4 Statistics and probability

A population is the full group a question concerns; a sample is the subset actually observed. Data may be qualitative or quantitative, discrete or continuous, and the type determines which summaries and graphs make sense.

Sampling method changes the route by which evidence enters the study. A random or stratified sample can reduce selection bias, while an opportunity sample is faster but may overrepresent whoever is available and willing.

If 40 of 200 sampled students prefer option A, the 20% is a sample proportion—not a fact about every student. Its usefulness depends on how the 200 were selected and whether the measurements were reliable.

A large sample does not repair a biased selection process. State the target population, sampling frame and likely source of error before generalising.

Sampling map: simple random gives every member an equal selection chance; systematic uses every $k$th member after a random start; stratified samples each subgroup in population proportion; quota fills category targets non-randomly; convenience uses readily available participants. Audit missing values and recording errors before analysis. An outlier lies more than $1.5\times IQR$ below $Q_1$ or above $Q_3$; investigate it rather than deleting it automatically.

Use a frequency table for discrete or grouped continuous data, an equal-class-width histogram for grouped continuous data, and a cumulative-frequency graph to read median, quartiles and percentiles. Frequency density histograms are not required at SL.

A box plot shows minimum, $Q_1$, median, $Q_3$ and maximum, with outliers marked separately by crosses. Compare two distributions using median, IQR, range and symmetry; a roughly symmetric box and whiskers can support, but not prove, possible normality.

Example

If a cumulative-frequency graph for 80 observations gives $Q_1=18$, median $=25$ and $Q_3=31$, then $IQR=13$. The outlier fences are $18-1.5(13)=-1.5$ and $31+1.5(13)=50.5$, so a value of 54 is plotted as an outlier cross.

Do not use bar gaps for continuous histogram classes or confuse cumulative frequency with class frequency. State units, preserve inequality class boundaries and compare both centre and spread.

The mean is the arithmetic average, the median is the middle ordered value, and the mode is the most frequent value. Range, variance and standard deviation describe how far scores vary around a centre.

The mean uses every value but is pulled by outliers; the median is more resistant. Standard deviation is small when scores cluster near the mean, but neither statistic alone tells you whether the measurement is valid.

For 2, 3, 3, 4 and 18, mean=6 while median=3. Reporting both exposes the outlier and prevents a reader from mistaking the arithmetic average for a typical observation.

Do not choose a statistic by habit. Match it to the measurement scale and distribution, and never delete an extreme value without explaining how it arose.

For grouped data, estimate the mean with class midpoints: $\bar x\approx\sum fm/\sum f$; the modal class has greatest frequency when class widths are equal. If every value becomes $aX+b$, the mean and median become $a$ times the old value plus $b$, while standard deviation becomes $|a|$ times the old standard deviation and variance becomes $a^2$ times the old variance. Quartile conventions can differ, so report the technology method when relevant.

A correlation coefficient summarises the direction and strength of an association between paired variables. A regression model uses an explanatory variable to estimate a response, with residuals showing what the model misses.

A positive correlation means larger values tend to occur together; a negative one means one tends to fall as the other rises. Strength concerns consistency around a pattern, not the slope's units or a causal mechanism.

If study time and score have r=0.82, a fitted line may predict scores within the observed range. It cannot show that study time alone caused the result: prior attainment, teaching and selection may also matter.

r=0 does not prove independence, and a strong r does not prove causation. Check outliers, range restriction, residual pattern and whether extrapolation leaves the observed data.

A by-eye best-fit line should pass through the mean point $(\bar x,\bar y)$. For $y=ax+b$, $a$ is the predicted change in $y$ per unit increase in $x$ and $b$ is predicted $y$ at $x=0$ when meaningful. A y-on-x regression predicts $y$ from $x$; rearranging it to predict $x$ is not generally valid. Pearson's $r$ measures only linear association.

A trial produces an outcome; the sample space $U$ lists all possible outcomes, and an event $A$ is a subset of $U$. For equally likely outcomes, $P(A)=n(A)/n(U)$ and $P(A')=1-P(A)$.

Represent $U$ with a complete list or outcome table. Relative frequency estimates probability from repeated trials and may differ from the theoretical value; with more trials it can become more stable but is not guaranteed to equal it.

Example

For two fair coins, $U=\{HH,HT,TH,TT\}$. The event of exactly one head is $\{HT,TH\}$, so its probability is $2/4=1/2$. If this probability applies to 128 trials, the expected number is $128(1/2)=64$.

The formula $n(A)/n(U)$ requires equally likely outcomes. An expected count such as 12.8 is a long-run average, not a claim that 0.8 of a person will occur in one trial.

Mutually exclusive events cannot occur together, so P(A∩B)=0. Independent events do not change one another's probability, so P(A∩B)=P(A)P(B) and P(A|B)=P(A) when defined.

Two non-impossible mutually exclusive events are automatically dependent: learning that A occurred makes B impossible. Independent events can occur together, such as separate results from a fair coin and die.

