Question 261478: Please help?
2. The mean amount of time a (unnamed) computer server is down is 18.3 minutes with a standard deviation of 5.7 minutes.
a. Assuming we know nothing about the shape of the data set, at least what percentage of times will the server be down between 8.325 and 28.275 minutes?
b. Assuming we can verify that the data set is approximately normally distributed, what percentage of times will the server be down less than 24 minutes?
c. Assuming we can verify that the data set is approximately normally distributed, 68% of the server downtimes will be within what range? (5 points)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The mean amount of time a (unnamed) computer server is down is 18.3 minutes with a standard deviation of 5.7 minutes.
a. Assuming we know nothing about the shape of the data set, at least what percentage of times will the server be down between 8.325 and 28.275 minutes?
Find a z-score for each of those numbers:
z(8.325) = (8.325-18.3)/5.7 = -1.75
z(28.275) = (28.275-18.3)/5.7 = 1.75
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According to Chebeshev at least (1-(1/1.75^2))% of the data lies
between those numbers.
Ans: (1-(1/1.75)^2) = 0.6734 = 67.34%
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b. Assuming we can verify that the data set is approximately normally distributed, what percentage of times will the server be down less than 24 minutes?
z(24) = (24-18.3)/5.7 = 1
P(x<24)=P(z<1) = 0.8413 = 84.13% of the time
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c. Assuming we can verify that the data set is approximately normally distributed, 68% of the server downtimes will be within what range?
68% of the population is within 1 std of the mean
Ans: 18.3 - 5.7 < x < 18.3 + 5.7
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Cheers,
Stan H.
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