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Index Page This is an index page, don't upload it. It has been known from the past experience that two percents of your emails are spam. Suppose you will receive 200 emails. (a) Find the mean and variance of the number of spam emails among those. (b) Using the Poisson approximation, find the probability that there will be at least 3 spam emails among those.

Paper For Above instruction

Introduction

The problem at hand involves modeling the number of spam emails received within a certain set of emails, using probabilistic methods. The given data suggest that 2% of emails are spam, with an expected total of 200 emails received. The problem requires calculating the mean and variance of the number of spam emails, and approximating the probability that at least three spam emails occur using the Poisson distribution. This study demonstrates the use of binomial distribution, its parameters, and the Poisson approximation in a practical context related to email spam detection.

Part A: Calculating Mean and Variance of Spam Emails

Given the information, the number of spam emails among the total emails can be modeled as a binomial random variable, which is suitable for fixed number of independent trials with the same probability of success in each trial. In this context:

- Number of trials, n = 200

- Probability of success (email being spam), p = 0.02

The binomial distribution is characterized by its parameters n and p, with the mean (expected value) and variance given respectively by:

\[ \text{Mean} = \mu = n \times p \]

\[ \text{Variance} = \sigma^2 = n \times p \times (1 - p) \]

Applying to this problem:

\[ \mu = 200 \times 0.02 = 4 \]

\[ \sigma^2 = 200 \times 0.02 \times 0.98 = 200 \times 0.0196 = 3.92 \]

Thus, the expected number of spam emails is 4, with a variance of approximately 3.92. These two measures describe the central tendency and dispersion of the number of spam emails among 200.

Part B: Using the Poisson Approximation

The binomial distribution can be approximated by the Poisson distribution when n is large and p is small, which is the case here (n=200, p=0.02). The parameter λ (lambda) of the Poisson distribution is the same as the mean of the binomial distribution:

\[ \lambda = n \times p = 4 \]

The Poisson distribution is characterized by the probability mass function:

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

To find the probability that at least 3 spam emails are received (i.e., \( P(X \geq 3) \)), it is easier to compute the complement:

\[ P(X \geq 3) = 1 - P(X

Using the Poisson probabilities for \( k=0,1,2 \):

\[

\begin{aligned}

P(X=0) &= \frac{4^0 e^{-4}}{0!} = e^{-4} \approx 0.0183 \\

P(X=1) &= \frac{4^1 e^{-4}}{1!} = 4 e^{-4} \approx 0.0733 \\

P(X=2) &= \frac{4^2 e^{-4}}{2!} = 8 e^{-4} \approx 0.1465 \\

\end{aligned}

\]

Summing these:

\[ P(X \leq 2) = P(0) + P(1) + P(2) \approx 0.0183 + 0.0733 + 0.1465 = 0.2381 \]

Thus, the probability that at least 3 spam emails occur is:

\[ P(X \geq 3) = 1 - 0.2381 = 0.7619 \]

Approximately 76.19% chance that 3 or more spam emails will be received among the 200.

Conclusion

This analysis demonstrates how the binomial distribution effectively models the number of spam emails; with large n and small p, the Poisson approximation provides a convenient and accurate method to estimate probabilities. The expected number of spam emails among 200 is 4, with a variance close to 3.92, highlighting variability consistent with the binomial behavior. The probability of receiving at least three spam emails is roughly 76.2%, emphasizing the non-negligible chance of multiple spam messages in a batch of 200. This approach underscores the practical application of probabilistic modeling in information filtering and email management.

References

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