Secondary 4 Additional Mathematics Statistics Probability Quiz

Secondary 4 Additional Mathematics Quiz - Statistics Probability

Answer Key and Marking Scheme

Total Marks: 40 marks
Duration: 40 minutes

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Section A: Standard Questions [1 – 10]

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Question 1 [2 marks]

(a) $P(\text{red}) = \frac{5}{5+7+3} = \frac{5}{15} = \frac{1}{3}$ [1 mark]

Teaching note: The probability of an event equals the number of favorable outcomes divided by the total number of equally likely outcomes. Here there are 15 marbles in total and 5 are red.

(b) $P(\text{not blue}) = 1 - P(\text{blue}) = 1 - \frac{7}{15} = \frac{8}{15}$ [1 mark]

Alternative: Count non-blue marbles directly: $\frac{5+3}{15} = \frac{8}{15}$

Common mistake: Forgetting that "not blue" includes both red and green marbles.

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Question 2 [2 marks]

(a) $P(\text{both}) = P(M) \times P(S) = \frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}$ [1 mark]

Teaching note: For independent events, $P(A \cap B) = P(A) \times P(B)$. The occurrence of one does not affect the other.

(b) $P(\text{fails at least one}) = 1 - P(\text{both pass}) = 1 - \frac{1}{2} = \frac{1}{2}$ [1 mark]

Alternative using complement: "At least one fail" is the complement of "both pass."

Common mistake: Trying to enumerate all cases (fail M pass S, pass M fail S, fail both) without using the complement rule, which is longer but also valid: $P(\text{fail M}) \times P(\text{pass S}) + P(\text{pass M}) \times P(\text{fail S}) + P(\text{fail M}) \times P(\text{fail S})$ $= \frac{1}{4} \times \frac{2}{3} + \frac{3}{4} \times \frac{1}{3} + \frac{1}{4} \times \frac{1}{3} = \frac{2+3+1}{12} = \frac{6}{12} = \frac{1}{2}$

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Question 3 [2 marks]

Sample space: 36 equally likely outcomes $(1,1)$ to $(6,6)$ when two dice are rolled.

(a) Sum equals 7: Favorable outcomes are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — six outcomes. [0.5 mark for identifying correct pairs]

$P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$ [0.5 mark]

(b) First score greater than second: Count outcomes where first > second:

• First = 2: second = 1 (1 way)
• First = 3: second = 1, 2 (2 ways)
• First = 4: second = 1, 2, 3 (3 ways)
• First = 5: second = 1, 2, 3, 4 (4 ways)
• First = 6: second = 1, 2, 3, 4, 5 (5 ways)

Total: $1+2+3+4+5 = 15$ outcomes [0.5 mark]

$P(\text{first} > \text{second}) = \frac{15}{36} = \frac{5}{12}$ [0.5 mark]

Teaching note: By symmetry, $P(\text{first} > \text{second}) = P(\text{second} > \text{first})$, and $P(\text{equal}) = \frac{6}{36} = \frac{1}{6}$. So each of the first two probabilities equals $\frac{1}{2}\left(1-\frac{1}{6}\right) = \frac{5}{12}$.

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Question 4 [2 marks]

Using the principle of inclusion-exclusion: $|P \cup C| = |P| + |C| - |P \cap C| = 45 + 30 - 15 = 60$ students study Physics or Chemistry or both. [0.5 mark for method]

(a) $P(\text{Physics or Chemistry}) = \frac{60}{80} = \frac{3}{4}$ [0.5 mark]

(b) $P(\text{neither}) = 1 - \frac{60}{80} = \frac{20}{80} = \frac{1}{4}$ [1 mark]

Alternative for (b): Or count directly: $80 - 60 = 20$ students study neither.

Teaching note: Venn diagrams help visualize this. The region "neither" is outside both circles.

