Question

A company is reviewing a batch of 24 products to determine if any are defective. On average, 3.2% of products are defective.

Answer 1

Using the binomial distribution, it is found that:

• 0.9599 = 95.99% probability that the company will find 2 or fewer defective products in this batch.
• 0.0066 = 0.66% probability that 4 or more defective products are found in this batch.

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

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials.

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

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by:

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

And p is the probability of a success on a single trial.

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• 24 products means that $n = 24$
• 3.2% are defective, thus $p = 0.032$

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The probability that 2 or fewer are defective is:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

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

$P(X = 0) = C_{24,0}.(0.032)^{0}.(0.968)^{24} = 0.4582$

$P(X = 1) = C_{24,1}.(0.032)^{1}.(0.968)^{23} = 0.3635$

$P(X = 2) = C_{24,2}.(0.032)^{2}.(0.968)^{22} = 0.1382$

Thus

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.4582 + 0.3635 + 0.1382 = 0.9599$

0.9599 = 95.99% probability that the company will find 2 or fewer defective products in this batch.

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The probability that 4 or more are defective is:

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

Then

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

$P(X = 0) = C_{24,0}.(0.032)^{0}.(0.968)^{24} = 0.4582$

$P(X = 1) = C_{24,1}.(0.032)^{1}.(0.968)^{23} = 0.3635$

$P(X = 2) = C_{24,2}.(0.032)^{2}.(0.968)^{22} = 0.1382$

$P(X = 3) = C_{24,3}.(0.032)^{3}.(0.968)^{21} = 0.0335$

Thus

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.4582 + 0.3635 + 0.1382 + 0.0335 = 0.9934$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.9934 = 0.0066$

0.0066 = 0.66% probability that 4 or more defective products are found in this batch.

A similar problem is given at brainly.com/question/23780714