A-Level Further Mathematics: Further Statistics
Exam Board Note
Further Statistics is an optional module available in all major UK specifications. Edexcel offers Further Statistics 1 (FS1/8FM0:27) and FS2. AQA covers these topics in its Statistics options. OCR A includes this material in Statistics A (Y532). CIE examines probability distributions and tests in Further Mathematics Papers 5 and 6. Spearman's rank correlation and the Wilcoxon test appear primarily on Edexcel; check your board's specification for exact topic requirements.
Further Statistics builds on the A-Level Statistics core with more sophisticated probability distributions, rigorous hypothesis testing techniques, and non-parametric methods. These tools are essential for data analysis in science, engineering, economics, and beyond.
1. Geometric Distribution
Definition: Geometric Distribution
If repeated independent Bernoulli trials each have probability $p$ of success, the number of trials $X$ needed to obtain the first success follows the geometric distribution $X \sim \text{Geo}(p)$:
$$P(X = r) = (1-p)^{r-1}p, \quad r = 1, 2, 3, \ldots$$
$$E(X) = \frac{1}{p}, \qquad \text{Var}(X) = \frac{1-p}{p^2}$$
The geometric distribution has the memoryless property: given that the first $k$ trials failed, the distribution of the remaining waiting time is still $\text{Geo}(p)$. Formally: $P(X > m + n \mid X > n) = P(X > m)$.
Worked Example 1.1 — Geometric Distribution
A biased coin has $P(\text{Heads}) = 0.3$. Let $X$ be the toss number of the first Head. Find (a) $P(X = 4)$, (b) $P(X \leq 3)$, (c) $E(X)$ and $\text{Var}(X)$.
Part (a) $P(X = 4) = (0.7)^3(0.3) = 0.343 \times 0.3 = 0.1029$
Part (b) $$P(X \leq 3) = P(X=1) + P(X=2) + P(X=3)$$ $$= 0.3 + 0.7(0.3) + 0.7^2(0.3) = 0.3 + 0.21 + 0.147 = 0.657$$ Alternatively: $P(X \leq 3) = 1 - P(X > 3) = 1 - (1-p)^3 = 1 - 0.343 = 0.657$ ✓
Part (c) $E(X) = \dfrac{1}{0.3} = \dfrac{10}{3} \approx 3.33$, $\quad\text{Var}(X) = \dfrac{0.7}{0.09} = \dfrac{70}{9} \approx 7.78$
2. Negative Binomial Distribution
Definition: Negative Binomial Distribution
The number of trials $X$ needed to obtain exactly $r$ successes (where each trial independently succeeds with probability $p$) follows the negative binomial distribution $X \sim \text{NB}(r, p)$:
$$P(X = n) = \binom{n-1}{r-1} p^r (1-p)^{n-r}, \quad n = r, r+1, r+2, \ldots$$
$$E(X) = \frac{r}{p}, \qquad \text{Var}(X) = \frac{r(1-p)}{p^2}$$
When $r = 1$, this reduces to the geometric distribution.
Worked Example 2.1 — Negative Binomial
A sharpshooter hits a target with probability $p = 0.6$ independently on each shot. Let $X$ be the number of shots to achieve the 3rd hit. Find $P(X = 5)$ and $E(X)$.
Step 1 $X \sim \text{NB}(3, 0.6)$. For $X = 5$: among the first 4 shots, exactly 2 must be hits, then the 5th shot must be a hit: $$P(X = 5) = \binom{4}{2}(0.6)^2(0.4)^2 \times 0.6 = 6 \times 0.36 \times 0.16 \times 0.6$$ $$= 6 \times 0.03456 = 0.20736 \approx 0.207$$
Step 2 $E(X) = \dfrac{3}{0.6} = 5$ shots.
3. Poisson Distribution
Definition: Poisson Distribution
The Poisson distribution $X \sim \text{Po}(\lambda)$ models the number of events in a fixed time or space, where events occur independently at a constant average rate $\lambda > 0$:
$$P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}, \quad x = 0, 1, 2, \ldots$$
$$E(X) = \lambda, \qquad \text{Var}(X) = \lambda$$
Key property: If $X \sim \text{Po}(\lambda)$ and $Y \sim \text{Po}(\mu)$ are independent, then $X + Y \sim \text{Po}(\lambda + \mu)$.
