Question

The computer that controls a bank's automatic teller machine crashes a mean of 0.5 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

Answer:

0.3216 = 32.16% probability that, in any seven-day week, the computer will crash less 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

Mean of 0.5

7-day week, so $\mu = 7*0.5 = 3.5$

What is the probability that, in any seven-day week, the computer will crash less than 3 times?

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

In which

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

$P(X = 0) = \frac{e^{-3.5}*(3.5)^{0}}{(0)!} = 0.0302$

$P(X = 1) = \frac{e^{-3.5}*(3.5)^{1}}{(1)!} = 0.1057$

$P(X = 2) = \frac{e^{-3.5}*(3.5)^{2}}{(2)!} = 0.1850$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0302 + 0.1057 + 0.1857 = 0.3216$

0.3216 = 32.16% probability that, in any seven-day week, the computer will crash less than 3 times