Question

A company manufacturers video games with a current defect rate of 0.95%. To make sure as few defective video games are delivered as possible, they are tested before delivery. The test is 98% accurate at determining if a video game is defective. If 100,000 products are manufactured and delivered in a month, Approximately how many defective products are expected to be delivered.

Answer 1

There would be about 19 defective products delivered.

First, let's start with the number of defective games. 
100000 x 0.0095 = 950

Now, the test will catch 98% of those defects. That means 2% of the defects will get through to the consumers.
0.02 x 950 = 19

Answer 2

Using the binomial distribution, it is found that 190 defective products are expected to be delivered.

For each piece delivered, there are only two possible outcomes. Either it is defective, or it is not. The probability of a piece delivered being defective is independent of any other piece, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

Probability of exactly x successes on n trials, with p probability.  

The expected value of the binomial distribution is:

$E(X) = np$

In this problem:

• 100,000 defective, thus $n = 100000$.
• 0.95% are defective, and of those, 2% are delivered, thus the probability of a defective delivery is $p = 0.095(0.02) = 0.0019$

Then:

$E(X) = np = 100000(0.0019) = 190$

190 defective products are expected to be delivered.

A similar problem is given at https://brainly.com/question/24261244