For quality control purposes, a company that manufactures sim chips for cell/smart phones routinely takes samples from its production process.Since it is important that these chips are nearly fault free, one inspection check involves using microscopic equipment to count the number of imperfections on each chip.Suppose the average number of imperfections per 1000 sim chips is 3.What is the probability that a sample of this size has no more than 2 imperfections?
A)0.4232
B)0.2510
C)0.2240
D)0.5689
E)0.0034

Question

Quizplus · Accepted Answer

This problem can be solved using the Poisson distribution formula, which is suitable for modeling the number of times an event happens in a fixed interval of time or space if these events happen with a known constant mean rate and independently of the time since the last event. The formula for the Poisson probability mass function is $ P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} $, where $ \lambda $ is the average rate (mean) of occurrence, $ k $ is the number of occurrences, and $ e $ is the base of the natural logarithm (approximately equal to 2.71828). Given that the average number of imperfections per 1000 sim chips is 3 ($ \lambda = 3 $), we want to find the probability of having no more than 2 imperfections. This means we need to calculate $ P(X=0) + P(X=1) + P(X=2) $.$ P(X=0) = \frac{3^0 e^{-3}}{0!} = e^{-3} $$ P(X=1) = \frac{3^1 e^{-3}}{1!} = 3e^{-3} $$ P(X=2) = \frac{3^2 e^{-3}}{2!} = \frac{9e^{-3}}{2} $Adding these probabilities together gives the total probability of having no more than 2 imperfections. The exact calculation of these values and their sum matches the option provided in the question, which is approximately 0.4232.

For Quality Control Purposes, a Company That Manufactures Sim Chips