Unit S1: Statistics 1

A mathematical model is a simplified representation of a real situation. In probability and statistics, it keeps the variables and relationships needed for a purpose while replacing some real-world complexity with stated assumptions.

A model is useful because it makes a complicated situation manageable and repeatable. The same structure can be applied to new data, relationships between variables can be made clearer, and assumptions can be changed to investigate whether a conclusion is robust.

Suppose a school models student journey times using ordinary weekday travel records. The model may deliberately ignore rare road closures and assume future weekdays resemble the recorded ones. It can help describe the usual pattern and plan an arrival time. If later observations consistently disagree, the school should reconsider the data period or assumptions rather than force the situation to fit the old model.

A statistical display should make the distribution's centre, spread, shape and unusual values easier to compare. Histograms, stem-and-leaf diagrams and box plots preserve different information, so interpret the feature each display actually encodes.

For classes $10\le x<15$ with frequency $20$ and $15\le x<25$ with frequency $30$, the densities are $20/5=4$ and $30/10=3$. The first bar is taller even though the second class contains more observations. To estimate a frequency in part of a class, use the corresponding fraction of its bar area, assuming values are evenly spread within that class.

Compare distributions in context: a larger median means a larger typical value, while a larger IQR means the central half is more variable. Quote the statistics or visible features that support the comparison; do not write only that one box or bar is 'bigger'.

A measure of location describes where a distribution is centred. The mean uses every value, the median divides ordered data into two halves, and the mode is the most frequent value or category.

For grouped continuous data, use each class midpoint as its representative value, so the resulting mean is an estimate. Locate a grouped median from cumulative frequency and use linear interpolation within its class:

Here $L$ is the lower class boundary, $C$ the cumulative frequency before the class, $f$ the class frequency, $w$ its width and $pN$ the target position. For example, if $N=80$ and the median class $40\le x<50$ has $C=30$ and $f=25$, then $Q_2=40+(40-30)10/25=44$. Use the position convention specified or established by the data.

Coding reduces arithmetic. If $y=(x-a)/b$, then $x=a+by$ and$$\bar x=a+b\bar y.$$For $n=20$, $a=50$, $b=5$ and $\sum y=12$, the original mean is $50+5(12/20)=53$. Decode the final value and restore its unit.

Dispersion measures how widely observations vary. Compare spread only after checking that the groups and units are meaningful; a smaller measure indicates greater consistency, not necessarily a better outcome.

For $2,4,4,6$, the mean is $4$ and$$\operatorname{Var}(X)=\frac{2^2+4^2+4^2+6^2}{4}-4^2=2.$$Hence $sigma=\sqrt2\approx1.41$ in the data's units, while the range is $6-2=4$. Variance is not reported in the original unit because its deviations were squared.

For a linear coding $X=a+bY$,$$\operatorname{Var}(X)=b^2\operatorname{Var}(Y),\qquad \sigma_X=|b|\sigma_Y.$$Adding a constant changes location but not spread; multiplying by $b$ multiplies range, IQR and standard deviation by $|b|$. For grouped data, midpoint-based variance and standard deviation are estimates.

Skewness describes asymmetry in a distribution. Name the direction and support it with the graph or statistic requested; the long or more spread-out tail gives the direction of skew.

If a numerical measure is specified, substitute consistently and use its sign. Common forms include$$\frac{Q_3-2Q_2+Q_1}{Q_3-Q_1},\qquad \frac{3(\bar x-Q_2)}{\sigma},\qquad \frac{\bar x-\text{mode}}{\sigma}.$$A positive result indicates positive skew and a negative result negative skew; a value near zero indicates little skew for that measure.

An outlier rule is supplied when required. For the common $1.5\times\mathrm{IQR}$ rule, calculate both fences:$$Q_1-1.5\mathrm{IQR},\qquad Q_3+1.5\mathrm{IQR}.$$Values strictly beyond the fences are outliers and are plotted separately on a box plot; whiskers extend to the most extreme non-outliers under this convention.

If $Q_1=12$ and $Q_3=20$, then $mathrm{IQR}=8$ and the fences are $0$ and $32$. A value of $35$ is an upper outlier, whereas $32$ lies on the fence and is not beyond it. The calculation identifies unusual position; context is still needed to interpret it.

Probability measures how likely an event is on a scale from 0 to 1. An impossible event has probability 0, a certain event probability 1, and all probabilities for a complete set of mutually exclusive outcomes add to 1.

Define the event before counting. Check that every outcome belongs to the same sample space, that totals use the correct population or trial count, and that outcomes are genuinely equally likely before using a simple favourable-over-total ratio.

