IBDP Maths SL 4.8 Binomial distribution, its mean and variance AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) [3 marks]
Number of rainy days: \( X \sim B(31, 0.2) \) (M1).
Probability: \( P(X = 10) = \binom{31}{10} (0.2)^{10} (0.8)^{21} \) (M1).
Compute: \( P(X = 10) \approx 0.0418894 \approx 0.0419 \) (A1).
(b) [3 marks]
Probability: \( P(X \geq 10) = 1 – P(X \leq 9) \) (M1).
Use binomial CDF: \( P(X \leq 9) \approx 0.925400 \) (M1).
\( P(X \geq 10) = 1 – 0.925400 \approx 0.0746 \) (A1).
(c) [2 marks]
First rain on day 10 (geometric distribution): \( P = (0.8)^9 \cdot 0.2 \) (M1).
Compute: \( (0.8)^9 \approx 0.134218 \), \( P \approx 0.134218 \cdot 0.2 \approx 0.0268435 \approx 0.0268 \) (A1).
Markscheme Answers:
(a) \( 0.0419 \) (M1M1A1)
(b) \( 0.0746 \) (M1M1A1)
(c) \( 0.0268 \) (M1A1)
Total [8 marks]
▶️ Answer/Explanation
(a) [2 marks]
Chocolate muffins: \( C \sim N(62, 2.9^2) \).
\( P(C < 61) = P\left(Z < \frac{61 – 62}{2.9}\right) = P(Z < -0.3448) \approx 0.3651 \approx 0.365 \) (M1A1).
(b) [3 marks]
\( X \sim B(12, 0.365) \) (M1).
\( P(X = 5) = \binom{12}{5} (0.365)^5 (0.635)^7 \approx 792 \times 0.00064757 \times 0.041569 \approx 0.2137 \approx 0.214 \) (M1A1).
(c)(i) [3 marks]
Banana muffins: \( B \sim N(68, 3.4^2) \).
\( P(B < 61) = P\left(Z < \frac{61 – 68}{3.4}\right) = P(Z < -2.0588) \approx 0.0198 \) (M1).
\( P(\text{<61g}) = (0.6 \times 0.365) + (0.4 \times 0.0198) \approx 0.219 + 0.00792 \approx 0.22697 \approx 0.227 \) (M1A1).
(c)(ii) [3 marks]
Conditional probability: \( P(\text{Chocolate} | \text{<61g}) = \frac{P(\text{Chocolate} \cap \text{<61g})}{P(\text{<61g})} \) (M1).
\( P(\text{Chocolate} \cap \text{<61g}) = 0.6 \times 0.365 \approx 0.219 \).
\( P(\text{Chocolate} | \text{<61g}) = \frac{0.219}{0.227} \approx 0.9652 \approx 0.965 \) (M1A1).
(d) [3 marks]
EITHER
New distribution: \( C \sim N(62, \sigma^2) \).
\( 0.6 \cdot P(C < 61) + 0.4 \cdot 0.0198 = 0.157 \) (M1).
\( 0.6 \cdot P(C < 61) + 0.00792 = 0.157 \Rightarrow P(C < 61) \approx 0.2485 \).
\( P\left(Z < \frac{61 – 62}{\sigma}\right) = P\left(Z < -\frac{1}{\sigma}\right) \approx 0.2485 \Rightarrow z \approx -0.6792 \).
\( -\frac{1}{\sigma} = -0.6792 \Rightarrow \sigma \approx 1.4723 \approx 1.47 \) (M1A1).
OR
\( P(C < 61) = 0.2485 \), use inverse normal: \( z = -0.6792 \) (M1).
\( \frac{61 – 62}{\sigma} = -0.6792 \Rightarrow \sigma = \frac{1}{0.6792} \approx 1.4723 \approx 1.47 \) (M1A1).
Markscheme Answers:
(a) 0.365 (M1A1)
(b) 0.214 (M1M1A1)
(c)(i) 0.227 (M1M1A1)
(c)(ii) 0.965 (M1M1A1)
(d) \( \sigma = 1.47 \, \text{g} \) (M1M1A1)
Total [14 marks]