A fair coin flipped \(N = 40\) times: the number of heads \(k\) that maximises the binomial probability is 20. The value decreases as it deviates from 20, and the curve looks like a bell, so we call it a bell curve.

Flip a coin 1,000 times: the mean of heads divided by 1,000 is approximately the probability of a head, 0.5. Do the whole experiment 1,000 times and plot the means: the value with the most experiments is around 0.5, and the plot forms a bell curve.

A single coin flip is a single probability; a few flips, the binomial distribution (a probability mass function); many flips, the normal distribution (a probability density function).

A normal distribution has two parameters: mean \(\mu\) and variance \(\sigma^2\). For any outcome \(x\), the quadratic difference from \(\mu\) is \((x - \mu)^2\), a U-shaped parabola with its minimum at \(\mu\).

Divide by the variance: \((x - \mu)^2/\sigma^2\). With \(\sigma^2 = 4\) the curve is wider than with \(\sigma^2 = 1\); \(\sigma^2 = 1/4\) gives a narrower curve; \(\sigma^2 = 0\) would be a spike.

Put a minus sign and a half in front, \(-\tfrac12 (x-\mu)^2/\sigma^2\): the parabola now opens downward with its maximum 0 at \(x = \mu\). Take the exponential:

\[ f(x) = e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}} \]

The exponent is always \(\le 0\), so \(f(x) \le 1\) with the maximum \(f(\mu) = 1\): the value of \(x\) that maximises \(f\) is \(\mu\). As \(x \to \pm\infty\) the exponent goes to \(-\infty\) and \(f \to 0\).

The area under this curve is \(\sqrt{2\pi\sigma^2}\). Probability must add up to 1, so the true normal distribution is normalised by that constant:

\[ N(x;\ \mu, \sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}}\,e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}} \]

This is the probability density function (pdf). \(E(X) = \mu\) and \(V(X) = \sigma^2\). The cumulative probability \(F(x)\) is the area under the pdf to the left of \(x\): an S-shaped curve from 0 to 1 with \(F(\mu) = 0.5\).

This distribution describes the limit of computing a mean over any set of experiments: as long as you repeat an experiment many times, the mean of the results has a distribution like this.

Quantitative data vary a lot, so we need a model to understand the pattern of variation, compare individuals to the group, and predict. The normal distribution is bell-shaped and symmetric, defined by the mean \(\mu\) (centre) and the standard deviation \(\sigma\) (spread). Many real-world variables (heights, test scores, measurement errors) approximate it; the central limit theorem says averages of large samples tend to be normal even if the data are not; and it lets us compute chances of events.

Practical benefits: standardisation (z-scores) to compare different datasets; confidence intervals; hypothesis testing; decisions such as "how many meals for 95% of expected participants?"

The empirical rule: no matter what \(\mu\) and \(\sigma\) are, the area between \(\mu - \sigma\) and \(\mu + \sigma\) is about 68%; between \(\mu \pm 2\sigma\) about 95%; between \(\mu \pm 3\sigma\) about 99.7%. Almost all values fall within 3 standard deviations.

Properties: mean, mode and median are all equal; the curve is symmetric about \(\mu\); exactly half the values are left of centre and half right; the total area is 1 (100%).

A standard normal model is a normal distribution with mean 0 and standard deviation 1.

\[ z = \frac{x - \mu}{\sigma} \]

\(z \gt 0\): above the mean. \(z \lt 0\): below the mean. \(|z|\): distance from the mean in standard-deviation units.

Comparing relative performance: maths 83% with class mean 81% and SD 7.3%; English 75% with class mean 55% and SD 8.5%.

\[ z_{\text{maths}} = \frac{83 - 81}{7.3} = 0.27,\qquad z_{\text{English}} = \frac{75 - 55}{8.5} = 2.35 \]

In maths you were only slightly above average; in English far above. You performed much better in English relative to the class.

Daily solar yield \(X \sim N(500, 40^2)\) kWh: \(x = 580\) gives \(z = (580 - 500)/40 = 2\), two standard deviations above the mean; \(P(X \gt 580) = 0.0228\), only about 2.28% of days.

Normalisation rescales into a fixed range; standardisation centres and rescales using mean and SD (s70).

The cdf of the standard normal, \(\Phi(z)\), is an integral that is not easy to evaluate, so it has been tabulated for \(\mu = 0\) and \(\sigma = 1\). That table serves every other \(\mu\) and \(\sigma\) through the z-score. By symmetry:

\[ \Phi(-a) = 1 - \Phi(a) \]

Reading the table: the row gives the first two digits of the z-score, the column the hundredths. Area to the left of \(z = 1.51\): 0.9345 (93.45%). For \(z = -0.33\): \(P(Z \lt -0.33) = 0.3707\), so \(P(Z \gt -0.33) = 0.6293\).

