Secondary 4 Additional Mathematics Statistics Probability Quiz

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Secondary 4 Additional Mathematics Quiz - Statistics Probability

Name: ____________________
Class: ____________________
Date: ____________________
Score: ________ / 60

Duration: 90 Minutes
Total Marks: 60

Instructions:

• Answer all questions.
• Show all necessary working.
• Use a scientific calculator where appropriate.
• Give your answers to 3 significant figures unless stated otherwise.

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Section A: Probability and Discrete Random Variables (Questions 1–10)

1. A bag contains 5 red balls and 7 blue balls. Two balls are drawn at random without replacement. Find the probability that both balls are of the same colour.
   
   
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2. The probability that a student passes Mathematics is 0.7, and the probability that they pass Physics is 0.6. Given that the probability of passing both is 0.4, find the probability that the student passes at least one of the subjects.
   
   
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3. A random variable $X$ has the probability distribution $P(X=x) = kx$ for $x = 1, 2, 3, 4$. Find the value of the constant $k$.
   
   
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4. For the distribution in Question 3, calculate the expected value $E(X)$.
   
   
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5. A discrete random variable $Y$ has a variance of 4 and a mean of 5. If a new variable $Z = 3Y + 2$ is created, find $E(Z)$ and $Var(Z)$.
   
   
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6. In a certain population, 30% of people are left-handed. If 5 people are chosen at random, find the probability that exactly 2 are left-handed.
   
   
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7. Two events $A$ and $B$ are independent. Given $P(A) = 0.4$ and $P(A \cup B) = 0.7$, find $P(B)$.
   
   
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8. A random variable $W$ has the following distribution: $P(W=1)=0.2, P(W=2)=0.5, P(W=3)=0.3$. Calculate the variance $Var(W)$.
   
   
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9. A fair coin is tossed 6 times. Find the probability of getting at least 4 heads.
   
   
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10. A box contains 10 light bulbs, 3 of which are defective. If 3 bulbs are selected at random without replacement, find the probability that at least one is defective.
    
    
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Section B: Normal Distribution (Questions 11–20)

11. A random variable $X$ follows a normal distribution with mean $\mu = 50$ and standard deviation $\sigma = 5$. Find $P(X < 42)$.
    
    
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12. For the distribution in Question 11, find $P(45 < X < 55)$.
    
    
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13. The weights of apples in a shipment are normally distributed with a mean of 120g. If 15% of the apples weigh more than 135g, find the standard deviation $\sigma$.
    
    
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14. A student's marks in a test are normally distributed with $\mu = 65$ and $\sigma = 10$. If the top 10% of students receive an 'A', find the minimum mark required for an 'A'.
    
    
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15. In a normal distribution, the area between $x = \mu - 1.2\sigma$ and $x = \mu + 1.2\sigma$ is approximately 0.77. Find the probability $P(X > \mu + 1.2\sigma)$.
    
    
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16. The heights of adult males in a city are normally distributed with $\mu = 175\text{cm}$ and $\sigma = 7\text{cm}$. Find the probability that a randomly selected male is between 168cm and 182cm.
    
    
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17. A machine fills bottles with liquid. The volume is normally distributed with $\mu = 500\text{ml}$ and $\sigma = 4\text{ml}$. A bottle is considered "underfilled" if it contains less than 495ml. Find the probability that a bottle is underfilled.
    
    
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18. For the machine in Question 17, if 1000 bottles are filled, how many are expected to be underfilled?
    
    
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19. A random variable $Y$ is normally distributed. Given $P(Y < 20) = 0.2$ and $P(Y > 30) = 0.1$, find the mean $\mu$ and standard deviation $\sigma$.
    
    
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20. The time taken to complete a puzzle is normally distributed with $\mu = 40$ mins and $\sigma = 6$ mins. Find the time $T$ such that only 2.5% of people take longer than $T$.
    
    
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Secondary 4 Additional Mathematics Quiz - Statistics Probability (Answers)

1. Same colour: $P(RR) + P(BB) = (\frac{5}{12} \times \frac{4}{11}) + (\frac{7}{12} \times \frac{6}{11}) = \frac{20}{132} + \frac{42}{132} = \frac{62}{132} = \frac{31}{66} \approx 0.470$. (M1 for each case, A1 for final answer)

2. At least one: $P(M \cup P) = P(M) + P(P) - P(M \cap P) = 0.7 + 0.6 - 0.4 = 0.9$. (M1 for formula, A2 for result)

3. Constant $k$: $\sum P(X=x) = 1 \implies k(1+2+3+4) = 1 \implies 10k = 1 \implies k = 0.1$. (M1 for sum=1, A2 for $k=0.1$)

4. $E(X)$: $E(X) = \sum x P(X=x) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0$. (M1 for formula, A2 for 3.0)