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Question 5 [2 marks]

(a) Using $P(A \cup B) = P(A) + P(B) - P(A \cap B)$: [0.5 mark for formula]

$0.7 = 0.4 + 0.5 - P(A \cap B)$

$P(A \cap B) = 0.9 - 0.7 = 0.2$ [0.5 mark]

(b) For independence, check if $P(A \cap B) = P(A) \times P(B)$ [0.5 mark for test]

$P(A) \times P(B) = 0.4 \times 0.5 = 0.2$ [0.25 mark]

Since $P(A \cap B) = 0.2 = P(A) \times P(B)$, events $A$ and $B$ are independent. [0.25 mark]

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Question 6 [2 marks]

(a) $P(X \leq 3) = P(X=1) + P(X=2) + P(X=3) = \frac{1}{15} + \frac{2}{15} + \frac{3}{15} = \frac{6}{15} = \frac{2}{5}$ [1 mark]

(b) $E(X) = \sum x \cdot P(X=x) = 1 \times \frac{1}{15} + 2 \times \frac{2}{15} + 3 \times \frac{3}{15} + 4 \times \frac{4}{15} + 5 \times \frac{5}{15}$ [0.5 mark for setting up]

$= \frac{1 + 4 + 9 + 16 + 25}{15} = \frac{55}{15} = \frac{11}{3}$ or $3\frac{2}{3}$ or $3.67$ (3 s.f.) [0.5 mark]

Teaching note: The expected value is the weighted average of all possible values, where weights are probabilities. It represents the long-run average if the experiment were repeated many times.

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Question 7 [2 marks]

(a) Modal class is $160 \leq h < 170$ (highest frequency of 64) [1 mark]

(b) Using mid-values: 145, 155, 165, 175, 185 [0.5 mark for mid-values]

Estimated mean $= \frac{24 \times 145 + 56 \times 155 + 64 \times 165 + 44 \times 175 + 12 \times 185}{200}$ [0.25 mark for formula]

$= \frac{3480 + 8680 + 10560 + 7700 + 2220}{200} = \frac{32640}{200} = 163.2$ cm [0.25 mark]

Teaching note: For grouped data, we use mid-values as estimates since we don't know exact values. This gives an estimate, not the exact mean.

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Question 8 [2 marks]

(a) Mean $\bar{x} = \frac{\sum x}{n} = \frac{250}{10} = 25$ [1 mark]

(b) Variance $= \frac{\sum x^2}{n} - \bar{x}^2 = \frac{6750}{10} - 25^2 = 675 - 625 = 50$ [0.5 mark]

Standard deviation $= \sqrt{50} = 5\sqrt{2} \approx 7.07$ (3 s.f.) [0.5 mark]

Teaching note: Standard deviation measures spread. The formula $\sigma = \sqrt{\frac{\sum(x-\bar{x})^2}{n}}$ is equivalent to $\sqrt{\frac{\sum x^2}{n} - \bar{x}^2}$ but the second is computationally easier when you have $\sum x^2$ and $\sum x$.

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Question 9 [2 marks]

(a) $E(X) = (-2) \times \frac{1}{4} + 0 \times \frac{1}{8} + 2 \times \frac{3}{8} + 4 \times \frac{1}{4}$ [0.5 mark]

$= -\frac{1}{2} + 0 + \frac{3}{4} + 1 = \frac{-2+0+3+4}{4} = \frac{5}{4} = 1.25$ or $1\frac{1}{4}$ [0.5 mark]

(b) $E(X^2) = (-2)^2 \times \frac{1}{4} + 0^2 \times \frac{1}{8} + 2^2 \times \frac{3}{8} + 4^2 \times \frac{1}{4}$ [0.5 mark]

$= 4 \times \frac{1}{4} + 0 + 4 \times \frac{3}{8} + 16 \times \frac{1}{4} = 1 + 0 + 1.5 + 4 = 6.5$ or $\frac{13}{2}$ [0.5 mark]

Teaching note: $E(X^2)$ is needed to find variance: $Var(X) = E(X^2) - [E(X)]^2$. Be careful: $E(X^2) \neq [E(X)]^2$.

