4.1 Statistics and probability - SL content

Sampling asks who was observed and how they were chosen.

A sample can describe a population only when the selection method limits systematic bias; random, stratified and quota designs answer different practical constraints.

Example

To estimate a school’s mean travel time, sample students from each year group rather than surveying only the nearest classroom.

Name the population, sampling frame and possible bias before calculating a statistic.

A large convenience sample can still be biased; sample size does not repair a distorted frame.

Boundary checklist: a population is the full group of interest and a sample is the observed subset; discrete data take countable values while continuous data can vary across an interval. An outlier lies more than $1.5\times IQR$ below $Q_1$ or above $Q_3$, but context decides whether it is valid or an error. Distinguish simple random, convenience, systematic, quota and stratified sampling, and inspect missing data and the sampling frame before generalising.

Choose a display that preserves distribution and comparison.

Frequency tables organise discrete or continuous data; equal-class-width histograms show continuous frequency, cumulative-frequency graphs locate medians, quartiles and percentiles, and box plots compare centre, spread, symmetry and outliers.

Example

If a cumulative-frequency graph for 80 values gives $Q_1=18$ and $Q_3=30$, then $IQR=12$ and the outlier fences are $0$ and $48$. Mark any value outside those fences with a cross on the box plot.

Read class inequalities without gaps, use the cumulative total to locate the required percentile, then compare box plots using median, IQR, range, symmetry and outliers.

For this syllabus section, frequency-density histograms are not required; a scatter plot belongs to bivariate correlation, not one-variable distribution display.

Summary statistics compress data without telling the whole story.

Mean, median, quartiles and standard deviation answer different questions about centre and spread, especially when outliers or skew are present.

Example

For 2,3,3,4,18 the mean is pulled upward while the median remains 3; median and IQR better describe a typical value.

Choose a resistant summary for skewed data and state what information the statistic loses.

Standard deviation is not a universal error bar and does not imply a normal distribution.

For grouped data, estimate the mean with $\bar x\approx\sum fm/\sum f$, where $m$ is each class midpoint. Adding a constant $c$ to every value adds $c$ to the mean and median but leaves standard deviation and IQR unchanged; multiplying every value by $k$ multiplies centre by $k$ and spread by $|k|$, while variance is multiplied by $k^2$. Technology and hand methods may use different quartile conventions.

Correlation measures association, not a causal mechanism.

A correlation coefficient describes the direction and strength of a linear association, while regression predicts one variable from another under model assumptions.

Example

Ice-cream sales and sunburn may rise together because temperature affects both; the correlation does not make sales the cause.

Plot the relationship, inspect outliers and direction, then state whether the regression is for prediction or explanation.

A strong correlation can be nonlinear or confounded; it does not prove causation.

Pearson's $r$ lies between $-1$ and $1$ and measures only linear association; its sign gives direction and $|r|$ gives linear strength. For a fitted line $y=ax+b$, $a$ is the predicted change in $y$ for one unit of $x$ and $b$ is the predicted value at $x=0$ when that interpretation is meaningful. Use the $y$-on-$x$ line to predict $y$ from $x$, avoid unjustified extrapolation, and never infer causation from correlation alone.

Probability starts with a defined sample space.

An event is a set of outcomes, and probabilities must be non-negative, sum to one across the sample space and reflect the stated model.

Example

For a fair die, P(even)=3/6=1/2 because the event {2,4,6} contains three equally likely outcomes.

Define the outcomes before using complements, unions or counting rules.

Equally likely is an assumption, not a property of every list of outcomes.

For equally likely outcomes in sample space $U$, $P(A)=n(A)/n(U)$ and $P(A')=1-P(A)$. Relative frequency estimates probability from observed trials. If an event has probability $p$ on each of $N$ comparable occasions, its expected number of occurrences is $Np$; for example, $128(0.1)=12.8$ means a long-run average, not that exactly 12.8 people can occur.

Combine events by matching the rule to their relationship.

Use addition for mutually exclusive alternatives, multiplication for a sequence, and conditional probability when the first event changes the sample space.

Example

If P(A)=0.4 and P(B|A)=0.5, then P(A∩B)=0.2; independence would require P(B|A)=P(B).

Draw a tree or table and label whether each branch is conditional or independent.

P(A∩B)=P(A)P(B) only when independence is justified, not merely because events are different.

General addition rule: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$; only mutually exclusive events have $P(A\cap B)=0$. Conditional probability is $P(A|B)=P(A\cap B)/P(B)$, so $P(A\cap B)=P(B)P(A|B)$. Independence requires $P(A\cap B)=P(A)P(B)$. Without replacement normally changes later branch probabilities; with replacement can preserve them.