FSTAT·1 — Poisson PMF $P(X=x) = \tfrac{e^{-\lambda}\lambda^x}{x!}$ for $\lambda=1$ (red) and $\lambda=3$ (blue). As $\lambda$ increases the distribution spreads right, with $E(X) = \lambda$ and $\text{Var}(X) = \lambda$.
Poisson approximation to Binomial: If $X \sim B(n, p)$ with $n$ large and $p$ small (typically $n > 50$, $p < 0.1$), then $X \approx \text{Po}(\lambda)$ where $\lambda = np$.
Worked Example 3.1 — Poisson Process
Cars arrive at a toll booth at an average rate of 3 per minute. Assuming a Poisson process, find (a) $P(X = 5)$ in a given minute, and (b) the probability that at least 2 cars arrive in a 30-second interval.
Part (a) $X \sim \text{Po}(3)$: $$P(X = 5) = \frac{e^{-3} \cdot 3^5}{5!} = \frac{e^{-3} \cdot 243}{120} = \frac{243e^{-3}}{120} \approx \frac{243 \times 0.04979}{120} \approx 0.1008$$
Part (b) In 30 seconds (half a minute), the rate is $\lambda = 1.5$. Let $Y \sim \text{Po}(1.5)$: $$P(Y \geq 2) = 1 - P(Y = 0) - P(Y = 1) = 1 - e^{-1.5} - 1.5e^{-1.5}$$ $$= 1 - 2.5e^{-1.5} \approx 1 - 2.5(0.2231) \approx 1 - 0.5578 = 0.442$$
4. Probability Generating Functions
Definition: Probability Generating Function
For a discrete random variable $X$ taking non-negative integer values, the probability generating function (PGF) is: $$G_X(t) = E(t^X) = \sum_{x=0}^{\infty} P(X = x)\,t^x$$
The PGF encodes all the probabilities: $P(X = r) = \dfrac{G^{(r)}(0)}{r!}$ where $G^{(r)}$ denotes the $r$th derivative.
Moments: $E(X) = G'(1)$ and $\text{Var}(X) = G''(1) + G'(1) - [G'(1)]^2$.
PGFs of Standard Distributions
Bernoulli $B(1,p)$: $G(t) = q + pt$ where $q = 1 - p$
Binomial $B(n,p)$: $G(t) = (q + pt)^n$
Geometric $\text{Geo}(p)$: $G(t) = \dfrac{pt}{1 - qt}$
Poisson $\text{Po}(\lambda)$: $G(t) = e^{\lambda(t-1)}$
Sums: If $X$ and $Y$ are independent, $G_{X+Y}(t) = G_X(t) \cdot G_Y(t)$.
Worked Example 4.1 — Using a PGF
A random variable $X$ has PGF $G(t) = \dfrac{1}{3-2t}$. Find $P(X = r)$ for $r \geq 0$ and compute $E(X)$.
Step 1 Rewrite as a power series. Factor out $\frac{1}{3}$: $$G(t) = \frac{1}{3\left(1 - \frac{2t}{3}\right)} = \frac{1}{3}\sum_{r=0}^{\infty}\left(\frac{2}{3}\right)^r t^r$$ $$= \sum_{r=0}^{\infty}\frac{1}{3}\left(\frac{2}{3}\right)^r t^r$$ So $P(X = r) = \dfrac{1}{3}\left(\dfrac{2}{3}\right)^r$ for $r = 0, 1, 2, \ldots$ This is a geometric-type distribution.
Step 2 $G'(t) = \dfrac{2}{(3-2t)^2}$, so $E(X) = G'(1) = \dfrac{2}{(3-2)^2} = 2$.
5. Central Limit Theorem
Theorem: Central Limit Theorem (CLT)
Let $X_1, X_2, \ldots, X_n$ be i.i.d. random variables with mean $\mu$ and variance $\sigma^2 < \infty$. Then as $n \to \infty$, the sample mean $\bar{X} = \dfrac{1}{n}\sum_{i=1}^n X_i$ satisfies:
$$\bar{X} \sim N\!\left(\mu, \frac{\sigma^2}{n}\right) \quad \text{approximately, for large } n$$
Equivalently, $Z = \dfrac{\bar{X} - \mu}{\sigma/\sqrt{n}} \sim N(0, 1)$ approximately.