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Question 10 [2 marks]

Given transformation $y = 2x + 5$:

(a) New mean $\bar{y} = 2\bar{x} + 5 = 2(12) + 5 = 29$ [1 mark]

(b) New standard deviation: Adding a constant (5) doesn't change spread, but multiplying by 2 scales spread by 2.

New standard deviation $= 2 \times 3 = 6$ [1 mark]

Teaching note: For $y = ax + b$: $E(Y) = aE(X) + b$ and $SD(Y) = |a| \times SD(X)$. The shift $b$ affects location (mean) but not spread (standard deviation). The scale factor $a$ affects both.

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Section B: Intermediate Questions [11 – 15]

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Question 11 [2 marks]

(a) $P(\text{both aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$ [1 mark]

Teaching note: This uses the multiplication rule for dependent events. After drawing one ace, only 3 aces remain from 51 cards.

(b) $P(\text{different suits}) = 1 - P(\text{same suit})$ [0.25 mark for strategy]

$P(\text{same suit}) = \frac{52}{52} \times \frac{12}{51} = \frac{12}{51} = \frac{4}{17}$ (first card any suit, second must match) [0.5 mark]

$P(\text{different suits}) = 1 - \frac{4}{17} = \frac{13}{17}$ [0.25 mark]

Alternative for (b): Direct count: $\frac{52 \times 39}{52 \times 51} = \frac{39}{51} = \frac{13}{17}$

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Question 12 [2 marks]

Let $D$ = "item is defective".

(a) Using law of total probability: [0.25 mark for method]

$P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)$

$= 0.02 \times 0.30 + 0.03 \times 0.50 + 0.05 \times 0.20$ [0.25 mark]

$= 0.006 + 0.015 + 0.010 = 0.031$ [0.5 mark]

(b) Using Bayes' theorem: $P(B|D) = \frac{P(D|B)P(B)}{P(D)}$ [0.25 mark for formula]

$= \frac{0.03 \times 0.50}{0.031} = \frac{0.015}{0.031} = \frac{15}{31} \approx 0.484$ (3 s.f.) [0.5 mark]

Teaching note: This is a classic Bayesian probability problem. Machine $B$ produces the most output, so even with moderate defect rate, it contributes significantly to defective items.

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Question 13 [2 marks]

Expected visual features for answer verification: The ogive passes through points (20, 8), (40, 32), (60, 78), (80, 108), (100, 120). Cumulative frequency ranges from 0 to 120.

(a) Median: $\frac{120}{2} = 60$th value. From curve, estimate mark ≈ 58±2 (reading across from 60 on y-axis to curve, down to x-axis) [0.67 mark, accept 56-60]

(b) Lower quartile $Q_1$: 30th value ≈ 42±2. Upper quartile $Q_3$: 90th value ≈ 72±2. [0.33 mark for each, accept range]

IQR = $Q_3 - Q_1 \approx 72 - 42 =$ 30±4 [0.33 mark]

(c) Students scoring > 70: Read cumulative frequency at mark = 70, approximately 95 students. [0.33 mark for method]

Students scoring more than 70 = $120 - 95 = 25$ students [0.17 mark]

Percentage = $\frac{25}{120} \times 100\% \approx 20.8\%$ or about 21% [0.17 mark, accept 15%-25%]

Teaching note: The interquartile range measures middle 50% spread. Reading from cumulative frequency curves requires careful estimation; small variations accepted due to graph reading.

