There is a 3% probability that a selected life insurance application contains an error. An auditor randomly selects100 applications. Using the Poisson approximation to the Binomial, calculate the probability that 95_(1)//% or less of the applications are error-free.

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There is a  3% probability that a selected life insurance application contains an error. An auditor randomly selects100 applications. Using the Poisson approximation to the Binomial, calculate the probability that  95_(1)//% or less of the applications are error-free.

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Understand the problem and parameters: Understand the problem and determine the parameters for the Poisson approximation. We are given that the probability of an error in a single application is 3%, or 0.03. We are also given that 100 applications are selected. We want to find the probability that 95% or less of these applications are error-free, which means 5% or more contain an error. To use the Poisson approximation, we need to calculate the mean (λ) of the distribution, which is the expected number of errors in the 100 applications. λ = n * p where n is the number of trials (applications) and p is the probability of success (error in an application).Calculate the mean: Calculate the mean (λ) of the Poisson distribution. λ = 100 * 0.03 λ = 3 The mean number of errors (λ) is 3.Use Poisson distribution: Use the Poisson distribution to find the probability that there are k errors, where k ranges from 0 to the number corresponding to 5% of the applications (since we want 95% or less to be error-free). First, we need to find the number of applications that correspond to 5% of 100, which is 5 applications. We will sum the probabilities of having 0, 1, 2, 3, 4, and 5 errors. The Poisson probability mass function is given by: P(k; λ) = (e^(-λ) * λ^k) / k! where k is the actual number of successes (errors), λ is the mean number of successes, and e is the base of the natural logarithm (approximately 2.71828).Calculate probabilities for k: Calculate the probabilities for k = 0 to k = 5 using the Poisson probability mass function. P(0; 3) = (e^(-3) * 3^0) / 0! = e^(-3) / 1 P(1; 3) = (e^(-3) * 3^1) / 1! = 3 * e^(-3) / 1 P(2; 3) = (e^(-3) * 3^2) / 2! = 9 * e^(-3) / 2 P(3; 3) = (e^(-3) * 3^3) / 3! = 27 * e^(-3) / 6 P(4; 3) = (e^(-3) * 3^4) / 4! = 81 * e^(-3) / 24 P(5; 3) = (e^(-3) * 3^5) / 5! = 243 * e^(-3) / 120Sum probabilities for 95%: Sum the probabilities from k = 0 to k = 5 to find the total probability that 95% or less of the applications are error-free. P(95% or less error-free) = P(0; 3) + P(1; 3) + P(2; 3) + P(3; 3) + P(4; 3) + P(5; 3) P(95% or less error-free) = e^(-3) + 3 * e^(-3) + 9 * e^(-3) / 2 + 27 * e^(-3) / 6 + 81 * e^(-3) / 24 + 243 * e^(-3) / 120Perform calculations for total probability: Perform the calculations to find the total probability. P(95% or less error-free) = e^(-3) * (1 + 3 + 4.5 + 4.5 + 3.375 + 2.025) P(95% or less error-free) = e^(-3) * (18.4) Now we need to calculate e^(-3) * 18.4 using a calculator or software that can handle the exponential function.Calculate final probability: Calculate e^(-3) * 18.4 to get the final probability. P(95% or less error-free) ≈ 0.0498 * 18.4 P(95% or less error-free) ≈ 0.91632

There is a $3 \%$ probability that a selected life insurance application contains an error. An auditor randomly selects $100$ applications. Using the Poisson approximation to the Binomial, calculate the probability that $95_{1} / \%$ or less of the applications are error-free.

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There is a $3 \%$ probability that a selected life insurance application contains an error. An auditor randomly selects $100$ applications. Using the Poisson approximation to the Binomial, calculate the probability that $95_{1} / \%$ or less of the applications are error-free.