The CLT is remarkable because it applies regardless of the distribution of the individual $X_i$, as long as the variance is finite. As a rule of thumb, $n \geq 30$ is usually sufficient for the approximation to be good, though fewer may suffice for near-symmetric distributions.
Worked Example 5.1 — Central Limit Theorem
The time to serve one customer at a bank follows an exponential distribution with mean 4 minutes. Find the probability that the total service time for 50 customers exceeds 215 minutes.
Step 1 Let $X_i \sim \text{Exp}(1/4)$, so $E(X_i) = 4$ and $\text{Var}(X_i) = 16$.
Step 2 Total time $T = X_1 + \cdots + X_{50}$. By the CLT: $$T \approx N(50 \times 4,\; 50 \times 16) = N(200,\; 800)$$
Step 3 $$P(T > 215) = P\!\left(Z > \frac{215 - 200}{\sqrt{800}}\right) = P\!\left(Z > \frac{15}{20\sqrt{2}}\right) = P(Z > 0.530)$$ $$\approx 1 - \Phi(0.530) \approx 1 - 0.7019 = 0.298$$
6. Confidence Intervals
A confidence interval is a range of values constructed from sample data that, with a stated probability (the confidence level), contains the true population parameter. A 95% CI does not mean "95% probability the parameter is in this interval"; rather, if the procedure is repeated many times, 95% of such intervals will contain the true parameter.
Confidence Interval for a Population Mean
For a sample of size $n$ with sample mean $\bar{x}$ from a $N(\mu, \sigma^2)$ population (or large $n$ by CLT):
$$\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$$
where $z_{\alpha/2}$ is the critical value from the standard normal: $z_{0.025} = 1.96$ for a 95% CI, $z_{0.005} = 2.576$ for a 99% CI.
If $\sigma$ is unknown and $n$ is small, use the $t$-distribution with $n-1$ degrees of freedom: $$\bar{x} \pm t_{n-1,\alpha/2} \cdot \frac{s}{\sqrt{n}}$$
Worked Example 6.1 — Confidence Interval
A sample of 36 measurements of a physical constant gives $\bar{x} = 8.42$ with standard deviation $s = 0.60$. Construct a 95% confidence interval for the true mean.
Step 1 $n = 36$ is large enough to use the normal approximation. Standard error $= \dfrac{s}{\sqrt{n}} = \dfrac{0.60}{6} = 0.10$.
Step 2 $z_{0.025} = 1.96$. The 95% CI is: $$8.42 \pm 1.96 \times 0.10 = 8.42 \pm 0.196$$ $$\Rightarrow (8.224,\; 8.616)$$
We are 95% confident that the true mean lies between 8.224 and 8.616.
7. Chi-Squared Tests
The chi-squared ($\chi^2$) test is used to test whether observed frequency data differs significantly from expected frequencies under some hypothesis. The test statistic is: $$\chi^2 = \sum_{\text{all cells}} \frac{(O - E)^2}{E}$$ where $O$ is the observed frequency and $E$ is the expected frequency. Under $H_0$, $\chi^2$ follows (approximately) a chi-squared distribution with degrees of freedom $\nu$.
Goodness-of-Fit Test
Tests whether a sample comes from a specified distribution. Degrees of freedom: $\nu = k - 1 - p$, where $k$ = number of categories and $p$ = number of parameters estimated from the data. Any expected frequency below 5 should be merged with an adjacent cell.
Worked Example 7.1 — Goodness-of-Fit Test
A die is rolled 120 times with the following results. Test at the 5% level whether the die is fair.
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Observed | 25 | 17 | 21 | 23 | 14 | 20 |
H$_0$ The die is fair (each face has probability 1/6).
Expected Each $E_i = 120 \times \frac{1}{6} = 20$.
Test Statistic $$\chi^2 = \frac{(25-20)^2}{20} + \frac{(17-20)^2}{20} + \frac{(21-20)^2}{20} + \frac{(23-20)^2}{20} + \frac{(14-20)^2}{20} + \frac{(20-20)^2}{20}$$ $$= \frac{25+9+1+9+36+0}{20} = \frac{80}{20} = 4.00$$
Critical Value $\nu = 6 - 1 = 5$ degrees of freedom. At 5% level, $\chi^2_{5, 0.05} = 11.07$.