Steps (s72): draw a bell curve and shade the area asked for; find the picture that matches; the row/column intersection is the probability.

Always standardise each boundary separately. Word problems involve the normal when they say "assume the variable is normally distributed" or "approximately normal"; then decide which type you have: between, more than / above, less than, or a lower cut-off.

In mathematics one counterexample disproves a universal claim ("all swans are white"). In statistics we almost never claim universals; we talk about tendencies: averages, probabilities, risk reductions. So one case, even a real one, cannot overturn a statement about the average effect in a population.

Claim in statistical form: "lower smoking prevalence increases average male lifespan". Aggregate evidence: 1950s–60s average male lifespan about 55–60 years (heavy-smoking era); today about 75. The mean moved up by 15–20 years. A 90-year-old heavy smoker is possible but rare, an outlier; outliers exist in almost every distribution and do not determine the mean. Letting one outlier overturn the mean would mean systematically bad decisions.

A distribution shows how values are spread across a population; a single point does not. Saying "smoking reduces average lifespan" does not mean every smoker dies early; it means the probability of dying earlier is much higher. In the 1950s distribution \(P(\text{live} \ge 90)\) was under 1%; today it is higher, but still not common.

Anecdote ≠ evidence. Look at the mean and the distribution, not the most memorable case.

Davos participation 2015–2024 suggests \(N \sim \text{Poisson}(\lambda)\) with \(\lambda \approx 3{,}150\). Expected participants \(E(N) = \lambda\); natural variability \(SD(N) = \sqrt\lambda \approx 56\). Capacity with a 10% buffer: \(C = 1.1\lambda = 3{,}465\).

The direct probability \(P(N = C)\) is the chance that exactly 3,465 show up, just one outcome. The cumulative probability \(P(N \le C)\) is the chance the meals are enough whether 3,465 or 3,400 or 3,000 show up. That is the correct service level, because as the manager you care whether stock suffices for all counts up to capacity.

\[ P(N \le C) = e^{-\lambda}\sum_{k=0}^{C}\frac{\lambda^k}{k!} \]

Term by term: \(P(N \le 0) = e^{-\lambda}\); \(P(N \le 1) = e^{-\lambda}(1 + \lambda)\); \(P(N \le 2) = e^{-\lambda}(1 + \lambda + \lambda^2/2!)\), and so on. In software: scipy.stats.poisson.cdf(C, λ), R ppois, Excel POISSON.DIST(C, λ, TRUE).

With \(\lambda = 3{,}150\) and \(C = 3{,}465\) the stock-out risk \(P(N \gt C)\) is below \(1.42 \times 10^{-8}\), far safer than needed.

Inferential statistics makes predictions or inferences about a population from a sample. What proportion of all HBKU students drink coffee? You could email everyone, or sample.

Population of 21 cups, 15 green (drinks coffee) and 6 red: the parameter is \(15/21 = 71\%\). A sample of 5 might give 1 green, statistic 20%; another sample of 5 might give 3 green, statistic 60%. The sampling distribution is the distribution of a statistic, any statistic: proportions, standard deviations, variances, differences in means.

N.B. the parameter is the numeric summary of a population and is a fixed value. Statistics change based on the sample you are looking at.

Two useful theorems: the law of large numbers and the central limit theorem (s94).

The CLT states that the distribution of the mean of a large number of independent, identically distributed (i.i.d.) random variables approaches a normal distribution, regardless of the shape of the original distribution, as the sample size becomes large. It applies to discrete and continuous distributions.

Conditions: independence; identically distributed; sample size sufficiently large, typically \(n \ge 30\), although this depends on the original distribution. If the population is already normal, the sample size does not need to be large.

\[ \bar X_n \sim N\!\left(\mu,\ \frac{\sigma^2}{n}\right) \]

\(n = 4\) gives SE \(= \sigma/2\); \(n = 64\) gives SE \(= \sigma/8\). Larger samples reduce the variability of the sample mean.

Proof of the mean (s99): by linearity of expectation,

\[ E[\bar X_n] = E\!\left[\frac{1}{n}\sum_{i=1}^{n} X_i\right] = \frac{1}{n}\sum_{i=1}^{n} E[X_i] = \frac{1}{n}\cdot n\mu = \mu \]

Proof of the variance (s100): the \(X_i\) are independent, so the variance of the sum is the sum of the variances, and \(\text{Var}(cX) = c^2\text{Var}(X)\):

\[ \text{Var}(\bar X_n) = \frac{1}{n^2}\sum_{i=1}^{n}\text{Var}(X_i) = \frac{1}{n^2}\cdot n\sigma^2 = \frac{\sigma^2}{n} \]

Importance (s101): lets us use the normal as an approximation for sums or averages even when the underlying distribution is unknown or non-normal; underpins hypothesis testing, confidence intervals, quality control and risk management; applications in polls, finance, machine learning.

