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Paper 4 (9164, Section A) · June 2007 · Poisson Distribution

Weekly supply to car dealer A follows a Poisson distribution with mean 43\frac{4}{3} cars, and weekly supply to car dealer B independently follows a Poisson distribution with mean 23\frac{2}{3} cars. What is the probability that, in a given week, fewer than three cars in total are supplied to A and B combined?

A0.4060.406
B0.6770.677
C0.8570.857
D0.9200.920

Explanation

The combined supply follows Po(4/3+2/3)=Po(2)\text{Po}(4/3+2/3)=\text{Po}(2). P(Y<3)=e−2[1+2+2]=5e−2≈0.677P(Y<3)=e^{-2}[1+2+2]=5e^{-2}\approx0.677. Omitting the P(2)P(2) term gives e−2(3)≈0.406e^{-2}(3)\approx0.406. Including P(3)P(3) as well (using Y≤3Y\leq3) gives ≈0.857\approx0.857. Averaging the two means instead of adding them, using Po(1)\text{Po}(1), gives e−1(2.5)≈0.920e^{-1}(2.5)\approx0.920.

Derived from ZIMSEC Maths Paper 1, June 2007, Q6 (Statistics)

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