Question 584836: This is from a normal distribution chapter: Not sure if I use the central limit theorem. Can anyone walk me through it? Trying to learn how to find the right clues to do problems such as these.
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?
b) Within what limits around the population mean will 90% of the samples fall? Min____ Max_____?
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..?)
d) If a random sample of 25 batteries is selected, what proportion of sample means will be between 100 and 115 hours?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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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