There is a 2% probability that a selected life insurance application contains an error. An auditor randomly selects 50 applications. Using the Poisson approximation to the Binomial, calculate the probability that 90% or less of the applications are error-free.

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Calculate Mean Errors: First, we need to determine the mean number of errors ($\lambda$) in 50 applications. The mean is calculated by multiplying the probability of an error in a single application by the number of applications. $\lambda = P(\text{error in one application}) \times \text{number of applications}$ $\lambda = 0.02 \times 50$ $\lambda = 1$Calculate Probability of Errors: Next, we need to calculate the probability that 90% or less of the applications are error-free. This means that 10% or more of the applications contain errors. Since we have 50 applications, 10% of 50 is 5 applications. Therefore, we are looking for the probability of having 5 or more errors.Use Poisson Distribution: Using the Poisson distribution, the probability of having exactly $k$ errors is given by the formula: $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ However, we need the cumulative probability of having 5 or more errors, which is the complement of the probability of having 4 or fewer errors. So, we will calculate the probability of having 0, 1, 2, 3, and 4 errors and subtract the sum from 1.Calculate Probability of 0 Errors: Calculating the probability of having 0 errors: $P(X=0) = \frac{e^{-1} \cdot 1^0}{0!} = \frac{e^{-1}}{1} = e^{-1}$Calculate Probability of 1 Error: Calculating the probability of having 1 error: $P(X=1) = \frac{e^{-1} \cdot 1^1}{1!} = \frac{e^{-1}}{1} = e^{-1}$Calculate Probability of 2 Errors: Calculating the probability of having 2 errors: $P(X=2) = \frac{e^{-1} \cdot 1^2}{2!} = \frac{e^{-1}}{2} = \frac{e^{-1}}{2}$Calculate Probability of 3 Errors: Calculating the probability of having 3 errors: $P(X=3) = \frac{e^{-1} \cdot 1^3}{3!} = \frac{e^{-1}}{6} = \frac{e^{-1}}{6}$Calculate Probability of 4 Errors: Calculating the probability of having 4 errors: $P(X=4) = \frac{e^{-1} \cdot 1^4}{4!} = \frac{e^{-1}}{24} = \frac{e^{-1}}{24}$Calculate Cumulative Probability: Now, we sum the probabilities of having 0, 1, 2, 3, and 4 errors to find the cumulative probability of having 4 or fewer errors: $P(X \leq 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)$ $P(X \leq 4) = e^{-1} + e^{-1} + \frac{e^{-1}}{2} + \frac{e^{-1}}{6} + \frac{e^{-1}}{24}$ $P(X \leq 4) = e^{-1}(1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24})$ $P(X \leq 4) = e^{-1}(\frac{24}{24} + \frac{24}{24} + \frac{12}{24} + \frac{4}{24} + \frac{1}{24})$ $P(X \leq 4) = e^{-1}(\frac{65}{24})$Calculate Probability of 5 or More Errors: Finally, we calculate the probability of having 5 or more errors, which is the complement of the probability of having 4 or fewer errors: $P(X \geq 5) = 1 - P(X \leq 4)$ $P(X \geq 5) = 1 - e^{-1}(\frac{65}{24})$Correct Probability Calculation: We can now compute the numerical value of $P(X \geq 5)$ using the value of $e^{-1}$ which is approximately 0.3679: $P(X \geq 5) = 1 - 0.3679(\frac{65}{24})$ $P(X \geq 5) = 1 - 0.3679 \times 2.7083$ $P(X \geq 5) = 1 - 0.9967$ $P(X \geq 5) = 0.0033$However, we made a mistake in the previous step. We were supposed to find the probability that 90% or less of the applications are error-free, which corresponds to the probability of having 4 or fewer errors, not 5 or more. Therefore, we should use the value we calculated for $P(X \leq 4)$ directly.

There is a $2 \%$ probability that a selected life insurance application contains an error. An auditor randomly selects $50$ applications. Using the Poisson approximation to the Binomial, calculate the probability that $90 \%$ or less of the applications are error-free.

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There is a $2 \%$ probability that a selected life insurance application contains an error. An auditor randomly selects $50$ applications. Using the Poisson approximation to the Binomial, calculate the probability that $90 \%$ or less of the applications are error-free.