Question

Lost-time accidents occur in a company at a mean rate of 0.8 per day. What is the probability that the number of lost-time accidents occurring over a period of 10 days will be no more than 2

Answer 1

0.01375 = 1.375% probability that the number of lost-time accidents occurring over a period of 10 days will be no more than 2.

Explanation:

We have the mean during the interval, which means that the Poisson distribution is used.

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.

Lost-time accidents occur in a company at a mean rate of 0.8 per day.

This means that
`\mu = 0.8n`, in which n is the number of days.

10 days:

This means that
`n = 10, \mu = 0.8(10) = 8`

What is the probability that the number of lost-time accidents occurring over a period of 10 days will be no more than 2?

This is:

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

In which

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

`P(X = 0) = (e^(-8)*8^(0))/((0)!) = 0.00034`

`P(X = 1) = (e^(-8)*8^(1))/((1)!) = 0.00268`

`P(X = 2) = (e^(-8)*8^(2))/((2)!) = 0.01073`

So

`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00034 + 0.00268 + 0.01073 = 0.01375`

0.01375 = 1.375% probability that the number of lost-time accidents occurring over a period of 10 days will be no more than 2.