Question

A manufacturer has determined that for every 1000 units it produces, 3 will be faulty. What is the probability that an order of 50 units will have one or more faulty units

Answer 1

0.1393 = 13.93% probability that an order of 50 units will have one or more faulty units.

Explanation:

Mean for a number of units, which means that the Poisson distribution is used to solve this question.

Poisson Distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

In which

x is the number of sucesses

e = 2.71828 is the Euler number

`\mu` is the mean in the given interval.

Mean:

3 defective for 1000, how many for 50?

3 - 1000

`\mu` - 50

Applying cross multiplication:

`\mu = (3*50)/(1000) = 0.15`

What is the probability that an order of 50 units will have one or more faulty units?

This is:

`P(X \geq 1) = 1 - P(X = 0)`

In which

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-0.15)*(0.15)^(0))/((0)!) = 0.8607`

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.8607 = 0.1393`

0.1393 = 13.93% probability that an order of 50 units will have one or more faulty units.