Question

Your office network has been measured to stay working an average of 2,200 hours with a standard deviation of 285 hours.

What is the probability that the network will stay up for 2,800 hours before it fails?

(Keep two decimal points in your percentage answer.)

Answer 1

1.74%

Step-by-step explanation:

Mean number of hours (μ) = 2,220 hours

Standard deviation (σ) = 285 hours

Assuming a normal distribution, the z-score for any number of hours a network stays up, X, is given by:

`z = (X-\mu)/(\sigma)`

For X = 2,800 hours, the z-score is:

`z=(2,800-2,200)/(285)\\z=2.11`

A z-score of 2.11 corresponds to the 98.26th percentile of a normal distribution. The probability that the network will stay up for 2,800 hours before it fails is:

`P(X>2,800) = 100\%-98.26\% = 1.74\%`

The probability is 1.74%.