A disk drive manufacturer has determined that his product has a “mean time to failure” of 10,000 hours, with a standard deviation of 500 hours. Due to the many factors involved, it is assumed that the time to failure is normally distributed. A network operations center purchases 200 of these disk drives and places them into service on the same day. They run continuously, 24/7. After how many hours can we reasonably expect that 99% of the drives are still running? At the moment the second drive (out of 200) fails, how much time do we have until we expect that 30% of the original 200 drives will have failed? Please explain your reasoning in obtaining the answer.

To find out after how many hours we can reasonably expect that 99% of the drives are still running, we need to calculate the time at which only 1% of the drives have failed.

First, let's calculate the Z-score corresponding to the desired probability. Since we want to find the time at which only 1% of the drives have failed, this corresponds to a probability of 0.01 on a standard normal distribution. Using a Z-table or a calculator, we can find that the Z-score corresponding to this probability is approximately 2.33.

Next, we can use the Z-score formula to find the corresponding value in hours:

Z = (X - μ) / σ

where Z is the Z-score, X is the value we are trying to find, μ is the mean time to failure, and σ is the standard deviation.

Plugging in the values, we have:

2.33 = (X - 10,000) / 500

Now, we can solve for X:

2.33 * 500 = X - 10,000
1165 + 10,000 = X
X = 11,165

Therefore, after approximately 11,165 hours, we can reasonably expect that 99% of the drives are still running.

Now, let's calculate the time until we expect that 30% of the original 200 drives will have failed.

First, let's find how many drives correspond to 30% of the original 200:

30% of 200 = 0.30 * 200 = 60 drives

Next, we need to find the corresponding Z-score for this probability. In this case, since we want to find the time at which 30% of the drives have failed, the probability would be 0.70 (since we are interested in the time when 70% of the drives are still running, which is the complement of 30%). Using a Z-table or a calculator, we can find that the Z-score corresponding to this probability is approximately -0.52.

Now, we can use the Z-score formula again to find the corresponding value in hours:

-0.52 = (X - 10,000) / 500

Solving for X:

-0.52 * 500 = X - 10,000
-260 + 10,000 = X
X = 9740

Therefore, at the moment the second drive fails, we can expect that 30% of the original 200 drives will have failed after approximately 9740 hours.