A table records 320 households, 200 with a driveway and 88 with exactly two cars. For a household chosen at random from this table,$$P(\text{driveway})=\frac{200}{320}=\frac58.$$Among the households with a driveway, suppose 40 have exactly one car. Then the joint event 'driveway and exactly one car' has probability $40/320=1/8$.

A sample space $\Omega$ contains every possible outcome. An event is a subset of that space; unions mean 'at least one', intersections mean 'both', and complements mean 'not'.

List outcomes or partition a Venn diagram into disjoint regions before adding. Every region of a Venn diagram must represent one exact combination of event membership, and all regions—including the outside region—must total 1.

Suppose $P(A)=0.55$, $P(B)=0.40$ and $P(A\cap B)=0.20$. Then$$P(A\cup B)=0.55+0.40-0.20=0.75,$$$$P((A\cup B)')=0.25,$$and, after restricting to $B$,$$P(A\mid B)=\frac{0.20}{0.40}=0.50.$$The denominator changes because the condition changes the relevant sample space.

Events $A$ and $B$ are independent when knowing that one occurred does not change the probability of the other.

To test independence, calculate both sides of one complete equality using the same data. Equality supports independence; a difference proves dependence. The conditional forms require the conditioning event to have positive probability.

For a fair six-sided die, let $A$ be 'even' and $B$ be 'a multiple of 3'. Then $P(A)=1/2$, $P(B)=1/3$ and $P(A\cap B)=P(6)=1/6$. Since $(1/2)(1/3)=1/6$, $A$ and $B$ are independent. If $C$ is 'greater than 3', then $P(A\cap C)=2/6=1/3$ but $P(A)P(C)=1/4$, so $A$ and $C$ are not independent.

Multi-stage probability uses the product law along one route and the sum law across distinct routes. A tree diagram records changing conditional probabilities; a Venn diagram partitions simultaneous event combinations.

At every node, outgoing branch probabilities total 1. Label branches with conditional probabilities for the situation at that node, not with unconditional probabilities copied from the start. In a Venn diagram, add only the disjoint regions that satisfy the event expression.

A bag contains 3 red and 2 blue counters. Two are selected without replacement. The probability of exactly one red is$$\frac35\frac24+\frac25\frac34=\frac35.$$The probability of at least one red is$$1-\frac25\frac14=\frac9{10}.$$With replacement, exactly one red instead has probability$$2\left(\frac35\right)\left(\frac25\right)=\frac{12}{25},$$because the second draw again starts from five counters.

When repeated routes have the same probability, count how many distinct orders produce the event before multiplying. When branch probabilities change without replacement, apparently similar orders may still need to be written separately so their numerators and denominators are correct.

A scatter diagram plots paired observations $(x,y)$ and reveals the direction, form and strength of their association. A linear regression line of $y$ on $x$ is the straight line used to model the average response $y$ for a given explanatory value $x$.

Calculate $S_{xy}$ and $S_{xx}$ from the data or supplied summaries, find $b$, then use the means to find $a$. The least-squares line passes through $(\bar x,\bar y)$; this gives a useful arithmetic check and a reliable point when drawing the line on the scatter diagram.

Suppose $\bar x=4$, $\bar y=11$, $S_{xx}=20$ and $S_{xy}=30$. Then $b=30/20=1.5$ and $a=11-1.5(4)=5$, so the regression line is$$y=5+1.5x.$$To draw it, calculate two fitted points within the plotted $x$-range, join them with a straight line, and check that it passes through $(4,11)$.

In a regression of $y$ on $x$, $x$ is the explanatory (independent) variable and $y$ is the response (dependent) variable. Substitute an observed-range value of $x$ to predict the corresponding $y$; use a regression of $x$ on $y$ only when predicting $x$ from $y$.

For $w=46.0+3.27h$, where height $h$ is in cm and weight $w$ is in kg, a 1 cm increase in height corresponds to an average predicted increase of $3.27$ kg in weight. A height of 153 cm may be substituted if it lies within the recorded height range; a prediction at 170 cm is extrapolation if 170 lies beyond that range.

A linear change of variable must be undone before interpreting or reporting the original quantity. For example, if $d=w/11.5$ and a fitted line is $w=-3.46t+30.9$, then dividing every term by 11.5 gives $d=-0.301t+2.69$. Keep the named variables and their units visible through the rearrangement.

The product moment correlation coefficient (PMCC), $r$, measures the direction and strength of a linear association between two quantitative variables.

