Question

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 this size​ (1000 chips) has 2​ imperfections? (Round to four decimal places.) Group of answer choices

Answer 1

Answer:

There is a 22.42% probability that a sample in this size has 2 imperfections.

Step-by-step explanation:

For each chip, there are only two possible outcomes. Either they are imperfect, or they are not.

This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem

A success is a chip being imperfect. Suppose the average number of imperfections per 1000 sim chips is 3. So $\pi = \frac{3}{1000} = 0.003$.

What is the probability that a sample this size​ (1000 chips) has 2​ imperfections?

The sample has 1000 chips, wo $n = 1000$

We want P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{1000,2}.(0.003)^{2}.(0.997)^{998} = 0.2242$

There is a 22.42% probability that a sample in this size has 2 imperfections.