Limitations (s102): needs a large sample (\(n \ge 30\)), especially for skewed or heavy-tailed data; needs independence (correlated time series may break it); needs identical distribution; does not apply to distributions with infinite variance (Cauchy); convergence can be slow for highly skewed data.

Energy-meter readings have \(\mu = 100\) and \(\sigma = 12\); a sample has \(n = 36\) readings.

\[ \sigma_{\bar X} = \frac{12}{\sqrt{36}} = 2 \] \[ z = \frac{104 - 100}{2} = 2 \] \[ P(\bar X \gt 104) = 1 - \Phi(2) = 1 - 0.9772 = 0.0228 \]

Only about 2.28% of samples of 36 readings have a mean above 104. Common error: use \(\sigma/\sqrt n\) for a sample mean, not \(\sigma\).

Dog show (s106): 250 dogs, mean weight 12 pounds, SD 8. Four dogs chosen at random; probability their average exceeds 25 pounds: SE \(= 8/\sqrt 4 = 4\), \(z = (25 - 12)/4 = 3.25\). Between 8 and 25 pounds (s108): \(z = -1\) and \(3.25\), so \(0.9994 - 0.1587 = 0.8407\).

When dealing with dependent random variables, we must account for their relationship through covariance when calculating variance and standard deviation.

\[ \text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) + 2\,\text{Cov}(X, Y) \] \[ \text{Var}\!\left(\sum_{i=1}^{n} X_i\right) = \sum_{i=1}^{n}\text{Var}(X_i) + 2\sum_{i \lt j}\text{Cov}(X_i, X_j) \] \[ \text{SD}(X + Y) = \sqrt{\text{Var}(X) + \text{Var}(Y) + 2\,\text{Cov}(X, Y)} \]

Example: \(\text{Var}(X) = 4\), \(\text{Var}(Y) = 9\), \(\text{Cov}(X, Y) = 2\):

\[ \text{Var}(X + Y) = 4 + 9 + 2(2) = 17,\qquad \text{SD}(X + Y) = \sqrt{17} = 4.12 \]

Key points: the variance for dependent variables includes the individual variances and the covariance between them. If \(\text{Cov}(X, Y) \gt 0\) they tend to move together; if \(\lt 0\), oppositely. These formulas adjust for dependencies when calculating the spread of a combination of variables.

For dependent \(X_1, \dots, X_n\), the variance of the sample mean reflects not only the variability of individual variables but also how they covary:

\[ \text{Var}(\bar X) = \frac{1}{n^2}\left(\sum_{i=1}^{n}\text{Var}(X_i) + 2\sum_{i \lt j}\text{Cov}(X_i, X_j)\right) \] \[ \text{SE}(\bar X) = \frac{1}{n}\sqrt{\sum_{i=1}^{n}\text{Var}(X_i) + 2\sum_{i \lt j}\text{Cov}(X_i, X_j)} \]

When variables are positively correlated, the standard error increases, making estimates less precise.

Example: three dependent variables with variances 4, 6, 5 and covariances \(\text{Cov}(X_1, X_2) = 1\), \(\text{Cov}(X_1, X_3) = 2\), \(\text{Cov}(X_2, X_3) = 1.5\):

\[ \text{Var}(\bar X) = \frac{1}{9}\big(4 + 6 + 5 + 2(1 + 2 + 1.5)\big) = \frac{24}{9} = 2.67 \] \[ \text{SE}(\bar X) = \sqrt{2.67} = 1.63 \]

Conclusion: with dependent variables the sampling distribution has higher variance and standard error than with independent ones, because the dependence (captured by covariance) increases variability in sample estimates.

The covariance between \(X\) and \(Y\) measures how much they vary together: positive when one increases as the other increases, negative when one increases as the other decreases. From a sample:

\[ \text{Cov}(X, Y) = \frac{1}{n-1}\sum_{i=1}^{n}(X_i - \bar X)(Y_i - \bar Y) \]

Five students' scores on two tests: \(X = 85, 78, 92, 70, 88\); \(Y = 90, 85, 95, 80, 89\).

\[ \bar X = \frac{413}{5} = 82.6,\qquad \bar Y = \frac{439}{5} = 87.8 \]

Products of deviations: \(2.4 \times 2.2 = 5.28\); \((-4.6)(-2.8) = 12.88\); \(9.4 \times 7.2 = 67.68\); \((-12.6)(-7.8) = 98.28\); \(5.4 \times 1.2 = 6.48\). Sum 190.6.

\[ \text{Cov}(X, Y) = \frac{190.6}{4} = 47.65 \]