Question

a certain pharmaceutical company know that, on average 4% of a certain type of pill has an ingredient that is below the minimum strength and thus unacceptable. what is the probability that fewer that 5 in a sample of 20 pills will be acceptable?

Answer 1

Using the binomial distribution, it is found that there is a 0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

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

Binomial probability distribution

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

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

The parameters are:

• x is the number of successes.

• n is the number of trials.

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

In this problem:

• The sample has 20 pills, hence $n = 20$.

• 100 - 4 = 96% are acceptable, hence $p = 0.96$

The probability that fewer that 5 in a sample of 20 pills will be acceptable is:

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

In which

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

$P(X = 0) = C_{20,0}.(0.96)^{0}.(0.04)^{20} = 0$

$P(X = 1) = C_{20,1}.(0.96)^{1}.(0.04)^{19} = 0$

$P(X = 2) = C_{20,2}.(0.96)^{2}.(0.04)^{18} = 0$

$P(X = 3) = C_{20,3}.(0.96)^{3}.(0.04)^{17} = 0$

$P(X = 4) = C_{20,4}.(0.96)^{4}.(0.04)^{16} = 0$

0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

A similar problem is given at brainly.com/question/24863377