IB Mathematics AHL 4.17 Poisson distribution -AI HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
5
Given: \( \log_{10} N = a – M \)
\( N = 100 \), \( M = 3 \)
\( \log_{10} 100 = a – 3 \)
\( \log_{10} 100 = 2 \)
\( 2 = a – 3 \)
\( a = 2 + 3 = 5 \)
Result: 5 [2]
(b)
100000
From (a): \( \log_{10} N = 5 – M \)
\( N = 10^{5 – M} = \frac{10^5}{10^M} \)
Given: \( N = \frac{b}{10^M} \)
Compare: \( \frac{b}{10^M} = \frac{10^5}{10^M} \)
\( b = 10^5 = 100000 \)
Alternative: \( 100 = \frac{b}{10^3} \), \( b = 100 \times 10^3 = 100000 \)
Result: 100000 [2]
(c)
0.00631
Use: \( N = \frac{10^5}{10^{7.2}} \)
\( = 10^{5 – 7.2} = 10^{-2.2} \)
\( 10^{2.2} \approx 100 \times 1.58489 \approx 158.489 \)
\( 10^{-2.2} \approx \frac{1}{158.489} \approx 0.0063096 \)
Rounded: 0.00631
Alternative: \( \log_{10} N = 5 – 7.2 = -2.2 \), \( N = 10^{-2.2} \approx 0.00631 \)
Result: 0.00631 [2]
(d)
0.532
\( Y \): Years until next earthquake of magnitude at least 7.2
Poisson: Mean \( N = 0.0063096 \)
In 100 years: \( X \sim \text{Po}(100 \times 0.0063096 = 0.63096) \)
\( P(Y > 100) = P(X = 0) \)
\( P(X = 0) = e^{-0.63096} \approx 0.532082 \)
Rounded: 0.532
Alternative (geometric): \( p \approx 0.0063096 \), \( P(Y > 100) = (1 – p)^{100} \approx e^{-0.63298} \approx 0.531 \)
Result: 0.532 [3]