Conclusion Since $4.00 < 11.07$, we do not reject $H_0$. There is insufficient evidence at the 5% level to conclude that the die is unfair.
Test of Independence (Contingency Table)
A contingency table chi-squared test examines whether two categorical variables are independent. For an $r \times c$ table:
- $E_{ij} = \dfrac{\text{(row } i \text{ total)} \times \text{(column } j \text{ total)}}{\text{grand total}}$
- Degrees of freedom: $\nu = (r-1)(c-1)$
Worked Example 7.2 — Contingency Table
Test at the 5% level whether grade achieved is independent of revision hours for the following data:
| Grade A | Grade B | Grade C | Row Total | |
|---|---|---|---|---|
| <10 hrs | 5 | 20 | 15 | 40 |
| 10–20 hrs | 15 | 25 | 10 | 50 |
| >20 hrs | 20 | 15 | 5 | 40 |
| Col Total | 40 | 60 | 30 | 130 |
Expected frequencies $E_{ij} = \dfrac{R_i C_j}{N}$:
| Grade A | Grade B | Grade C | |
|---|---|---|---|
| <10 hrs | 12.31 | 18.46 | 9.23 |
| 10–20 hrs | 15.38 | 23.08 | 11.54 |
| >20 hrs | 12.31 | 18.46 | 9.23 |
Test Statistic $\chi^2 = \dfrac{(5-12.31)^2}{12.31} + \dfrac{(20-18.46)^2}{18.46} + \cdots \approx 16.76$
Critical Value $\nu = (3-1)(3-1) = 4$. $\chi^2_{4, 0.05} = 9.488$.
Conclusion $16.76 > 9.488$. Reject $H_0$. There is significant evidence at the 5% level that grade and revision hours are not independent.
8. Non-Parametric Tests
Non-parametric tests make fewer assumptions about the population distribution. They are used when data is ordinal, or when the normality assumption for parametric tests is not justified.
Spearman's Rank Correlation Coefficient
Spearman's Rank Correlation
For $n$ pairs of observations, rank each variable separately. If $d_i$ is the difference between the ranks of the $i$th pair:
$$r_s = 1 - \frac{6\sum d_i^2}{n(n^2-1)}$$
$r_s$ ranges from $-1$ (perfect negative association) to $+1$ (perfect positive association). To test whether the population rank correlation $\rho_s = 0$, compare $r_s$ with critical values from statistical tables.
Worked Example 8.1 — Spearman's Rank Correlation
Eight students' scores in Mathematics and Physics are ranked below. Calculate $r_s$ and test at the 5% level (one-tailed) whether there is positive association.
| Student | Maths Rank | Physics Rank | $d$ | $d^2$ |
|---|---|---|---|---|
| A | 1 | 2 | -1 | 1 |
| B | 2 | 1 | 1 | 1 |
| C | 3 | 4 | -1 | 1 |
| D | 4 | 3 | 1 | 1 |
| E | 5 | 7 | -2 | 4 |
| F | 6 | 5 | 1 | 1 |
| G | 7 | 6 | 1 | 1 |
| H | 8 | 8 | 0 | 0 |
| Total | 10 |
$$r_s = 1 - \frac{6 \times 10}{8(64-1)} = 1 - \frac{60}{504} = 1 - 0.119 = 0.881$$
From tables (one-tailed, 5%, $n = 8$): critical value $= 0.6429$. Since $0.881 > 0.6429$, we reject $H_0$ and conclude there is significant positive association.
Wilcoxon Signed-Rank Test
The Wilcoxon signed-rank test is a non-parametric alternative to the paired $t$-test. It tests whether the median difference between paired observations is zero.
Procedure:
- Compute the differences $d_i = x_i - y_i$. Discard any $d_i = 0$.
- Rank the absolute values $|d_i|$ from smallest to largest. Assign average ranks for ties.
- Attach the sign of $d_i$ to each rank.
- Compute $W^+ = $ sum of positive ranks, $W^- = $ sum of negative ranks. The test statistic is $W = \min(W^+, W^-)$.