If $S_{xy}=-91.55$, $S_{xx}=445.57$ and $S_{yy}=26.43$, then$$r=\frac{-91.55}{\sqrt{445.57(26.43)}}=-0.844\ \text{(approximately)}.$$In context, larger values of one variable tend to occur with smaller values of the other. Because $r$ is fairly close to $-1$, a negative linear model may be useful, but the scatter diagram should still be checked.

Adding a constant to either variable or multiplying it by a positive constant does not change $r$; multiplying one variable by a negative constant reverses the sign. This is why converting units by a positive linear coding preserves the strength and direction of correlation.

A discrete random variable assigns a numerical value to each outcome of a random process and can take only separate, countable values. Write the random variable with a capital letter, such as $X$, and a possible value with a lower-case letter, such as $x$.

If three T-shirts are selected and $X$ is their median selling price, then repeated selections may produce many ordered combinations, but $X$ can still take only the listed price values. Different combinations that give the same median belong to the same event $X=x$.

The probability function gives the probability of each possible value, $p(x)=P(X=x)$. Every probability is between 0 and 1, and the probabilities over the complete support of $X$ add to 1.

To construct a distribution, list every possible value, combine the probabilities of all disjoint routes that produce that value, and check the final probabilities total 1. To recover a probability from a cumulative distribution, subtract adjacent cumulative totals: here $P(X=1)=F(1)-F(0)=0.70-0.20=0.50$.

Translate an event before reading the table. For example, $P(3X-3<X+2)=P(X<2.5)=P(X\le2)=1$ for this integer-valued support. A non-integer cutoff is handled by including precisely the supported values that satisfy the inequality.

The expectation $E(X)$ is the probability-weighted long-run mean of a discrete random variable. Variance measures its expected squared spread about that mean.

For $X=0,1,2$ with probabilities $0.20,0.50,0.30$,$$E(X)=0(0.20)+1(0.50)+2(0.30)=1.10,$$$$E(X^2)=0^2(0.20)+1^2(0.50)+2^2(0.30)=1.70.$$Therefore $\operatorname{Var}(X)=1.70-1.10^2=0.49$, and the standard deviation is $\sqrt{0.49}=0.70$.

If $Y=5-3X$, then $E(Y)=5-3(1.10)=1.70$ and $\operatorname{Var}(Y)=(-3)^2(0.49)=4.41$. Adding a constant shifts every value without changing spread; multiplying by $a$ multiplies every deviation by $a$, so variance is multiplied by $a^2$.

A discrete uniform distribution has a finite set of possible values that are all equally likely. If there are $n$ values, each has probability $1/n$.

Suppose $B$ is equally likely to be $1,3,5$ or $7$. Then$$E(B)=\frac{1+3+5+7}{4}=4,$$$$E(B^2)=\frac{1^2+3^2+5^2+7^2}{4}=21,$$so $\operatorname{Var}(B)=21-4^2=5$. Symmetry also shows immediately that the mean is the midpoint, 4.

For equally likely consecutive integers $a,a+1,\ldots,b$, the number of values is $n=b-a+1$ and the mean is $(a+b)/2$. The variance may be found from the same $E(X^2)-[E(X)]^2$ method, which remains valid for any explicitly listed uniform support.

A Normal distribution is a continuous, bell-shaped model that is symmetric about its mean $\mu$. Write $X\sim N(\mu,\sigma^2)$, where $\sigma^2$ is the variance and $\sigma>0$ is the standard deviation. Symmetry makes the mean, median and mode coincide, and probabilities are areas under the curve.

Here $\Phi(z)=P(Z\le z)$ is read from the cumulative Normal table or calculator. For $X\sim N(210,25^2)$,$$P(190<X<240)=\Phi(1.2)-\Phi(-0.8)=0.8849-0.2119=0.6730.$$Draw the event mentally before using the table: a right tail needs a complement, while an interval needs a difference.

For an unknown boundary or parameter, turn the stated cumulative probability into its $z$-value and use $x=\mu+z\sigma$. If $P(X<152)=0.05$ and $P(X<180)=0.60$, then$$152=\mu-1.6449\sigma,\qquad180=\mu+0.2533\sigma.$$Subtracting gives $\sigma=28/1.8982\approx14.8$, then $\mu\approx176$. The same method finds percentiles, quartiles and symmetric limits.

After finding a single-observation probability $p$, ordinary probability rules still apply. For independent observations, use products or a complement such as $1-(1-p)^n$. For a conditional event, divide the joint probability by the probability of the condition; the Normal table supplies the component areas, not the final conditional ratio.