Statistics and Probability

Statistics and probability carry about 3 questions (our estimate).

The booklet's model question — a lottery with 10 prizes and 25 blanks — shows the usual level: a direct probability, a conditional or Bayes calculation, a binomial, or the mean / standard deviation / regression of a small data set.

All are computed from short formulas; the risk is arithmetic, not concepts.

7.1 Measures of location and dispersion

Location. Mean \bar x=\frac{\sum x}{n} (grouped: \frac{\sum fx}{\sum f}, or step deviation A+h\frac{\sum fu}{\sum f}). Median: middle value of ordered data (average of the two middle ones if n is even); grouped: L+\dfrac{n/2-F}{f}h.

Mode: the most frequent value; grouped: L+\dfrac{f_1-f_0}{2f_1-f_0-f_2}h. Empirical relation \text{mode}=3\,\text{median}-2\,\text{mean}. Quartiles Q_1,Q_2,Q_3; n-th percentile.

Combined mean of two groups: \dfrac{n_1\bar x_1+n_2\bar x_2}{n_1+n_2}. For y=ax+b: \bar y=a\bar x+b.

Dispersion.

• Range = max - min; quartile deviation \frac{Q_3-Q_1}{2}; mean deviation \frac{\sum|x-\bar x|}{n}.
• Variance \sigma^2=\dfrac{\sum(x-\bar x)^2}{n}=\dfrac{\sum x^2}{n}-\bar x^2; standard deviation \sigma=\sqrt{\sigma^2}.
• For y=ax+b: \sigma_y=|a|\sigma_x (adding a constant does not change the spread).
• Coefficient of variation \text{CV}=\dfrac{\sigma}{\bar x}\times100\%: a smaller CV means a more consistent set.
• The sample variance divides by n-1; the paper's data sets are treated as the whole population unless stated.

7.2 Correlation and regression

Karl Pearson's coefficient r=\dfrac{\text{cov}(x,y)}{\sigma_x\sigma_y}=\dfrac{n\sum xy-\sum x\sum y}{\sqrt{n\sum x^2-(\sum x)^2}\sqrt{n\sum y^2-(\sum y)^2}}, with -1\le r\le1 (r=\pm1 perfect linear, 0 no linear relation).

Spearman's rank \rho=1-\dfrac{6\sum d^2}{n(n^2-1)}.

Regression lines. y on x: y-\bar y=b_{yx}(x-\bar x) with b_{yx}=r\dfrac{\sigma_y}{\sigma_x}; x on y: x-\bar x=b_{xy}(y-\bar y), b_{xy}=r\dfrac{\sigma_x}{\sigma_y}.

Both lines pass through (\bar x,\bar y), so the means are found by solving the two lines together. b_{yx}b_{xy}=r^2, and b_{yx},b_{xy},r share one sign.

The angle between the lines is \tan\theta=\dfrac{1-r^2}{|r|}\cdot\dfrac{\sigma_x\sigma_y}{\sigma_x^2+\sigma_y^2}.

7.3 Probability: rules, conditional, Bayes, binomial

P(A)=\dfrac{\text{favourable}}{\text{total}} for equally likely outcomes; 0\le P\le1; P(A')=1-P(A). Two dice: 36 outcomes; sum 7 in 6 of them (\frac{1}{6}); sum \ge10 in 6 (\frac{1}{6}). A leap year has 52 weeks and 2 extra days; P(53\text{ Sundays})=\frac{2}{7}.

• Addition: P(A\cup B)=P(A)+P(B)-P(A\cap B); for mutually exclusive events the last term is 0.
• Conditional: P(A|B)=\dfrac{P(A\cap B)}{P(B)}. Multiplication: P(A\cap B)=P(A)P(B|A).
• Independent: P(A\cap B)=P(A)P(B). (Mutually exclusive events with non-zero probabilities are never independent.)
• Total probability P(B)=\sum P(A_i)P(B|A_i). Bayes: P(A_k|B)=\dfrac{P(A_k)P(B|A_k)}{\sum P(A_i)P(B|A_i)}.
• Without replacement: two reds from 3 red, 5 blue: \frac{3}{8}\cdot\frac{2}{7}=\frac{3}{28}.

Binomial distribution

For n independent trials with success probability p (q=1-p): P(X=r)={}^nC_rp^rq^{n-r}; mean np, variance npq, SD \sqrt{npq}; "at least one success" =1-q^n. Expected value E(X)=\sum xP(x). Four tosses of a fair coin, exactly two heads: \frac{6}{16}=\frac{3}{8}.

