Barry has a medical plan with a $1,200 deductible, 20% coinsurance, and a $5,000 coinsurance cap. His allowable medical expenses
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We assume here that <em>the probability for an office network to fail</em> follows a <em>normal distribution</em> with a <em>population mean of 2,200 hours</em> and a <em>population standard deviation of 285 hours</em>.
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 <em>normally distributed data</em>, 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 <em>z-score</em> for the raw value of x = 2,800 hours, to later consult a <em>standard cumulative normal table </em>and find the probability associated with this z-score. To calculate it, we can use the z-score formula:
Where
And <em>x</em> is the raw score or the 2,800 hours of operation for the office network.
Thus
Having a z = 2.11 (approximately) and consulting a <em>standard cumulative normal table, </em>we have that<em> </em>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 <em>has failed by this time</em>.
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.