- Compare $W$ with the critical value from the Wilcoxon table. Reject $H_0$ if $W \leq $ critical value.
Worked Example 8.2 — Wilcoxon Signed-Rank Test
Before/after scores for 7 participants in a training programme:
| Participant | Before | After | $d$ | $|d|$ | Rank | Signed Rank |
|---|---|---|---|---|---|---|
| 1 | 65 | 70 | +5 | 5 | 4 | +4 |
| 2 | 72 | 80 | +8 | 8 | 6 | +6 |
| 3 | 58 | 55 | -3 | 3 | 2 | -2 |
| 4 | 84 | 90 | +6 | 6 | 5 | +5 |
| 5 | 61 | 64 | +3 | 3 | 2 | +2 |
| 6 | 79 | 88 | +9 | 9 | 7 | +7 |
| 7 | 50 | 51 | +1 | 1 | 1 | +1 |
Note: participants 3 and 5 both have $|d| = 3$, so they share ranks 2 and 3 → each gets rank $(2+3)/2 = 2.5$. Recalculating:
$W^+ = 4 + 6 + 5 + 2.5 + 7 + 1 = 25.5$, $\quad W^- = 2.5$
$W = \min(25.5, 2.5) = 2.5$
From tables ($n = 7$, 5% one-tailed), the critical value is 3. Since $W = 2.5 \leq 3$, we reject $H_0$ and conclude the training programme significantly improved scores.
9. Practice Problems
Problem 1 — Geometric Distribution
A student passes each driving theory test with probability 0.7, independently. Find the probability that the student passes within the first 3 attempts, and find the expected number of attempts needed.
Show Solution
$X \sim \text{Geo}(0.7)$ where $X$ = number of attempts to first pass.
$P(X \leq 3) = 1 - (1-0.7)^3 = 1 - 0.3^3 = 1 - 0.027 = 0.973$
$E(X) = \dfrac{1}{0.7} = \dfrac{10}{7} \approx 1.43$ attempts.
Problem 2 — Negative Binomial
A quality control inspector tests items from a production line. Each item is defective independently with probability 0.2. What is the probability that the 3rd defective item is found on the 10th item tested?
Show Solution
$X \sim \text{NB}(3, 0.2)$. We need $P(X = 10)$: among the first 9 items, exactly 2 are defective, and item 10 is defective.
$$P(X = 10) = \binom{9}{2}(0.2)^2(0.8)^7 \times 0.2 = 36 \times 0.04 \times 0.2097 \times 0.2$$
$$= 36 \times 0.04 \times 0.2097 \times 0.2 \approx 36 \times 0.001678 \approx 0.0604$$
Problem 3 — PGF
The PGF of $X$ is $G(t) = \dfrac{(1 + t)^4}{16}$. Identify the distribution of $X$, find $E(X)$ and $\text{Var}(X)$.
Show Solution
We have $G(t) = \dfrac{(1+t)^4}{16} = \left(\dfrac{1+t}{2}\right)^4 = \left(\dfrac{1}{2} + \dfrac{1}{2}t\right)^4$.
This matches the binomial PGF $(q + pt)^n$ with $n = 4$ and $p = \frac{1}{2}$. So $X \sim B(4, \frac{1}{2})$.
$E(X) = G'(1) = \dfrac{4(1+t)^3}{16}\Big|_{t=1} = \dfrac{4 \times 8}{16} = 2$
$G''(t) = \dfrac{12(1+t)^2}{16}$, so $G''(1) = \dfrac{12 \times 4}{16} = 3$.
$\text{Var}(X) = G''(1) + G'(1) - [G'(1)]^2 = 3 + 2 - 4 = 1 = np(1-p) = 4 \times \frac{1}{4}$ ✓
Problem 4 — Central Limit Theorem
The lifetimes (in hours) of a certain brand of battery are independently distributed with mean 50 and standard deviation 8. A sample of 64 batteries is selected. Find the probability that the sample mean lifetime exceeds 52 hours.
Show Solution
By the CLT, $\bar{X} \approx N\!\left(50, \dfrac{64}{64}\right) = N(50, 1)$.
