Question

The computer that controls a bank's automatic teller machine crashes a mean of 0.6 times per day. What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.

Answer 1

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

Explanation:

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 of 0.6 times a day

7 day week, so
`\mu = 7*0.6 = 4.2`

What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)`

In which

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

`P(X = 0) = (e^(-4.2)*(4.2)^(0))/((0)!) = 0.0150`

`P(X = 1) = (e^(-4.2)*(4.2)^(1))/((1)!) = 0.0630`

`P(X = 2) = (e^(-4.2)*(4.2)^(2))/((2)!) = 0.1323`

So

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0150 + 0.0630 + 0.1323 = 0.2103`

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.