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Chapter 5: Problem 111

A baker who makes fresh cheesecakes daily sells an average of five such cakes per day. How many cheesecakes should he make each day so that the probability of running out and losing one or more sales is less than . 10 ? Assume that the number of cheesecakes sold each day follows a Poisson probability distribution. You may use the Poisson probabilities table from Appendix \(\mathrm{C}\).

Short Answer

Expert verified
The baker should make \(x\) cheesecakes, where \(x\) is the smallest integer which satisfies the cumulative probability condition in step 2. The exact number can be identified using the Poisson Probabilities table, as described in step 3.

Step by step solution

01

Understand the Poisson Probability Distribution

The Poisson Probability Distribution is used for count-based distributions where events are rare and occur independently of each other. Here, the 'events' are the sales of cheesecakes. It is given by the formula \[P(X=k)= \frac{{e^{-\lambda} \cdot \lambda^k}}{k!}\] where \(P(X=k)\) is the probability of \(k\) events happening in an interval, \(e\) is a constant approximately equal to 2.71828, \(\lambda\) is the average rate of value (in this case, it's the average sales - 5) and \(k\) is the actual number of events.
02

Calculate the Required Number of Cheesecakes

For the baker to not run out of cheesecakes, the cumulative probability of selling up to \(x\) cakes must be at least 0.9. More formally, \[\sum_{i=0}^{x} P(X=i) \geq 0.9\] Start by calculating \(P(X=i)\) for \(i\) ranging from 0 to a reasonable upper limit (say 10), and then sum the probabilities until it exceeds the target of 0.9. The corresponding \(i\) will be the number of cheesecakes the baker should make.
03

Use the Poisson Probabilities Table (Table C)

Making use of the Poisson Probabilities table, locate \(\lambda=5\) row and find the smallest number of cakes (the column number) where the cumulative probability is greater than or equal to 0.9. That's the answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cumulative Probability
Cumulative probability is a crucial concept when dealing with probability distributions like the Poisson distribution. Instead of looking at the probability of a single outcome, cumulative probability considers the probability of all outcomes up to a certain point. It answers questions like, "What is the probability of selling up to or fewer than a certain number of cheesecakes?"

In the context of our baker, cumulative probability is used to ensure he doesn't run out of cheesecakes. By calculating cumulative probabilities, he can decide how many cakes to bake such that the chance of running out is below 10%. This is done by summing the probabilities of selling 0, 1, 2, ..., and so on, up to a number where the total probability meets or exceeds 90%. This means, with 90% certainty, the baker will sell that many cakes and no more.
Average Rate
In a Poisson probability distribution, the parameter \( \lambda \) represents the average rate of occurrence of an event within a fixed interval. For our baker, this average rate, \( \lambda \), is 5 cheesecakes per day. This average rate becomes crucial in calculating the probability of different numbers of sales in a day using the Poisson formula:
  • \( \lambda \) is typically the known average or expected number of occurrences (here, sales).
  • The variability in the actual daily sales will still center around this average over the long run.
Understanding the average rate helps translate real-world averages into statistical models that can reliably predict probabilities of future occurrences.
Rare Events
The Poisson distribution often deals with rare events. A rare event in this context is something that's not only infrequent but also spreads out over time. For the baker, this might mean days when he sells significantly more or fewer than the average number of cakes.

Rare events highlight the importance of planning for extremes, not just averages. While the average sale rate is 5, the rare occurrences of significantly more sales must be considered to avoid shortages. By planning for these unusual but possible sales spikes, the baker can better serve his customers without losing potential sales.
Independent Events
A key assumption in using the Poisson distribution is that events happen independently. This means that the sale of one cheesecake doesn't affect the likelihood of another cheesecake being sold. Each sale is its own isolated event when considering probabilities.
  • Independence ensures that past events don鈥檛 skew current probabilities.
  • It simplifies calculations because the probability structure doesn鈥檛 change from sale to sale.
By knowing that cheesecake sales are independent events, the baker can confidently use the Poisson distribution to forecast sales. This helps in making informed decisions about production without external sales influences fluctuating his results.

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Most popular questions from this chapter

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