Question 584836
Assume that a lifetime (in hrs) of a particular type of transistor battery is approximately mound shaped, not necessarily normal, with a mean of 100 hrs and standard deviation of 20 hrs.
a) What proportion of the batteries will last between 100 and 115 hrs?
z(100) = 0
z(115) = (115-100)/20 = 15/20 = 3/4
P(100< x <115) = P(0< z <3/4) = normalcdf(0,3/4) = 0.2734
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b) Within what limits around the population mean will 90% of the samples fall?
Find the z-value with a left-tail of 0.05: -1.645
Then the z-value with a right-tail of 0.05 is: +1.645
Min = -1.645*20 + 100 = 67.1
Max = +1.645*20 + 100 = 132.9
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c) If a random sample of 49 batteries is selected, what "proportion" of sample means will be between 100 and 115 hrs? (THIS THROWS ME off since I thought proportions were done later with confidence intervals..?)
z(100) = 0
z(115) = (115-100)/[20/sqrt(49)] = 5.25
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P(100< x-bar <115) = P(0< z < 5.25) = 0.5000
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d) If a random sample of 25 batteries is selected, what proportion of sample means will be between 100 and 115 hours?
z(100) = 0
z(115) = (115-100)/[20/sqrt(25)] = 15/(20/5) = 15/4
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P(100< x-bar <115) = P(0< z <15/4) = 0.4999
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Cheers,
Stan H.
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