CIE A level Math -Probability & Statistics 2: 6.1 The Poisson distribution: approximation : Exam Style Questions Paper 6

solution:- 

Step 1: Define the Poisson Distributions

Each entrance follows an independent Poisson distribution for a 10-second period:

$$X_1 \sim Po(0.4), \quad X_2 \sim Po(0.1), \quad X_3 \sim Po(0.2), \quad X_4 \sim Po(0.5)$$

The total number of arrivals in 10 seconds is:

$$X = X_1 + X_2 + X_3 + X_4$$

Since the sum of independent Poisson distributions is also Poisson, we get:

$$X \sim Po(0.4 + 0.1 + 0.2 + 0.5) = Po(1.2)$$

Step 2: Scale to 1 Minute (60 Seconds)

Since Poisson processes scale linearly with time, in 60 seconds (which is 6 times a 10-second period), the total number of passengers follows:

$$Y \sim Po(6 \times 1.2) = Po(7.2)$$

We need to find:

$$P(Y > 3) = 1 – P(Y \leq 3)$$

Step 3: Compute

The Poisson cumulative probability formula is:

$$P(Y = k) = \frac{e^{-\lambda} \lambda^k}{k!}$$

where

$\lambda = 7.2$

.

We calculate:

$$P(Y = 0) = \frac{e^{-7.2} 7.2^0}{0!} = e^{-7.2}$$ $$P(Y = 1) = \frac{e^{-7.2} 7.2^1}{1!} = e^{-7.2} \times 7.2$$ $$P(Y = 2) = \frac{e^{-7.2} 7.2^2}{2!} = e^{-7.2} \times \frac{7.2^2}{2}$$ $$P(Y = 3) = \frac{e^{-7.2} 7.2^3}{3!} = e^{-7.2} \times \frac{7.2^3}{6}$$

Approximating using a calculator:

$$e^{-7.2} \approx 0.0007466$$ $$P(Y = 0) \approx 0.0007466$$ $$P(Y = 1) \approx 0.0007466 \times 7.2 = 0.005375$$ $$P(Y = 2) \approx 0.0007466 \times \frac{7.2^2}{2} = 0.01935$$ $$P(Y = 3) \approx 0.0007466 \times \frac{7.2^3}{6} = 0.04644$$

Summing these:

$$P(Y \leq 3) \approx 0.0007466 + 0.005375 + 0.01935 + 0.04644 = 0.07191$$

Thus:

$$P(Y > 3) = 1 – 0.07191 = 0.9281$$

Final Answer:

\(0.928\) (rounded to 3 decimal places)

Markscheme———–

$\lambda=7.2$

$P(X>3)=1-e^{-7.2}(1+7.2+\frac{7.2^{2}}{2!}+\frac{7.2^{3}}{3!})$

or $1-e^{-7.2}(1+7.2+25.92+62.21)$

or $1-(0.0007466+0.005375+0.01935+0.04644)$

= 0.928 (3sf)