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Question 14 [2 marks]

(a) Using $\sum P(X=x) = 1$: $0.1 + p + 0.3 + q + 0.2 = 1$, so $p + q = 0.4$ [0.25 mark]

Using $E(X) = 2.2$: $0 \times 0.1 + 1 \times p + 2 \times 0.3 + 3 \times q + 4 \times 0.2 = 2.2$ [0.25 mark] $p + 0.6 + 3q + 0.8 = 2.2$ $p + 3q = 0.8$ [0.25 mark]

Solving: $(p + 3q) - (p + q) = 0.8 - 0.4 = 0.4$, so $2q = 0.4$, thus $q = 0.2$ and $p = 0.2$ [0.25 mark]

(b) $E(X^2) = 0^2 \times 0.1 + 1^2 \times 0.2 + 2^2 \times 0.3 + 3^2 \times 0.2 + 4^2 \times 0.2$ [0.25 mark]

$= 0 + 0.2 + 1.2 + 1.8 + 3.2 = 6.4$ [0.25 mark]

$Var(X) = E(X^2) - [E(X)]^2 = 6.4 - 2.2^2 = 6.4 - 4.84 =$ $1.56$ [0.5 mark]

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Question 15 [2 marks]

Given: $X \sim N(150, 20^2)$

(a) $Z = \frac{130-150}{20} = -1$ [0.25 mark]

$P(X < 130) = P(Z < -1) = 1 - 0.8413 = 0.1587$ [0.25 mark]

Number of apples = $500 \times 0.1587 =$ 79 or 80 apples [0.5 mark, accept 79-80]

(b) $P(X > k) = 0.05$, so $P(X \leq k) = 0.95$ [0.25 mark]

$P(Z \leq z) = 0.95$ gives $z \approx 1.645$ [0.25 mark]

$\frac{k - 150}{20} = 1.645$

$k = 150 + 32.9 =$ $182.9$ g or 183 g (3 s.f.) [0.25 mark for equation, 0 mark already counted]

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Section C: Advanced Questions [16 – 20]

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Question 16 [2 marks]

(a) No restrictions: $\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$ ways [0.67 mark]

(b) At least 2 women: Cases are (2 women, 3 men), (3 women, 2 men), (4 women, 1 man)

$= \binom{4}{2}\binom{6}{3} + \binom{4}{3}\binom{6}{2} + \binom{4}{4}\binom{6}{1}$ [0.33 mark for correct approach]

$= 6 \times 20 + 4 \times 15 + 1 \times 6 = 120 + 60 + 6 = 186$ ways [0.33 mark]

(c) More men than women: Cases are (3 men, 2 women), (4 men, 1 woman), (5 men, 0 women)

$= \binom{6}{3}\binom{4}{2} + \binom{6}{4}\binom{4}{1} + \binom{6}{5}\binom{4}{0}$ [0.33 mark]

$= 20 \times 6 + 15 \times 4 + 6 \times 1 = 120 + 60 + 6 = 186$ ways [0.33 mark]

Teaching note: Combinations $\binom{n}{r}$ count unordered selections. The key is identifying which cases satisfy the condition and summing them.

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Question 17 [2 marks]

(a) For a valid probability distribution, $\sum P(X=r) = 1$: [0.33 mark for condition]

$k \times 1 \times 4 + k \times 2 \times 3 + k \times 3 \times 2 + k \times 4 \times 1 = 1$ [0.33 mark]

$4k + 6k + 6k + 4k = 20k = 1$ [0.17 mark]

Therefore $k = \frac{1}{20} = 0.05$ [0.17 mark]

Note: Original question stated $\frac{1}{10}$ which is incorrect based on working; correct value is $\frac{1}{20}$.

(b) $P(1 \leq X < 3) = P(X=1) + P(X=2) = \frac{4}{20} + \frac{6}{20} = \frac{10}{20} = \frac{1}{2}$ [0.33 mark]

(c) First find $E(X) = 1 \times \frac{4}{20} + 2 \times \frac{6}{20} + 3 \times \frac{6}{20} + 4 \times \frac{4}{20} = \frac{4+12+18+16}{20} = \frac{50}{20} = 2.5$ [0.33 mark for method]

$E(3X-2) = 3E(X) - 2 = 3(2.5) - 2 = 7.5 - 2 = 5.5$ or $\frac{11}{2}$ [0.33 mark]

Teaching note: Linearity of expectation: $E(aX+b) = aE(X) + b$. We don't need to find the full distribution of $3X-2$.

