Question

On the average, a grocer sells three of a certain article per week. How many of these should he have in stock so that the chance of his running out within a week is less than \(0.01 ?\) Assume a Poisson distribution.

Short answer

The grocer needs to have at least \(k\) articles in stock, where \(k\) is the smallest integer such that the cumulative Poisson probability \(P(X \leq k) \geq 0.99\). The exact value of \(k\) would depend on calculations or a lookup in a Poisson distribution table. This ensures that the probability of running out of stock in any given week is less than 0.01.

Step-by-step solution

Calculate cumulative probabilities

Using the formula for the Poisson probability, we calculate the cumulative probability of selling \(k\) articles in a week. We start from \(k = 0\) and continue until \(P(X \leq k) \geq 0.99\). We have \(\lambda = 3\).

Identify the minimum stock

The number at which the cumulative probability equals or surpasses 0.99 will be the minimum stock. This means that the grocer needs to have at least this number of articles in stock to ensure that the probability of running out within the week is less than 0.01.

Computation

Here we note the variation of \(P(X \leq k)\) with \(k\) and try to bring it closer to 0.99. We need a computation tool or a Poisson table to look for the smallest value of \(k\) such that \(P(X \leq k) \geq 0.99\).