(a) A Galton board, the pegs and the bins

(b) 10 rows, 2,000 balls: simulation vs C(10,k)/1024

Chapter summary

• Mean, median, mode answer "where"; \sigma and CV answer "how spread out"; spread ignores a shift.
• Both regression lines meet at the means; r^2=b_{yx}b_{xy}.
• For "at least one", use the complement.
• Bayes' theorem reverses a conditional: weigh each cause by prior × likelihood, then normalise.
• Binomial needs a fixed n, independent trials and a constant p.

Practice questions

Set A — Recall (10 questions)

1. The mean of 2,4,6,8,10 isa5b6c7d8
2. The median of 3,7,5,9,11 isa5b9c7d6.5
3. If the variance of a data set is 16, its standard deviation isa4b8c16d2
4. If the mean is 50 and the standard deviation 5, the coefficient of variation isa1%b10%c0.1%d25%
5. Karl Pearson's coefficient of correlation lies betweena0 and 1b0 and \inftyc-1 and 1d-\infty and \infty
6. When two dice are thrown, the probability that the sum is 7 isa1/12b1/9c5/36d1/6
7. If P(A)=0.5, P(B)=0.4 and P(A\cap B)=0.2, then P(A\cup B) isa0.7b0.9c0.1d0.2
8. The mean of a binomial distribution with n=10, p=0.3 isa7b2.1c3d0.3
9. If A and B are independent, P(A)=\frac{1}{2} and P(B)=\frac{1}{3}, then P(A\cap B) isa1/5b2/3c5/6d1/6
10. If b_{yx}=0.8 and b_{xy}=0.45, the correlation coefficient r isa0.36b0.6c0.8d0.45

Answers and explanations — Set A

1. (b) — 30/5.

2. (c) — ordered: 3, 5, 7, 9, 11; the middle one is 7.

3. (a) — \sqrt{16}.

4. (b) — \frac{5}{50}\times100.

5. (c) — |r|\le1.

6. (d) — six of 36 outcomes: \frac{1}{6}.

7. (a) — 0.5+0.4-0.2.

8. (c) — np=3 (variance would be 2.1).

9. (d) — multiply: \frac{1}{2}\cdot\frac{1}{3}.

10. (b) — r^2=0.8\times0.45=0.36, r=0.6 (both coefficients positive).

Set B — Understanding (7 questions)

1. The mean of 10 observations is 20. Each is multiplied by 3 and then 5 is added. The new mean isa60b65c75d105
2. Five observations have mean 4 and another ten have mean 7. The combined mean isa5.5b5c6d6.5
3. If the mean of a moderately skewed distribution is 24 and the median is 22, the mode is abouta20b22c24d18
4. For n=5 and \sum d^2=10, Spearman's rank correlation isa0.5b0.4c0.6d-0.5
5. A bag has 3 red and 5 blue balls. Two are drawn without replacement. The probability that both are red isa9/64b3/28c3/8d1/4
6. Machine A makes 60% of the items, 2% of them defective; machine B makes 40%, 5% defective. A defective item is picked. The probability it came from A isa3/5b5/8c3/8d1/4
7. A fair coin is tossed 4 times. The probability of exactly two heads isa3/8b1/4c1/2d5/16

Answers and explanations — Set B

1. (b) — 3\times20+5=65.

2. (c) — \frac{5\cdot4+10\cdot7}{15}=6.

3. (d) — 3(22)-2(24)=18.

4. (a) — 1-\frac{6\cdot10}{5\cdot24}=0.5.

5. (b) — \frac{3}{8}\times\frac{2}{7}=\frac{3}{28}. (a) is the with-replacement value.

6. (c) — \frac{0.6\times0.02}{0.6\times0.02+0.4\times0.05}=\frac{0.012}{0.032}=\frac{3}{8}.

7. (a) — ^4C_2/16=\frac{6}{16}.

Set C — Application (6 questions)

1. The standard deviation of 2,4,4,4,5,5,7,9 isa4b2c\sqrt{2}d3
2. The regression lines are 3x+2y=26 and 6x+y=31. The means (\bar x,\bar y) area(4,7)b(7,4)c(5,6)d(3,8)
3. A die is rolled twice. The probability that the sum is at least 10 isa1/12b1/9c1/6d5/36
4. A and B are independent, P(A)=0.4 and P(A\cup B)=0.7. Then P(B) isa0.3b0.5c0.75d0.35
5. A fair coin is tossed three times. The probability of at least one head isa3/8b1/2c1/8d7/8
6. The probability that a randomly chosen leap year has 53 Sundays isa1/7b3/7c2/7d5/7

Answers and explanations — Set C

1. (b) — mean 5, \sum(x-5)^2=32, variance 4.

