Welcome to Statistical Distributions!
Have you ever wondered how insurance companies decide on premiums, or how video game designers calculate the drop rate of a rare item? The answer lies in Statistical Distributions. In this chapter of AS 2: Applied Mathematics, you will learn how to describe chance using mathematical models. Don't worry if probability felt tricky in the past — we are going to break everything down step-by-step with clear real-world examples, memory tricks, and easy-to-follow rules.
1. Discrete Random Variables (DRVs)
Before diving into fancy formulas, let's understand what a Discrete Random Variable is:
• Variable: A quantity that can take different values.
• Random: The value it takes is determined by chance.
• Discrete: It can only take distinct, separate values (usually whole numbers, like the number of goals scored or heads flipped), with no values in between.
We usually denote a random variable using a capital letter like \(X\), and the specific value it takes with a lowercase letter like \(x\). So, \(\text{P}(X = x)\) means "the probability that the random variable \(X\) equals the specific number \(x\)".
The Two Golden Rules of Probability Distributions
For any valid discrete random variable \(X\), two crucial rules must always hold:
1. Every individual probability must be between 0 and 1 inclusive: \(0 \le \text{P}(X = x) \le 1\)
2. The sum of all probabilities must equal 1: \(\sum \text{P}(X = x) = 1\)
Worked Example: Finding an Unknown Constant
A discrete random variable \(X\) has the following probability distribution:
\(\text{P}(X = 1) = k\)
\(\text{P}(X = 2) = 2k\)
\(\text{P}(X = 3) = 0.4\)
\(\text{P}(X = 4) = 0.1\)
Find the value of the constant \(k\).
Step-by-step solution:
1. Use the rule that all probabilities add up to 1:
\(k + 2k + 0.4 + 0.1 = 1\)
2. Simplify the equation:
\(3k + 0.5 = 1\)
3. Solve for \(k\):
\(3k = 0.5 \implies k = \frac{0.5}{3} = \frac{1}{6}\)
Quick Review & Key Takeaway: Whenever you see an unknown constant like \(k\) or \(c\) in a probability distribution table, always add up all the probabilities and set the total equal to \(1\).
2. Expected Value \(\text{E}(X)\) and Variance \(\text{Var}(X)\)
Just like standard data sets have a mean and a spread, a probability distribution has an expected value (the theoretical mean) and a variance (a measure of spread).
Expected Value: \(\text{E}(X)\) or \(\mu\)
The expected value represents the long-run average value if you repeated the random experiment millions of times. It is defined as:
\(\text{E}(X) = \mu = \sum x \cdot \text{P}(X = x)\)
Analogy: Think of \(\text{E}(X)\) as a seesaw balance point. You multiply each value \(x\) by its "weight" (its probability), then add them all together.
Variance: \(\text{Var}(X)\) or \(\sigma^2\)
Variance measures how spread out the possible values are around the mean. The working formula you must know is:
\(\text{Var}(X) = \sigma^2 = \text{E}(X^2) - [\text{E}(X)]^2\)
Where \(\text{E}(X^2) = \sum x^2 \cdot \text{P}(X = x)\).
The standard deviation \(\sigma\) is simply the square root of the variance: \(\sigma = \sqrt{\text{Var}(X)}\).
Memory Aid for Variance
Remember this simple phrase: "The mean of the squares minus the square of the mean."
That is: \(\text{Var}(X) = \text{E}(X^2) - (\text{E}(X))^2\).
