Question

The computer that controls a bank's automatic teller machine crashes a mean of 0.60.6 times per day. What is the probability that, in any seven-day week, the computer will crash more than 33 times? Round your answer to four decimal places.

Answer 1

Answer:

0.6045 = 60.45% probability that, in any seven-day week, the computer will crash more than 3 times.

Step-by-step explanation:

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

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

The computer that controls a bank's automatic teller machine crashes a mean of 0.6 per 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 more than 3 times?

Either it crashes 3 or less times, or it crashes more than 3 times. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 3) + P(X > 3) = 1$

We want P(X > 3). So

$P(X > 3) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-4.2}*(4.2)^{0}}{(0)!} = 0.0150$

$P(X = 1) = \frac{e^{-4.2}*(4.2)^{1}}{(1)!} = 0.0630$

$P(X = 2) = \frac{e^{-4.2}*(4.2)^{2}}{(2)!} = 0.1323$

$P(X = 3) = \frac{e^{-4.2}*(4.2)^{3}}{(3)!} = 0.1852$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0150 + 0.0630 + 0.1323 + 0.1852 = 0.3955$

$P(X > 3) = 1 - P(X \leq 3) = 1 - 0.3955 = 0.6045$

0.6045 = 60.45% probability that, in any seven-day week, the computer will crash more than 3 times.