Statistics and probability flashcards

A population is every member of the group being studied. A sample is a smaller subset chosen to represent it. Sample statistics are used to estimate population values.

Simple random: every member equally likely. Systematic: every \(k\)th member. Stratified: proportional from each subgroup. Quota: set numbers per group, chosen non-randomly. Convenience: easiest to reach, often biased.

Discrete data can take only separate values, usually counts (shoes owned). Continuous data can take any value in a range, usually measurements (height, time).

When the data is skewed or has outliers. The median is not affected by extreme values, while the mean is pulled towards them.

\(Q_1\), \(Q_2\), \(Q_3\) split ordered data into quarters; \(IQR=Q_3-Q_1\). A value is an outlier if it is below \(Q_1-1.5\times IQR\) or above \(Q_3+1.5\times IQR\).

Minimum (or lowest non-outlier), \(Q_1\), median, \(Q_3\) and maximum (or highest non-outlier). Outliers are marked separately as crosses.

Plot cumulative frequency against the upper class boundary. For \(n\) values, read across at \(\frac12n\) for the median, \(\frac14n\) for \(Q_1\), \(\frac34n\) for \(Q_3\), and \(\frac{p}{100}n\) for the \(p\)th percentile, then read down to the axis.

It measures spread about the mean. Adding \(k\) leaves it unchanged. Multiplying every value by \(k\) multiplies it by \(|k|\) (the variance by \(k^2\)). The mean changes with both.

Mean \(=5\). Squared deviations: 9, 1, 1, 1, 16 sum to 28. Variance \(=\frac{28}{5}=5.6\), so \(\sigma=\sqrt{5.6}\approx2.37\).

\(r\) lies between \(-1\) and \(1\). Its sign gives the direction of the linear relationship and its size gives the strength (close to \(\pm1\) is strong, near 0 is weak). It does not show causation.

It models \(y\) as a linear function of \(x\), to predict \(y\) from a given \(x\). It passes through the mean point \((\bar x,\bar y)\).

Interpolation predicts within the range of the data, where the trend is observed. Extrapolation predicts outside it, assuming the trend continues, which may not hold.

The gradient \(m\) is the average change in \(y\) for each one-unit increase in \(x\). The intercept \(c\) is the predicted \(y\) when \(x=0\), meaningful only if \(x=0\) makes sense in context.

\(0\le P(A)\le1\), and \(P(A')=1-P(A)\). For equally likely outcomes, \(P(A)=\frac{n(A)}{n(U)}\).

\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). For mutually exclusive events, \(P(A\cap B)=0\) so it simplifies to \(P(A)+P(B)\).

\(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\): the probability of \(A\) given that \(B\) has occurred. On a tree diagram it is the probability on the second branch.

Independent: one does not affect the other, \(P(A\cap B)=P(A)P(B)\). Mutually exclusive: they cannot both occur, \(P(A\cap B)=0\). Events with non-zero probabilities cannot be both.

The probabilities on the second set of branches depend on the first outcome, because the numbers remaining have changed. Multiply along branches, add across paths.

Each \(P(X=x)\) lies between 0 and 1 and \(\sum P(X=x)=1\). The mean is \(E(X)=\sum x\,P(X=x)\).

Fixed number of trials \(n\); two outcomes (success or failure); constant probability \(p\) of success; independent trials. Mean \(np\), variance \(np(1-p)\).

\(P(X=2)=\binom52(0.4)^2(0.6)^3=10\times0.16\times0.216=0.3456\).

Symmetric bell shape about the mean \(\mu\), with mean = median = mode. About 68% of values lie within \(1\sigma\), 95% within \(2\sigma\) and 99.7% within \(3\sigma\) of the mean.

\(P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}\), where \(P(B)=P(B\mid A)P(A)+P(B\mid A')P(A')\).

\(P(+)=0.95(0.01)+0.05(0.99)=0.059\). \(P(D\mid+)=\frac{0.0095}{0.059}\approx0.161\). Most positives are false.

\(\mathrm{Var}(X)=E(X^2)-[E(X)]^2\), where \(E(X^2)=\sum x^2P(X=x)\). Equivalently \(\sum(x-\mu)^2P(X=x)\). The standard deviation is its square root.

\(E(aX+b)=aE(X)+b\) and \(\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)\). Adding the constant \(b\) shifts the mean but does not change the spread.

\(f(x)\ge0\) for all \(x\) and \(\int_{-\infty}^{\infty}f(x)\,dx=1\) (total area is 1). Probabilities are areas: \(P(a

\(\int_0^2kx\,dx=2k=1\), so \(k=\frac12\). \(E(X)=\int_0^2x\cdot\frac x2\,dx=\left[\frac{x^3}{6}\right]_0^2=\frac43\).

Mean \(E(X)=\int x f(x)\,dx\). Median \(m\) solves \(\int_{-\infty}^{m}f(x)\,dx=\frac12\). Mode is the \(x\)-value where \(f(x)\) is greatest.