Worked Example: Calculating \(\text{E}(X)\) and \(\text{Var}(X)\)
Let \(X\) be the score on a biased 4-sided spinner with probabilities:
\(\text{P}(X = 1) = 0.1\)
\(\text{P}(X = 2) = 0.3\)
\(\text{P}(X = 3) = 0.4\)
\(\text{P}(X = 4) = 0.2\)
Step 1: Calculate \(\text{E}(X)\)
\(\text{E}(X) = (1)(0.1) + (2)(0.3) + (3)(0.4) + (4)(0.2)\)
\(\text{E}(X) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7\)
Step 2: Calculate \(\text{E}(X^2)\)
Square each \(x\) value first, then multiply by the probability:
\(\text{E}(X^2) = (1^2)(0.1) + (2^2)(0.3) + (3^2)(0.4) + (4^2)(0.2)\)
\(\text{E}(X^2) = (1)(0.1) + (4)(0.3) + (9)(0.4) + (16)(0.2)\)
\(\text{E}(X^2) = 0.1 + 1.2 + 3.6 + 3.2 = 8.1\)
Step 3: Calculate \(\text{Var}(X)\)
\(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\)
\(\text{Var}(X) = 8.1 - (2.7)^2 = 8.1 - 7.29 = 0.81\)
Standard deviation: \(\sigma = \sqrt{0.81} = 0.9\)
Common Mistake to Avoid: Never confuse \(\text{E}(X^2)\) with \([\text{E}(X)]^2\). Calculate \(\text{E}(X^2)\) by squaring the \(x\) values first. Then subtract the square of your earlier answer for \(\text{E}(X)\).
Key Takeaway: \(\text{E}(X)\) gives the central balance point, while \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2\) measures the dispersion of the distribution.
3. The Binomial Distribution
The Binomial Distribution is one of the most famous and useful probability distributions. It models experiments where you repeat a trial multiple times and count how many "successes" you get.
When can you use a Binomial Model? (The BINS Criteria)
You can model a random variable \(X\) using a binomial distribution if and only if all four conditions are met. Use the mnemonic BINS to remember them:
• B – Binary outcomes: Each trial has only two possible outcomes, conventionally called success and failure.
• I – Independent trials: The outcome of any one trial does not affect the outcome of any other trial.
• N – Number of trials fixed: There is a fixed, predetermined number of trials, denoted by \(n\).
• S – Same probability of success: The probability of success, denoted by \(p\), is constant for every trial.
We write this in statistical shorthand as: \(X \sim \text{B}(n, p)\)
The Binomial Probability Formula
If \(X \sim \text{B}(n, p)\), the probability of getting exactly \(r\) successes out of \(n\) trials is:
\(\text{P}(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\)
Where:
• \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\) is the number of ways to choose \(r\) successes from \(n\) trials (use the \(^n\text{C}_r\) button on your calculator).
• \(p^r\) is the probability of getting \(r\) successes.
• \((1-p)^{n-r}\) is the probability of getting the remaining \((n-r)\) failures. Sometimes \((1-p)\) is written as \(q\).
Mean and Variance of a Binomial Distribution
For a binomial distribution \(X \sim \text{B}(n, p)\), you don't need to build a massive table to find the mean and variance. You can use these simple shortcuts:
• Mean: \(\text{E}(X) = \mu = np\)
• Variance: \(\text{Var}(X) = \sigma^2 = np(1-p) = npq\)
Example: If you flip a fair coin 100 times (\(n = 100, p = 0.5\)), the expected number of heads is \(\text{E}(X) = 100 \times 0.5 = 50\), and the variance is \(\text{Var}(X) = 100 \times 0.5 \times 0.5 = 25\).
Worked Example: Exact Binomial Probability
A fair 6-sided die is rolled 8 times. What is the probability of rolling exactly 3 sixes?
Step-by-step solution:
1. Check conditions: Fixed trials \(n = 8\), constant probability of success \(p = \frac{1}{6}\), independent rolls, only two outcomes (a six or not a six).
2. State the model: \(X \sim \text{B}\left(8, \frac{1}{6}\right)\), where \(X\) is the number of sixes.
3. Apply the formula for \(r = 3\):
\(\text{P}(X = 3) = \binom{8}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^{8 - 3}\)
4. Calculate values:
\(\binom{8}{3} = 56\)
\(\text{P}(X = 3) = 56 \times \left(\frac{1}{6}\right)^3 \times \left(\frac{5}{6}\right)^5 \approx 56 \times 0.004630 \times 0.40188 \approx 0.104\) (to 3 significant figures).
Key Takeaway: Check BINS before assuming a distribution is Binomial. Use \(\text{P}(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\) for individual probabilities.
