Answer:
1.74%
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:
[tex]z = \frac{X-\mu}{\sigma}[/tex]
For X = 2,800 hours, the z-score is:
[tex]z=\frac{2,800-2,200}{285}\\z=2.11[/tex]
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:
[tex]P(X>2,800) = 100\%-98.26\% = 1.74\%[/tex]
The probability is 1.74%.
