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.

View step-by-step help

Home

Math Problems

Precalculus

Find values of normal variables

Full solution

Q. 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.

Understand the problem and parameters: Understand the problem and determine the parameters for the Poisson approximation. $\newline$ 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 $(\lambda)$ of the distribution, which is the expected number of errors in the $100$ applications. $\newline$ $\lambda = n \times p$ $\newline$ 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 $\lambda$ of the Poisson distribution. $\lambda = 100 \times 0.03$ $\lambda = 3$ The mean number of errors $\lambda$ 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). $\newline$ 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 $0$ , $1$ , $k$ $0$ , $k$ $1$ , $k$ $2$ , and $5$ errors. $\newline$ The Poisson probability mass function is given by: $\newline$ $k$ $4$ $\newline$ where $k$ is the actual number of successes (errors), $k$ $6$ is the mean number of successes, and $k$ $7$ is the base of the natural logarithm (approximately $k$ $8$ ).
Calculate probabilities for k: Calculate the probabilities for $k = 0$ to $k = 5$ using the Poisson probability mass function. $P(0; 3) = \frac{e^{-3} \cdot 3^0}{0!} = \frac{e^{-3}}{1}$ $P(1; 3) = \frac{e^{-3} \cdot 3^1}{1!} = \frac{3 \cdot e^{-3}}{1}$ $P(2; 3) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{9 \cdot e^{-3}}{2}$ $P(3; 3) = \frac{e^{-3} \cdot 3^3}{3!} = \frac{27 \cdot e^{-3}}{6}$ $P(4; 3) = \frac{e^{-3} \cdot 3^4}{4!} = \frac{81 \cdot e^{-3}}{24}$ $P(5; 3) = \frac{e^{-3} \cdot 3^5}{5!} = \frac{243 \cdot e^{-3}}{120}$
Sum 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. $P(95\% \text{ or less error-free}) = P(0; 3) + P(1; 3) + P(2; 3) + P(3; 3) + P(4; 3) + P(5; 3)$ $P(95\% \text{ or less error-free}) = e^{-3} + 3 \cdot e^{-3} + \frac{9 \cdot e^{-3}}{2} + \frac{27 \cdot e^{-3}}{6} + \frac{81 \cdot e^{-3}}{24} + \frac{243 \cdot e^{-3}}{120}$
Perform calculations for total probability: Perform the calculations to find the total probability. $\newline$ $P(95\% \text{ or less error-free}) = e^{-3} \times (1 + 3 + 4.5 + 4.5 + 3.375 + 2.025)$ $\newline$ $P(95\% \text{ or less error-free}) = e^{-3} \times (18.4)$ $\newline$ Now we need to calculate $e^{-3} \times 18.4$ using a calculator or software that can handle the exponential function.
Calculate final probability: Calculate $e^{-3} \times 18.4$ to get the final probability. $\newline$ $P(95\% \text{ or less error-free}) \approx 0.0498 \times 18.4$ $\newline$ $P(95\% \text{ or less error-free}) \approx 0.91632$