A car-hire firm has two cars, which it hires out day by day. The number of demand for a car on each day follows Poissondistribution with mean 1.5. Calculate the proportion of days on which neither car is used
Question
A car-hire firm has two cars, which it hires out day by day. The number of demand for a car on each day follows Poissondistribution with mean 1.5. Calculate the proportion of days on which neither car is used
Solution
To solve this problem, we need to use the formula for the Poisson distribution, which is:
P(X=k) = λ^k * e^-λ / k!
where:
- P(X=k) is the probability of k events in an interval,
- λ is the average rate of value or mean of distribution,
- e is the base of the natural logarithm (approximately equal to 2.71828),
- k! is the factorial of k.
In this case, the mean (λ) is 1.5 (the average number of cars demanded per day), and we want to find the probability that no cars are demanded (k=0).
So, we substitute these values into the formula:
P(X=0) = 1.5^0 * e^-1.5 / 0! = 1 * e^-1.5 / 1 = e^-1.5 = 0.2231
So, the proportion of days on which neither car is used is approximately 0.2231, or 22.31%.
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