$P(\bar{X} > 52) = P\!\left(Z > \dfrac{52-50}{1}\right) = P(Z > 2) = 1 - \Phi(2) \approx 1 - 0.9772 = 0.0228$
Problem 5 — Chi-Squared Goodness of Fit
The number of defects per batch is modelled as Poisson. A sample of 100 batches gives: 0 defects: 40, 1 defect: 35, 2 defects: 16, 3 defects: 7, $\geq 4$ defects: 2. Test at the 5% level whether a Poisson model is appropriate.
Show Solution
Estimate $\hat{\lambda} = (0\times40 + 1\times35 + 2\times16 + 3\times7 + 4\times2)/100 = (0+35+32+21+8)/100 = 0.96$.
Expected frequencies with $\lambda = 0.96$:
$E_0 = 100e^{-0.96} = 38.3$, $E_1 = 100 \times 0.96e^{-0.96} = 36.8$, $E_2 = 100\times\frac{0.96^2}{2}e^{-0.96} = 17.6$, $E_3 = 5.7$, $E_{\geq4} = 1.6$.
Since $E_3 = 5.7 < 5$ and $E_{\geq 4} = 1.6 < 5$, merge categories $\geq 3$: observed $= 9$, expected $= 7.3$.
$\chi^2 = \dfrac{(40-38.3)^2}{38.3} + \dfrac{(35-36.8)^2}{36.8} + \dfrac{(16-17.6)^2}{17.6} + \dfrac{(9-7.3)^2}{7.3}$
$\approx 0.075 + 0.088 + 0.145 + 0.396 = 0.704$
Degrees of freedom: $\nu = 4 - 1 - 1 = 2$ (4 groups, 1 parameter estimated). Critical value $\chi^2_{2,0.05} = 5.991$.
Since $0.704 < 5.991$, do not reject $H_0$. The Poisson model is a good fit.
Problem 6 — Confidence Interval
A random sample of 10 measurements of resistance (in ohms) from a normal distribution gives: $\bar{x} = 4.82$, $s = 0.35$. Construct a 99% confidence interval for the population mean.
Show Solution
$n = 10$ is small, so use $t_{9, 0.005}$ for a 99% CI (two-tailed, so $\alpha/2 = 0.005$). From $t$-tables: $t_{9, 0.005} = 3.250$.
Margin of error $= 3.250 \times \dfrac{0.35}{\sqrt{10}} = 3.250 \times 0.1107 = 0.360$.
99% CI: $(4.82 - 0.360,\; 4.82 + 0.360) = (4.460,\; 5.180)$.
Problem 7 — Contingency Table
A study investigates whether smoking status (smoker / non-smoker) is independent of developing a certain condition (yes / no). Results: smoker-yes: 30, smoker-no: 70, non-smoker-yes: 20, non-smoker-no: 80. Test at the 5% level.
Show Solution
Row totals: smokers = 100, non-smokers = 100. Column totals: yes = 50, no = 150. Grand total = 200.
Expected: $E_{11} = \dfrac{100 \times 50}{200} = 25$, $E_{12} = 75$, $E_{21} = 25$, $E_{22} = 75$.
$\chi^2 = \dfrac{(30-25)^2}{25} + \dfrac{(70-75)^2}{75} + \dfrac{(20-25)^2}{25} + \dfrac{(80-75)^2}{75}$
$= 1 + \frac{1}{3} + 1 + \frac{1}{3} = 2.667$
$\nu = (2-1)(2-1) = 1$. Critical value $\chi^2_{1, 0.05} = 3.841$.
Since $2.667 < 3.841$, do not reject $H_0$. No significant evidence of association between smoking and condition at the 5% level.
Problem 8 — Spearman's Rank
Six judges rank 6 wines (A–F). Judge 1's ranks: 1, 2, 3, 4, 5, 6. Judge 2's ranks: 2, 1, 4, 3, 6, 5. Compute $r_s$ and comment.
Show Solution
Differences $d$: $-1, 1, -1, 1, -1, 1$. So $d^2 = 1$ for each, $\sum d^2 = 6$.
$r_s = 1 - \dfrac{6 \times 6}{6(36-1)} = 1 - \dfrac{36}{210} = 1 - 0.171 = 0.829$
This is a strong positive correlation, suggesting the two judges largely agree on the ranking order, despite some swapping of adjacent pairs.