5. Probability & Statistics 1

Identify variable type, sample size, grouping and comparison question. Choose a display whose encoding answers that question, label scale/units/key and state one relevant advantage and disadvantage.

For raw numerical data, a stem-and-leaf diagram may preserve more information than grouped displays. Grouping improves compactness but loses within-class positions.

Use the same scale/class boundaries where direct comparison is intended; otherwise visual differences may be caused by encoding choices.

For back-to-back stems, use one shared key and order leaves away from the stem consistently on each side so both datasets can be reconstructed.

State comparisons using numerical features: median/centre, IQR/range, shape or modal class. Label axes, units and sample identity.

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.

To estimate the number/proportion below value $x$, read cumulative count $C(x)$, then use $C(x)/N$. Above $x$: $[N-C(x)]/N$.

With $N=80$, $Q_3$ is the x-value at cumulative frequency $60$. If $C(50)=62$ and $C(30)=18$, the estimated proportion between $30$ and $50$ is $(62-18)/80=0.55$.

Build/use $n,\sum x,\sum x^2$ (or $\sum f,\sum fx,\sum fx^2$). Given totals can be inserted directly; do not reconstruct fictional raw values.

For two datasets, add $n$, $\sum x$ and $\sum x^2$, then recompute the combined mean and SD. Do not average the two means/SDs unless sample sizes and formulas justify it.

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.

For 5 distinct books with two specified together: $4!\times2!$. For them not together: $5!-4!\times2!$. State whether ends/particular seats create separate cases.

If rows or seats are distinguishable, allocate occupants with combinations or direct seat choices, then arrange each row. Avoid counting the same final seating through different allocation orders.

For two fair dice, use 36 ordered pairs. A total of 8 occurs for $(2,6),(3,5),(4,4),(5,3),(6,2)$, so the probability is $5/36$.

From 5 red and 3 blue balls, two chosen together have $\binom82$ equiprobable unordered pairs. Exactly one of each has $\binom51\binom31$ pairs, giving $15/28$.

A bag has 3 red and 2 blue balls. Two are drawn without replacement. The path red then blue has probability $3/5\times2/4=3/10$; blue then red has probability $2/5\times3/4=3/10$. Therefore one of each has probability $3/10+3/10=3/5$.

Without replacement, update both the favourable count and total after the first draw. With replacement, the second-stage probabilities return to their original values.

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.

A fair die is known to show more than 2. The restricted outcomes are $\{3,4,5,6\}$. Given this information, the probability of an even result is $2/4=1/2$, not $3/6$.

A bag contains 5 red and 3 blue balls. If the first ball drawn without replacement is known to be red, then $P(\text{second red}\mid\text{first red})=4/7$. Equivalently, a joint probability can be built as $P(A\cap B)=P(B)P(A\mid B)$.

Thus $E(X)=1.1$, $E(X^2)=1.7$ and $\operatorname{Var}(X)=1.7-1.1^2=0.49$. A probability table may come from enumeration or from combining outcomes that give the same value of $X$.

If 8 independent items are defective with probability 0.1, the probability of exactly 2 defective items is $\binom82(0.1)^2(0.9)^6$. For ranges such as at least 2, sum the relevant values or use a shorter complement.

If each attempt succeeds with probability 0.2, then the first success on attempt 4 has probability $(0.8)^3(0.2)$. ‘After attempt 4’ means four failures, so its probability is $(0.8)^4$.

These are long-run summaries, not necessarily possible single outcomes. A binomial expectation can be non-integer; a geometric expectation is the average trial number on which the first success occurs.

For $X\sim B(20,0.3)$, $E(X)=20(0.3)=6$ and $\operatorname{Var}(X)=20(0.3)(0.7)=4.2$. The standard deviation, if requested, is $\sqrt{4.2}$.

For $Y\sim Geo(0.2)$, $E(Y)=1/0.2=5$: over many repetitions, the first success occurs on trial 5 on average.

For $X\sim N(10,4)$, the standard deviation is 2, so $P(X<12)=P(Z<(12-10)/2)=P(Z<1)=0.8413$.

If $X\sim N(50,6^2)$, then $$P(X>62)=P\left(Z>\frac{62-50}{6}\right)=P(Z>2)=1-0.9772=0.0228.$$ This displays the full standardisation required.

If $P(X<x)=0.90$, tables give $z\approx1.282$, so $(x-\mu)/\sigma=1.282$ and $x=\mu+1.282\sigma$. If $x$ is known, the same equation gives a relationship between $\mu$ and $\sigma$.

For a central probability, split the excluded area equally between two tails before finding $z$. For example, central 90% leaves 0.05 per tail and uses bounds $\mu\pm1.645\sigma$.

If $X\sim B(100,0.4)$, then $np=40$ and $nq=60$, so the conditions hold and $Y\sim N(40,24)$. To approximate $P(X\le45)$, use $$P(Y<45.5)=P\left(Z<\frac{45.5-40}{\sqrt{24}}\right).$$

The correction gives each integer value its full unit-wide bar: the bar for $X=45$ extends from 44.5 to 45.5. Sketching the included integer bars makes the correct boundary visible.