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

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

Question

A bus station has exactly four entrances. In the morning the numbers of passengers arriving at these entrances during a 10-second period have the independent distributions $Po(0.4)$, $Po(0.1)$, $Po(0.2)$ and $Po(0.5)$.

Find the probability that the total number of passengers arriving at the four entrances to the bus station during a randomly chosen 1-minute period in the morning is more than 3.

▶️Answer/Explanation

solution:- 

Step 1: Define the Poisson Distributions

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

X1Po(0.4),X2Po(0.1),X3Po(0.2),X4Po(0.5)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=X1+X2+X3+X4X = X_1 + X_2 + X_3 + X_4

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

XPo(0.4+0.1+0.2+0.5)=Po(1.2)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:

YPo(6×1.2)=Po(7.2)Y \sim Po(6 \times 1.2) = Po(7.2)

We need to find:

P(Y>3)=1P(Y3)P(Y > 3) = 1 – P(Y \leq 3)


Step 3: Compute

P(Y3)P(Y \leq 3)

The Poisson cumulative probability formula is:

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

where

λ=7.2\lambda = 7.2

.

We calculate:

P(Y=0)=e7.27.200!=e7.2P(Y = 0) = \frac{e^{-7.2} 7.2^0}{0!} = e^{-7.2} P(Y=1)=e7.27.211!=e7.2×7.2P(Y = 1) = \frac{e^{-7.2} 7.2^1}{1!} = e^{-7.2} \times 7.2 P(Y=2)=e7.27.222!=e7.2×7.222P(Y = 2) = \frac{e^{-7.2} 7.2^2}{2!} = e^{-7.2} \times \frac{7.2^2}{2} P(Y=3)=e7.27.233!=e7.2×7.236P(Y = 3) = \frac{e^{-7.2} 7.2^3}{3!} = e^{-7.2} \times \frac{7.2^3}{6}

Approximating using a calculator:

e7.20.0007466e^{-7.2} \approx 0.0007466 P(Y=0)0.0007466P(Y = 0) \approx 0.0007466 P(Y=1)0.0007466×7.2=0.005375P(Y = 1) \approx 0.0007466 \times 7.2 = 0.005375 P(Y=2)0.0007466×7.222=0.01935P(Y = 2) \approx 0.0007466 \times \frac{7.2^2}{2} = 0.01935 P(Y=3)0.0007466×7.236=0.04644P(Y = 3) \approx 0.0007466 \times \frac{7.2^3}{6} = 0.04644

Summing these:

P(Y3)0.0007466+0.005375+0.01935+0.04644=0.07191P(Y \leq 3) \approx 0.0007466 + 0.005375 + 0.01935 + 0.04644 = 0.07191

Thus:

P(Y>3)=10.07191=0.9281P(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)

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