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?

Answer 1

We assume here that the probability for an office network to fail follows a normal distribution with a population mean of 2,200 hours and a population standard deviation of 285 hours.

Answer:

The probability that the network will stay up for 2,800 hours before it fails is about 1.743%.

Explanation:

According to the question that the office network "has been measured to stay working an average of 2,200 hours", we can conclude that, for normally distributed data, at this working time, the office network has a probability of failure of 50% and a probability of being working of 50%, too.

As the office network still operates, the probability of failure increases following a normal distribution. So, for 2,800 hours of operation, we need to calculate the probability of failure for this network.

For this, we need to determine the z-score for the raw value of x = 2,800 hours, to later consult a standard cumulative normal table and find the probability associated with this z-score. To calculate it, we can use the z-score formula:

$z\;score = \frac{x - \mu}{\sigma}$

Where

$\\ \mu\;is\;the\;population\;mean$

$\\ \sigma\;is\;the\;population\;standard\;deviation$

And x is the raw score or the 2,800 hours of operation for the office network.

Thus

$z = \frac{2800 - 2200}{285}$

$z = 2.105 \approx 2.11$

Having a z = 2.11 (approximately) and consulting a standard cumulative normal table, we have that P(z<2.11) = 0.98257.

In other words, for 2,800 hours of operation for the office network, there is a probability of about 98.257% that this network has failed by this time.

Therefore, the probability that the network will stay up for 2,800 hours is 1 - 0.98257 = 0.01743 or about 1.743% of being working before it fails (or for only about 1.743% of the cases, the office network stays working for 2,800 hours).

The graph below has the shaded area that represents this probability.