5. Linear Transformation: $E(Z) = 3E(Y) + 2 = 3(5) + 2 = 17$. $Var(Z) = 3^2 Var(Y) = 9(4) = 36$. (M1 E(Z), A1 17, M1 Var(Z), A1 36)

6. Binomial: $n=5, p=0.3, r=2$. $P(X=2) = \binom{5}{2} (0.3)^2 (0.7)^3 = 10 \times 0.09 \times 0.343 = 0.3087 \approx 0.309$. (M1 formula, M1 substitution, A2 result)

7. Independent Events: $P(A \cup B) = P(A) + P(B) - P(A)P(B)$ $0.7 = 0.4 + P(B) - 0.4P(B) \implies 0.3 = 0.6P(B) \implies P(B) = 0.5$. (M1 formula, M1 algebra, A1 result)

8. Variance $Var(W)$: $E(W) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1.0 + 0.9 = 2.1$. $E(W^2) = 1^2(0.2) + 2^2(0.5) + 3^2(0.3) = 0.2 + 2.0 + 2.7 = 4.9$. $Var(W) = 4.9 - (2.1)^2 = 4.9 - 4.41 = 0.49$. (M1 $E(W)$, M1 $E(W^2)$, A2 result)

9. At least 4 heads: $P(X \ge 4) = P(X=4) + P(X=5) + P(X=6)$ $\binom{6}{4}(0.5)^6 + \binom{6}{5}(0.5)^6 + \binom{6}{6}(0.5)^6 = (15 + 6 + 1) \times \frac{1}{64} = \frac{22}{64} = \frac{11}{32} \approx 0.344$. (M1 identifying cases, M1 binomial terms, A2 result)

10. At least one defective: $1 - P(\text{none defective}) = 1 - (\frac{7}{10} \times \frac{6}{9} \times \frac{5}{8}) = 1 - \frac{210}{720} = 1 - \frac{7}{24} = \frac{17}{24} \approx 0.708$. (M1 complement, M1 calculation, A2 result)

11. Normal $P(X < 42)$: $z = \frac{42-50}{5} = -1.6$. $P(Z < -1.6) = 0.0548$. (M1 z-score, A2 result)

12. Normal $P(45 < X < 55)$: $z_1 = \frac{45-50}{5} = -1, z_2 = \frac{55-50}{5} = 1$. $P(-1 < Z < 1) = 0.6827$. (M1 z-scores, A2 result)

13. Find $\sigma$: $P(X > 135) = 0.15 \implies P(Z > z) = 0.15 \implies z \approx 1.036$. $1.036 = \frac{135-120}{\sigma} \implies \sigma = \frac{15}{1.036} \approx 14.5\text{g}$. (M1 z-value, M1 formula, A2 result)

14. Min mark for 'A': Top 10% $\implies P(Z > z) = 0.1 \implies z \approx 1.282$. $x = 65 + 1.282(10) = 77.82 \approx 77.8$. (M1 z-value, M1 formula, A2 result)

15. Symmetry: $P(\mu - 1.2\sigma < X < \mu + 1.2\sigma) = 0.77$. $P(X > \mu + 1.2\sigma) = \frac{1 - 0.77}{2} = \frac{0.23}{2} = 0.115$. (M1 symmetry logic, A2 result)

16. Height: $z_1 = \frac{168-175}{7} = -1, z_2 = \frac{182-175}{7} = 1$. $P(-1 < Z < 1) = 0.683$. (M1 z-scores, A2 result)

17. Underfilled: $z = \frac{495-500}{4} = -1.25$. $P(Z < -1.25) = 0.1056 \approx 0.106$. (M1 z-score, A2 result)

18. Expected count: $1000 \times 0.1056 = 105.6 \approx 106$ bottles. (M1 multiplication, A2 result)

19. Simultaneous $\mu, \sigma$: $P(Z < \frac{20-\mu}{\sigma}) = 0.2 \implies \frac{20-\mu}{\sigma} = -0.842$ (1) $P(Z > \frac{30-\mu}{\sigma}) = 0.1 \implies \frac{30-\mu}{\sigma} = 1.282$ (2) Subtract (1) from (2): $\frac{10}{\sigma} = 2.124 \implies \sigma \approx 4.71$. Substitute into (1): $20 - \mu = -0.842(4.71) \implies \mu = 20 + 3.97 = 23.97 \approx 24.0$. (M1 z-values, M1 equations, M1 solving $\sigma$, M1 solving $\mu$, A2 final values)

20. Time $T$: Top 2.5% $\implies P(Z > z) = 0.025 \implies z = 1.96$. $T = 40 + 1.96(6) = 40 + 11.76 = 51.76 \approx 51.8$ mins. (M1 z-value, M1 formula, A2 result)