5. Probability & Statistics 1

Choose a display suited to the data: bar charts for categories, histograms for continuous intervals, and scatter plots for paired numerical variables. Axes need labels, units and honest scales.

Keep class widths visible in histograms and avoid implying continuity for categorical bars. A graph is a model of the data, not decoration.

A histogram with unequal class widths uses frequency density so each bar area represents frequency.

A histogram’s bar height is not always frequency; unequal widths require density.

A chart should encode the intended quantity with a consistent scale, labelled axes and a legend when needed. Pie charts show parts of a whole; box plots show distribution summaries.

Check that categories do not overlap, totals match the denominator and truncated axes are clearly marked. Use the chart to compare evidence, not to infer causation.

A box plot’s median line and quartiles compare centre and spread without displaying every observation.

A visually larger sector or bar is meaningful only when the scale and category totals are comparable.

Mean, median and mode describe location. Range, interquartile range, variance and standard deviation describe spread; each responds differently to outliers and skew.

Use the mean when all values and squared deviations are meaningful, the median for skewed or ordinal data, and state which spread measure matches the centre.

One extreme value can raise the mean and standard deviation while leaving the median and IQR almost unchanged.

A larger mean does not imply greater variability, and “average” is not automatically the arithmetic mean.

Cumulative frequency totals observations up to each boundary. On a cumulative-frequency graph, read the median at N/2, quartiles at N/4 and 3N/4, and use differences to estimate an interquartile range.

Use class boundaries, draw a smooth monotone curve only as an approximation, and read values against the horizontal axis carefully.

For N=80, the upper quartile is the x-value at cumulative frequency 60; IQR is Q3−Q1.

Cumulative frequency itself is not a percentile value, and a graph gives an estimate rather than an exact raw-data quartile.

For grouped data, estimate the mean with Σfx/Σf using class midpoints x, and use Σfx² to estimate variance. The result depends on treating each class as concentrated at its midpoint.

Keep a table of f, x, fx and fx², use the same units, and describe the answer as an estimate because within-class positions are unknown.

A class 10≤x<20 contributes frequency×15 to the estimated total, not frequency×10 or ×20.

Grouped-data statistics are not exact raw-data statistics; changing class widths or boundaries changes the estimate.

nPr counts ordered arrangements of r objects from n; nCr counts unordered selections. Use nPr=n!/(n−r)! and nCr=n!/[r!(n−r)!].

Decide whether positions matter before choosing a formula, and avoid counting the same outcome under different descriptions.

Choosing president and secretary from 8 people uses 8P2; choosing a two-person committee uses 8C2.

The same two people can form two ordered roles but only one unordered committee.

For arrangements with restrictions, count valid positions directly, treat linked objects as a block, or subtract forbidden arrangements from the total.

State whether objects are distinct, whether rotations are identified, and whether repetition is allowed before counting.

Arranging 5 books with two specified books together gives 4!×2!, treating the pair as one block and ordering the pair internally.

A block has internal arrangements; forgetting them undercounts, while counting them twice overcounts.

For repeated identical items, divide by factorials for duplicate arrangements; for selections with repetition, distinguish stars-and-bars style counting from ordinary combinations.

Write a small case first to check what an outcome means, then generalise. Position restrictions can change the formula entirely.

The distinct arrangements of A,A,B,C are 4!/2!, because swapping the A’s creates no new arrangement.

Treating identical objects as labelled creates artificial outcomes and inflates the count.

P(A∪B)=P(A)+P(B)−P(A∩B), P(Aᶜ)=1−P(A), and P(A|B)=P(A∩B)/P(B) when P(B)>0.

Draw a Venn diagram or tree when events overlap, and identify the denominator in conditional probability before substituting numbers.

If P(A)=0.6, P(B)=0.5 and P(A∩B)=0.2, then P(A∪B)=0.9.

Adding P(A) and P(B) without subtracting overlap can exceed 1.

Events are mutually exclusive when A∩B=∅. They are independent when P(A∩B)=P(A)P(B), equivalently P(A|B)=P(A) when defined.

Disjoint non-zero events cannot be independent: learning that one occurred makes the other impossible. Test the stated relationship with the correct equation.

A single die roll being even and odd is mutually exclusive; two independent coin tosses are not mutually exclusive across different tosses.

“Independent” does not mean unrelated in everyday language, and mutually exclusive events are not independent unless one has probability zero.

For equally likely outcomes, P(A)=number favourable/number total. For unequal outcomes, use the given probabilities and ensure they sum to one.

Define the sample space, avoid counting outcomes with different probabilities as equal, and use complements when the direct event is cumbersome.

The probability of at least one six in two fair rolls is 1−(5/6)²=11/36.

“At least one” includes two occurrences; subtracting only the probability of exactly one misses cases.

A discrete random variable takes separate values with probabilities p(x)≥0 and Σp(x)=1. Its probability table defines the sample space and supports expectation calculations.

Check the support and total probability before using formulas. A probability mass at one value is not a continuous density over an interval.

For X=0,1,2 with probabilities 0.2,0.5,0.3, the probabilities sum to 1 and describe every possible outcome.

A discrete variable can have many values; “discrete” means countable separated outcomes, not necessarily only two.

If independent trials have constant success probability p, the number X of the first success has P(X=r)=(1−p)^{r−1}p for r=1,2,… .

Confirm that trials are independent and p is constant. “First success” counts trials including the successful one; a waiting-time variant may count failures instead.

With p=0.2, P(X=4)=0.8³×0.2=0.1024.

A geometric model does not describe the number of successes in a fixed number of trials; that is binomial.

For discrete X, E(X)=Σxp(x), E(X²)=Σx²p(x), and Var(X)=E(X²)−[E(X)]². Standard deviation is the square root of variance.

Variance is measured in squared units; standard deviation returns to the original units. For aX+b, E=aE(X)+b and Var=a²Var(X).

If E(X)=4 and Var(X)=9, then Var(2X+1)=36 and E(2X+1)=9.

Adding a constant changes the mean but not the variance; multiplying by a changes variance by a², not a.

X~N(μ,σ²) is centred at μ with standard deviation σ. Standardise using Z=(X−μ)/σ, where Z~N(0,1), then use symmetry or normal tables/calculator.

State whether a probability is a lower tail, upper tail or interval. A continuous variable has P(X=a)=0, so endpoint inclusion does not change a probability.

For X~N(10,4), P(X<12)=P(Z<1)=0.8413.

σ is the standard deviation, not the variance; writing N(μ,σ) in a convention that expects variance changes the model.

Use the normal CDF for P(X≤x), complements for upper tails and differences for intervals. Inverse normal questions solve for x from a stated probability.

Draw the tail or interval, standardise consistently and undo the transformation only after finding the z-value. Check whether the resulting value lies in the physical domain.

The central 90% interval for a normal variable uses z≈±1.645, so bounds are μ±1.645σ.

“At least” and “more than” are upper tails, while “between” requires subtracting two cumulative probabilities.

If X counts successes in n independent trials with constant success probability p, X~B(n,p) and P(X=r)=C(n,r)p^r(1−p)^{n−r}.

Check fixed n, two outcomes per trial, independence and constant p. Use complements or cumulative probabilities for “at least” and “at most” wording.

For X~B(10,0.3), P(X=2)=C(10,2)0.3²0.7⁸.

Binomial trials need not be equally likely across outcomes, but p must remain constant; sampling without replacement may break independence.