A discrete random variable assigns values to countable outcomes.

A discrete probability distribution lists countable values $x$ and probabilities $P(X=x)$ that sum to 1. Its expected value is $E(X)=\sum xP(X=x)$, a weighted long-run mean rather than a guaranteed single outcome.

Example

If X is 0,1,2 with probabilities .2,.5,.3, E(X)=0(.2)+1(.5)+2(.3)=1.1.

Check that all probabilities are non-negative and sum to 1, then calculate $E(X)=\sum xP(X=x)$. If $X$ is a player's gain, $E(X)=0$ describes a fair game in the long run.

Expected value need not be attainable in one play, and variance is not a required calculation in this specific SL 4.7 objective.

The binomial model counts successes in fixed independent trials.

It requires a fixed n, two outcomes per trial, constant success probability p and independence; then P(X=k)=nCk pᵏ(1−p)ⁿ⁻ᵏ.

Example

For five independent shots with p=.4, P(X=2)=10(.4)²(.6)³≈.3456.

Test the four assumptions before using the formula or technology.

Repeated trials from a changing population or with different p are not binomial just because they count successes.

If $X\sim B(n,p)$, then $E(X)=np$ and $Var(X)=np(1-p)$, so $SD(X)=\sqrt{np(1-p)}$. For the five-shot example, the mean number of successes is $5(0.4)=2$ and the variance is $5(0.4)(0.6)=1.2$. Technology may calculate cumulative or interval probabilities, but the fixed-$n$, two-outcome, constant-$p$ and independence assumptions must be checked first.

The normal model turns a continuous measurement into areas.

A normal variable is described by mean μ and standard deviation σ; probabilities are areas between cut-offs, not point probabilities.

Example

If $X\sim N(70,5^2)$, approximately 68% of values lie from 65 to 75, 95% from 60 to 80 and 99.7% from 55 to 85. For other probabilities or an inverse-normal cut-off, enter the mean and standard deviation directly in technology.

Sketch and shade the required area, use normal or inverse-normal technology with the stated $\mu$ and $\sigma$, and check that the output lies on the expected side of the mean.

The 68–95–99.7 rule is approximate. In this SL 4.9 objective, inverse-normal calculations use the given mean and standard deviation without first transforming to $z$; standardisation belongs to SL 4.12.

Regression direction matters when predicting one variable from another.

The regression of $x$ on $y$ predicts $x$ from a known value of $y$ and minimises squared residuals measured in the $x$-direction. It is not interchangeable with the regression of $y$ on $x$.

Example

If $x$ is fuel used and $y$ is distance, an $x$-on-$y$ line may predict fuel from a given distance. It should not be rearranged to predict distance from fuel; use the separately fitted $y$-on-$x$ line for that direction.

State which variable is being predicted, choose that variable 'on' the known variable, and keep prediction within a defensible data range.

A high correlation does not make the two regression lines identical.

Conditional probability restricts the denominator.

P(A|B)=P(A∩B)/P(B) describes A inside the cases where B has occurred, so the reference population changes.

Example

Among 100 tested devices, 20 are defective and 8 of those are overheating: P(overheating|defective)=8/20, not 8/100.

Write the conditioning event under the denominator and use a two-way table when possible.

P(A|B) and P(B|A) answer different questions.

Equivalent multiplication form: $P(A\cap B)=P(B)P(A|B)$. Events are independent exactly when conditioning does not change the probability, for example $P(A|B)=P(A)$ (with $P(B)>0$); the syllabus also permits comparison with $P(A|B')$. A numerical equality must be shown rather than assuming independence from the wording.

Standardisation measures distance from the mean in standard deviations.

For X with mean μ and standard deviation σ, z=(X−μ)/σ lets normal probabilities be read on a common scale.

Example

A score 85 from N(70,10²) has z=1.5, so it is 1.5 standard deviations above the mean.

Keep the sign: negative z-values lie below the mean, and convert back with X=μ+σz.

A z-score is relative position, not a percentage or a guarantee of rarity.

Unknown-parameter example: suppose a normal model assigns $z=-1$ to $x=70$ and $z=2$ to $x=100$. Then $(70-\mu)/\sigma=-1$ and $(100-\mu)/\sigma=2$. Subtracting gives $30/\sigma=3$, so $\sigma=10$ and then $\mu=80$. Use technology to obtain the needed inverse-normal cut-offs, then solve the resulting z-equations with $\sigma>0$.