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Question 18 [2 marks]

Using principle of inclusion-exclusion: $|H \cup G| = 40 - 8 = 32$ students study History or Geography. [0.25 mark]

$|H \cap G| = |H| + |G| - |H \cup G| = 18 + 15 - 32 = 1$ student studies both. [0.25 mark]

(a) $P(\text{both}) = \frac{1}{40}$ [0.5 mark]

(b) $P(G|H) = \frac{P(G \cap H)}{P(H)} = \frac{|H \cap G|}{|H|} = \frac{1}{18}$ [1 mark]

Teaching note: Conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$. "Given History" restricts our sample space to just the 18 History students.

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Question 19 [2 marks]

(a) $P(X < 25) = 0.0668 \Rightarrow P\left(Z < \frac{25-\mu}{\sigma}\right) = 0.0668$

Since $P(Z < -1.5) \approx 0.0668$ from tables, $\frac{25-\mu}{\sigma} = -1.5$ [0.25 mark]

$P(X > 35) = 0.0228 \Rightarrow P\left(Z > \frac{35-\mu}{\sigma}\right) = 0.0228$

Since $P(Z > 2) = 0.0228$, $\frac{35-\mu}{\sigma} = 2$ [0.25 mark]

Solving: $25 - \mu = -1.5\sigma$ and $35 - \mu = 2\sigma$

Subtracting: $10 = 3.5\sigma$, so $\sigma = \frac{20}{7} \approx 2.857$ or $2.86$ (3 s.f.) [0.25 mark]

Then $\mu = 25 + 1.5 \times \frac{20}{7} = 25 + \frac{30}{7} = \frac{205}{7} \approx$ $29.3$ (3 s.f.) [0.25 mark]

(b) $P(28 < X < 40)$: Convert to Z-scores

$Z_1 = \frac{28 - \frac{205}{7}}{\frac{20}{7}} = \frac{196-205}{20} = \frac{-9}{20} = -0.45$ [0.17 mark]

$Z_2 = \frac{40 - \frac{205}{7}}{\frac{20}{7}} = \frac{280-205}{20} = \frac{75}{20} = 3.75$ [0.17 mark]

$P(-0.45 < Z < 3.75) \approx P(Z < 3.75) - P(Z < -0.45)$

$\approx 0.9999 - (1 - 0.6736) = 0.9999 - 0.3264 \approx$ $0.673$ or $0.674$ [0.33 mark for method and answer]

Using more precise values: approximately 0.6736 or accept 0.674 or 0.67 (2 s.f.)

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Question 20 [2 marks]

(a) P(same colour) = P(both red) + P(both blue) [0.25 mark]

$= \frac{4}{10} \times \frac{3}{9} + \frac{6}{10} \times \frac{5}{9}$ [0.25 mark]

$= \frac{12}{90} + \frac{30}{90} = \frac{42}{90} = \frac{7}{15}$ [0.5 mark]

(b) Exactly 2 red out of 3 drawn: Cases are RRD, RDR, DRR (where D = not red = blue)

Each has probability $\frac{4}{10} \times \frac{3}{9} \times \frac{6}{8} = \frac{72}{720} = \frac{1}{10}$ [0.5 mark for one case or using combinations]

Number of arrangements = $\binom{3}{2} = 3$ [0.25 mark]

Total probability = $3 \times \frac{4 \times 3 \times 6}{10 \times 9 \times 8} = 3 \times \frac{72}{720} = 3 \times \frac{1}{10} = \frac{3}{10}$ [0.25 mark]

Alternative using combinations: $\frac{\binom{4}{2}\binom{6}{1}}{\binom{10}{3}} = \frac{6 \times 6}{120} = \frac{36}{120} = \frac{3}{10}$

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END OF ANSWER KEY

Mark summary: Section A (20 marks) + Section B (10 marks) + Section C (10 marks) = 40 marks total