Question

Let the number of chocolate drops in a certain type of cookie have a Poisson distribution. We want the probability that a cookie of this type contains at least two chocolate drops to be greater than $0.99.$ Find the smallest value of the mean that the distribution can take.

Short answer

The smallest mean (lambda) satisfying the condition is approximately 6.64.

Step-by-step solution

Start with the Cumulative Distribution Function

The cumulative distribution function of a Poisson distribution is given by, $P(x;\lambda )=e^{-\lambda }\sum _{k=0}^x\frac{{\lambda }^k}{k!}$. This gives the probability of having up to 'x' events.

Transform the inequality

We want P(X >= 2) > 0.99, but it's easier to work with the cumulative distribution function if we transform this into 1 - P(X < 2) > 0.99, or equivalently, P(X < 2) < 0.01.

Define the left-hand side in terms of lambda

The left-hand side where x = 1, is $e^{-\lambda }(1+\lambda )$. So we have that $e^{-\lambda }(1+\lambda )<0.01$.

Solve for lambda

Solving this inequality for $\lambda$ is not straightforward analytically, so you would use a numerical method or trial and error. Using a calculator or software, we find that the smallest $\lambda$ that satisfies the inequality is approximately 6.64.

Key concepts

Cumulative Distribution Function

A Cumulative Distribution Function (CDF) is an essential concept in probability theory. It helps us understand the probability of a random variable being equal to or less than a certain value. For the Poisson distribution, specifically, the CDF computes the total probability of having up to a specific number of events in a fixed interval of time or space.
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The formula for the CDF in a Poisson distribution is $P(x;\lambda )=e^{-\lambda }\sum _{k=0}^x\frac{{\lambda }^k}{k!}$. Here, $\lambda$ represents the average rate or mean, and $x$ is the number of events.
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When solving problems using the CDF, like in our original exercise, it is often necessary to manipulate the function to find probabilities different from the standard less-than format. Transforming it allows for ease in solving inequalities, which is key in probability calculations.

Poisson inequality

In the context of Poisson distributions, inequalities often come in handy to determine thresholds or specific characteristics in probability scenarios.
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The exercise involves transforming the original probability objective $P(X\ge 2)>0.99$ into an inequality involving the CDF, specifically $P(X<2)<0.01$. This transformation helps in simplifying the calculation.
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By rewriting $P(X<2)$ in terms of $\lambda$, it becomes easier to find the necessary conditions that $\lambda$ must satisfy. Such transformations provide a clearer pathway to solve the problem due to the simplified algebraic manipulation.
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Remember, inequalities in Poisson distributions ensure we meet desired probability criteria by adjusting $\lambda$, the mean.

Numerical Methods

Numerical methods are powerful tools in most mathematics fields, particularly when analytical solutions are hard to find or verify. In our specific exercise involving Poisson distributions, although we have an inequality formulated, solving it analytically can be challenging.
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The inequality $e^{-\lambda }(1+\lambda )<0.01$ involves exponential and factorial terms. This complexity makes direct solving nearly impossible, hence the need for numerical methods.
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• Tools like calculators or programming software can iterate over possible values for $\lambda$ to find an approximate solution.
• Trial and error might sometimes work, but software greatly speeds up the process and increases accuracy.

In our exercise, this approach helps locate that the smallest $\lambda$ satisfying the transformed inequality is about 6.64. Thus, numerical methods provide a concrete way to solve real-world problems accurately.

Mean Value of Distribution

The mean value of a Poisson distribution, denoted by $\lambda$, is a measure of the average number of occurrences or events over a specified interval. It's fundamental in characterizing and manipulating the distribution.
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The mean plays a crucial role in defining how the distribution behaves and, consequently, how the probabilities are calculated. In our example, the mean affects the likelihood of having at least a given number of events, such as chocolate drops in a cookie.
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There are a few key points to consider about the mean of a Poisson distribution:

• A larger $\lambda$ typically increases the chance of observing more events within the interval.
• The exercise hinges on adjusting the mean to meet a probability criterion – the minimum mean that still guarantees the desired probability for the number of cookie drops.

Therefore, understanding the mean's impact on the distribution provides insightful controls over how probabilities are formulated and achieved in various statistical challenges.