2. (a) — from y=31-6x: 3x+62-12x=26\Rightarrow x=4, y=7.

3. (c) — (4,6),(5,5),(6,4),(5,6),(6,5),(6,6): \frac{6}{36}.

4. (b) — 0.7=0.4+P(B)-0.4P(B)\Rightarrow0.6P(B)=0.3.

5. (d) — 1-(\frac{1}{2})^3.

6. (c) — 366 days = 52 weeks + 2 days; the pair contains a Sunday in 2 of 7 equally likely cases.

Group D — Model answers for long questions (10 marks each, for deeper revision)

1. Find the mean, median, mode, variance and standard deviation of grouped data, and the regression line and correlation of paired data. [10]
2. Solve problems using Bayes' theorem and the binomial distribution. [10]

Model answer D1: Dispersion, correlation and regression (10 marks)

Introduction. Averages describe the centre of data and the standard deviation describes its spread. The coefficient of variation is CV=\frac{\sigma}{\bar x}\times100\%.

Raw data: 2,4,4,4,5,5,7,9. Mean =\frac{40}{8}=5; median =4.5; mode =4; the squared deviations are 9,1,1,1,0,0,4,16 (sum 32), so variance =\frac{32}{8}=4, \sigma=2, and CV=40\%.

Grouped data (50 values):

Class	f	Mid x	fx	fx^2	Cumulative f
0–10	5	5	25	125	5
10–20	8	15	120	1800	13
20–30	15	25	375	9375	28
30–40	16	35	560	19600	44
40–50	6	45	270	12150	50

1. Mean =\frac{1350}{50}=27.
2. Variance =\frac{43050}{50}-27^2=861-729=132, so \sigma=\sqrt{132}=11.49.
3. Median: \frac N2=25 falls in 20–30, so median =20+\frac{25-13}{15}\times10=28.
4. Mode: modal class 30–40, so mode =30+\frac{16-15}{2(16)-15-6}\times10=30.91.

Correlation and regression. For x=1,2,3,4,5 and y=2,4,5,4,5: \bar x=3, \bar y=4, \sum(x-\bar x)(y-\bar y)=6, \sum(x-\bar x)^2=10, \sum(y-\bar y)^2=6.

1. r=\frac{6}{\sqrt{10\times6}}=0.775.
2. Slope b_{yx}=\frac{6}{10}=0.6; the line is y-4=0.6(x-3), i.e. y=0.6x+2.2.
3. Prediction at x=6: y=5.8.

Conclusion. Spread and relationship are summarised by \sigma, r and the regression line, all computed from simple sums.

Model answer D2: Probability, Bayes' theorem and the binomial distribution (10 marks)

Introduction. P(A\cup B)=P(A)+P(B)-P(A\cap B), P(A\mid B)=\frac{P(A\cap B)}{P(B)}, and independent events satisfy P(A\cap B)=P(A)P(B).

Bayes' theorem. Machines A, B, C make 50\%, 30\% and 20\% of the items, and 1\%, 2\% and 3\% of their items are defective. An item is found defective: which machine made it?

1. P(D)=0.5(0.01)+0.3(0.02)+0.2(0.03)=0.005+0.006+0.006=0.017 (total probability).
2. P(A\mid D)=\frac{0.005}{0.017}=0.294.
3. P(B\mid D)=\frac{0.006}{0.017}=0.353.
4. P(C\mid D)=\frac{0.006}{0.017}=0.353.
5. The three posterior probabilities add up to 1, and machine A, though it makes half the items, is least likely to be the source.

Binomial distribution. P(X=r)={}^nC_rp^rq^{n-r}, with mean np and variance npq.

Case	Calculation	Value
10 fair-coin tosses, exactly 5 heads	\frac{{}^{10}C_5}{1024}=\frac{252}{1024}	0.246
At least 8 heads	\frac{45+10+1}{1024}	0.0547
n=6, p=\frac13: mean	np	2
n=6, p=\frac13: variance	npq=6\cdot\frac13\cdot\frac23	\frac43

Conclusion. Bayes' theorem reverses a conditional probability using the total probability, and the binomial model describes repeated independent trials.