4. Cumulative Probabilities and Inequalities
In exam questions, you are rarely asked only for an exact value like \(\text{P}(X = 3)\). More often, you will need to find the probability of a range, such as "at least 4" or "fewer than 7".
Because \(X\) only takes whole number values (\(0, 1, 2, 3, \dots, n\)), translating the words into correct mathematical inequalities is essential.
Translating Words to Inequalities
Here is your translation guide for a discrete variable:
• "At most \(k\)" / "No more than \(k\)": \(\text{P}(X \le k)\)
• "Fewer than \(k\)" / "Less than \(k\)": \(\text{P}(X < k) = \text{P}(X \le k - 1)\)
• "At least \(k\)" / "No fewer than \(k\)": \(\text{P}(X \ge k) = 1 - \text{P}(X \le k - 1)\)
• "More than \(k\)" / "Greater than \(k\)": \(\text{P}(X > k) = 1 - \text{P}(X \le k)\)
• "Between \(a\) and \(b\) inclusive": \(\text{P}(a \le X \le b) = \text{P}(X \le b) - \text{P}(X \le a - 1)\)
The Number Line Visual Trick
When in doubt, write down a small list of numbers and circle what you want!
Suppose \(n = 6\) and you want "at least 4 successes" (\(X \ge 4\)):
Possible outcomes: \(0, 1, 2, 3, \mathbf{4, 5, 6}\)
You want the outcomes \(\{4, 5, 6\}\).
The total probability of everything is \(1\). The outcomes you do not want are \(\{0, 1, 2, 3\}\), which is \(\text{P}(X \le 3)\).
Therefore: \(\text{P}(X \ge 4) = 1 - \text{P}(X \le 3)\).
Worked Example: Handling Inequalities
A manufacturing process produces light bulbs, of which \(10\%\) are defective. A quality inspector selects a random sample of 20 bulbs.
Let \(X\) be the number of defective bulbs: \(X \sim \text{B}(20, 0.1)\).
Question (a): Find the probability that there are at most 2 defective bulbs.
"At most 2" means \(\text{P}(X \le 2)\):
\(\text{P}(X \le 2) = \text{P}(X = 0) + \text{P}(X = 1) + \text{P}(X = 2)\)
Using calculator cumulative binomial function (or summing the terms):
\(\text{P}(X = 0) = \binom{20}{0} (0.1)^0 (0.9)^{20} \approx 0.1216\)
\(\text{P}(X = 1) = \binom{20}{1} (0.1)^1 (0.9)^{19} \approx 0.2702\)
\(\text{P}(X = 2) = \binom{20}{2} (0.1)^2 (0.9)^{18} \approx 0.2852\)
\(\text{P}(X \le 2) \approx 0.1216 + 0.2702 + 0.2852 = 0.677\) (to 3 significant figures).
Question (b): Find the probability that there are at least 3 defective bulbs.
"At least 3" means \(\text{P}(X \ge 3)\):
\(\text{P}(X \ge 3) = 1 - \text{P}(X \le 2)\)
Using our answer from part (a):
\(\text{P}(X \ge 3) = 1 - 0.677 = 0.323\) (to 3 significant figures).
Key Takeaway: Cumulative tables and standard calculator functions give \(\text{P}(X \le k)\). Always rewrite \(\text{P}(X \ge k)\) as \(1 - \text{P}(X \le k - 1)\) before calculating.
5. Quick Summary & Exam Checklist
Before sitting your AS 2 Applied Mathematics exam, make sure you can answer YES to each of these:
• Can you find unknown constants by using \(\sum \text{P}(X = x) = 1\)?
• Can you compute \(\text{E}(X) = \sum x \text{P}(X = x)\) and \(\text{Var}(X) = \text{E}(X^2) - (\text{E}(X))^2\)?
• Can you state and verify the BINS criteria for a binomial model?
• Can you use the formula \(\text{P}(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\) accurately?
• Do you remember that for \(X \sim \text{B}(n, p)\), the mean is \(np\) and the variance is \(np(1-p)\)?
• Can you confidently convert phrases like "fewer than", "at least", and "more than" into cumulative \(\le\) expressions?