Question

On the average a grocer sells 3 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 will be less than 0.01? Assume a Poisson distribution.

Short answer

The minimum number of articles the grocer should stock depends on the calculations in Step 3. However, it should be the smallest value of k that causes the cumulative probability to be at least 0.99.

Step-by-step solution

Understanding the Problem

We want the probability of selling up to and including k articles to be at least 0.99 (1-0.01), where the number of articles sold follows a Poisson distribution with average rate (lambda) of 3 articles per week. So the problem is expressed as \( P(X \leq k) \geq 0.99 \).

Cumulative Poisson Distribution

The cumulative Poisson distribution function defines the probability that a random variable is less than or equal to a certain value, k, for a given mean rate (lambda). It is given as: \( P(X \leq k) = \sum_{i=0}^{k} \frac{e^{-\lambda} \lambda^i}{i!} \), where e is Euler's number (approximately equal to 2.71828), and i! is the factorial of i.

Calculate Cumulative Probability for Different Values of k

Start calculating the cumulative probability for different values of k starting from 0 until you find a k for which the cumulative probability is greater than or equal to 0.99. Since calculations involving factorials and exponentials can become involved, it's practical to do this calculation with the help of a software or a scientific calculator that can perform these calculations.

Determine the Minimum Stock Quantity

The minimum number of articles the grocer needs to stock is the smallest value of k that make the cumulative probability greater than or equal to 0.99.

Key concepts

Cumulative Probability in Poisson Distribution

Cumulative probability is a key concept to understand when dealing with Poisson distributions. In our grocer's problem, we are interested in finding how likely it is for the grocer to sell a certain number of articles in a week. This likelihood is expressed as a probability. Cumulative probability specifically looks at the chance of selling up to and including a certain number of articles, not exceeding it.

For instance, the probability that the grocer sells at most "k" articles is denoted by the cumulative probability \( P(X \leq k) \). This calculation sums up the probabilities of all possible outcomes from zero articles up to "k."

To solve our problem with the Poisson distribution, we utilize the formula:

• \( P(X \leq k) = \sum_{i=0}^{k} \frac{e^{-\lambda} \lambda^i}{i!} \)

Here, \( e \) is Euler's number, approximately 2.71828, and \( \lambda \) (lambda) is the average rate of occurrence, which is 3 in the grocer's example. We compute the probability for each number of articles sold and aggregate these for the cumulative probability.

Probability Calculation Using Poisson Distribution

The process for calculating probabilities using a Poisson distribution involves understanding how individual probabilities contribute to a bigger picture—in our case, the cumulative probability.

A Poisson distribution is particularly useful when dealing with events that happen independently over a fixed period of time. For our grocer, the event is selling articles in a week. The average number "lambda" is 3. This distribution helps calculate the probability of selling exactly "k" articles, which is done using the formula:

• \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \)

Imagine you calculate this probability for 0 articles, 1 article, 2 articles, etc. Each of these numbers uses the individual probability formula shared above.

To optimize our work, we use tools like scientific calculators or software that already integrate these computations. These assist in performing the multiplications and factorial calculations needed to find each probability up to our target of reaching \( P(X \leq k) \geq 0.99 \). The cumulative probability combines these individual probabilities.

Finding the Minimum Stock Quantity

Once we have a handle on cumulative probabilities, our task turns into a search for the minimum stock needed to satisfy a specific condition. In our problem, the condition is that the probability of running out of stock remains less than 0.01, thus ensuring we aim for our cumulative probability \( P(X \leq k) \geq 0.99 \).

Effectively, this involves trying out different "k" values, where "k" represents the number of articles. For each value of "k," we calculate the cumulative Poisson probability. If the probability with a certain "k" is \( \geq 0.99 \), we consider "k" as a viable stock quantity. The smallest "k" that satisfies this condition is our answer.

In practical scenarios, these calculations become easier using computational tools that handle iteration. In our grocer's example, start testing with minimum possible stock values and increment until you achieve the desired cumulative probability. This systematic trial will lead to acquiring the minimum stock quantity, optimizing costs and ensuring customer satisfaction.