For a fair coin and die, P(head and six)=1/2×1/6=1/12. For one die roll, ‘odd’ and ‘even’ are mutually exclusive and their intersection is zero; they are not independent because P(odd|even)=0.

‘Independent’ does not mean unrelated in every philosophical sense, and ‘exclusive’ does not mean equally likely. Use the intersection or conditional rule that matches the claim.

Use a Venn diagram for overlapping sets, a tree for sequential conditions, and a table or sample-space diagram for paired outcomes. Without replacement, branch probabilities change because the remaining total and composition change; with replacement they reset. Independence must be checked from $P(A\cap B)=P(A)P(B)$ or $P(A|B)=P(A)$, not assumed from the wording.

A discrete random variable $X$ takes countable numerical values with probabilities $P(X=x)$ satisfying $0\le P(X=x)\le1$ and $\sum P(X=x)=1$. Its expected value is $E(X)=\sum xP(X=x)$.

Complete a missing probability by making the total equal 1, then multiply each outcome by its probability and add. In a gain model, positive values are winnings and negative values are losses; $E(X)=0$ describes a fair game in the long run.

Example

A game pays \4$with probability$0.2$and loses$$1$with probability$0.8$. Then$E(X)=4(0.2)-1(0.8)=0$, so it is fair by expected gain, although any individual play still wins$4 or loses $1.

Expected value need not be an attainable outcome and does not predict one play. At this SL objective, do not import variance transformations from the separate AHL random-variable content.

X~B(n,p) models the number of successes in n trials when each trial has two outcomes, trials are independent, n is fixed and the success probability p stays constant. P(X=r)=C(n,r)p^r(1−p)^(n−r).

The modelling decision comes before the calculator command. If trials affect one another, have more than two outcomes, or p changes, the binomial assumptions fail even if the question uses the word ‘success’.

If a sensor fails with probability 0.1 independently on each of 20 checks, the number of failures is B(20,0.1). ‘At least two’ means P(X≥2), not P(X=2), so the event wording controls the calculation.

A binomial distribution is not any count between 0 and n. State the trial, success definition, independence and constant-p assumption explicitly.

For $X\sim B(n,p)$, $E(X)=np$ and $Var(X)=np(1-p)$, so $SD(X)=\sqrt{np(1-p)}$. In the sensor example, $E(X)=20(0.1)=2$ failures and $Var(X)=20(0.1)(0.9)=1.8$. Use technology for exact, cumulative or interval probabilities and translate phrases such as 'at most' and 'more than' into inequalities before entering values.

A normal distribution $N(\mu,\sigma^2)$ is continuous, bell-shaped and symmetric about mean $\mu$; mean, median and mode coincide, and $\sigma$ controls spread. Total area under the curve is 1 and an exact point has probability 0.

Approximately 68%, 95% and 99.7% of values lie within $1\sigma$, $2\sigma$ and $3\sigma$ of $\mu$. Use technology for lower-tail, upper-tail and interval probabilities, and inverse normal to find a boundary from a given cumulative probability.

Example

If $X\sim N(300,40^2)$, then about 68% of observations lie from $260$ to $340$. An inverse-normal calculation for cumulative probability $0.95$ returns the value $x$ satisfying $P(X\le x)=0.95$.

At this SL objective, standardized $z$ transformation and continuity correction are not required. Check that a normal model is contextually reasonable, and enter standard deviation $\sigma$, not variance $\sigma^2$, when technology requests it.

Spearman's rank correlation converts paired observations to ranks and measures whether high ranks tend to accompany high or low ranks. With no ties, r_s=1−6Σd²/[n(n²−1)], where d is the rank difference.

Rank ties need an agreed average-rank procedure, and a significance decision depends on the sample size and test rule. The coefficient describes a monotonic pattern, not necessarily a straight-line relationship.

If study hours and score have r_s=0.80, students with higher hours generally rank higher in score. A third variable, restricted range or a few influential pairs could still explain or distort the pattern.

Ranking removes units but not bias. A high r_s is not proof of causation, and a low value can hide a curved relationship that a rank coefficient does not capture.

State $H_0$ and $H_1$, choose significance level $\alpha$, and calculate a test statistic or p-value. Reject $H_0$ when $p\le\alpha$ (or the statistic enters the supplied critical region); otherwise fail to reject $H_0$, always in context.

For a $\chi^2$ independence test, expected frequency is $(\text{row total})(\text{column total})/(\text{grand total})$ and $df=(r-1)(c-1)$. For goodness of fit at SL, $df=n-1$. Expected frequencies exceed 5 in examinations; technology gives $\chi^2$ and the upper-tail p-value.

For the SL t-test, compare two unpaired population means with unknown variance, assume normal underlying variables and equal group variances, and use the pooled two-sample test. Choose one- or two-tailed $H_1$ from the claim, not after viewing the result.

Do not say 'accept $H_0$' or treat significance as practical importance. Poisson-mean tests and Type I/II error calculations belong to AHL 4.18, not this SL Objective.