multiplying by a constant \(a\) and adding a constant \(b\) (creating a new variable \(Y = aX + b\)), the mean and variance change according to strict algebraic rules.
Transformation Rules
For any constants \(a\) and \(b\):
• Mean / Expectation: Affected by both scaling and shifting:
\(E(aX + b) = aE(X) + b\)
• Variance: Affected only by the square of the scaling factor (adding a constant does not change spread!):
\(\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X)\)
• Standard Deviation:
\(\sigma_{aX+b} = |a|\sigma_X\)
Why Doesn't Adding \(b\) Change the Variance?
Think of your entire class standing in a line along a measuring tape. If everyone takes 5 steps forward (\(+5\)), the average position shifts by 5, but the distances between students stay exactly the same. Therefore, the spread (variance) is completely unchanged by \(+b\)!
Worked Example: Coding
Suppose a random variable \(X\) has \(E(X) = 12\) and \(\operatorname{Var}(X) = 4\). Let \(Y = 3X - 5\).
• \(E(Y) = E(3X - 5) = 3E(X) - 5 = 3(12) - 5 = 36 - 5 = 31\)
• \(\operatorname{Var}(Y) = \operatorname{Var}(3X - 5) = 3^2 \operatorname{Var}(X) = 9 \times 4 = 36\)
• Standard deviation of \(Y = \sqrt{36} = 6\) (or \(|3| \times \sqrt{4} = 3 \times 2 = 6\)).
Key Takeaway: Never add \(b\) to variance, and always square the multiplier \(a\) when finding \(\operatorname{Var}(aX + b)\).
6. Common Exam Pitfalls & Examiner Tips
Make sure to review these common mistakes before sitting paper SFM21:
• Strict vs Non-Strict Geometric Tail Inequalities:
Remember that \(P(X > r) = (1-p)^r\). For example, \(P(X > 4) = (1-p)^4\). If a question asks for \(P(X \ge 4)\), rewrite it as \(P(X > 3) = (1-p)^3\). Do not confuse the two!
• Forgetting to Scale the Poisson Rate \(\lambda\):
Always check the time or space frame in multi-part questions. If \(\lambda = 2\) per minute, and part (b) asks about a 5-minute interval, you must use \(\lambda = 10\).
• Sign Error in the Variance Formula:
Double-check your algebra: it is \(\operatorname{Var}(X) = E(X^2) - [E(X)]^2\), with a subtraction, not an addition.
• Coding Variance Errors:
Writing \(\operatorname{Var}(3X + 4) = 3\operatorname{Var}(X) + 4\) is an instant loss of marks. The correct calculation is \(3^2 \operatorname{Var}(X) = 9\operatorname{Var}(X)\).
• Single Point Probabilities in Continuous Contexts:
Remember that for any continuous random variable, \(P(X = k) = 0\). If asked to find \(P(X = 3)\) for a continuous variable, the answer is immediately \(0\).
Quick Summary Checklist
• Discrete General: \(\sum P(X=x) = 1\), \(E(X) = \sum x P(X=x)\), \(\operatorname{Var}(X) = E(X^2) - \mu^2\).
• Geometric \(\operatorname{Geo}(p)\): Trials to 1st success, \(P(X=r)=(1-p)^{r-1}p\), \(P(X > r)=(1-p)^r\), \(E(X)=\frac{1}{p}\), \(\operatorname{Var}(X)=\frac{1-p}{p^2}\).
• Poisson \(\operatorname{Po}(\lambda)\): Independent random event counts at constant rate, \(P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}\), \(E(X)=\lambda\), \(\operatorname{Var}(X)=\lambda\).
• Continuous General: \(\int f(x)dx = 1\), \(F(x) = \int_{-\infty}^x f(t)dt\), \(f(x) = F'(x)\), \(E(X) = \int x f(x)dx\).
• Linear Transformations: \(E(aX+b) = aE(X)+b\), \(\operatorname{Var}(aX+b) = a^2\operatorname{Var}(X)\).