(SHOW ALL WORK) Light Bulbs: The mean lifespan of a standard 60 watt incandescen
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Question
(SHOW ALL WORK) Light Bulbs: The mean lifespan of a standard 60 watt incandescent light bulb is 875 hours with a standard deviation of 80 hours. The mean lifespan of a standard 14 watt compact fluorescent light bulb (CFL) is 10,000 hours with a standard deviation of 1,500 hours. These two bulbs put out about the same amount of light. Assume the lifespan’s of both types of bulbs are normally distributed to answer the following questions.
(a) I select one incandescent light bulb and put it in my barn. It seems to last forever and I estimate that it has lasted more than 2000 hours. What is the probability of selecting a random incandescent light bulb and having it last 2000 hours or more. Did something unusual happen here?
(b) I select one CFL bulb and put it in the bathroom. It doesn’t seem to last very long and I estimate that it has lasted less than 5,000 hours. What is the probability of selecting a random CFL and having it last less than 5,000 hours. Did something unusual happen here?
(c) Compare the the lifespan of the middle 99% of all incandescent and CFL light bulbs.
(d) Is there much of a chance that I happen to buy an incandescent light bulb that lasts longer than a randomly selected CFL?
Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 2000
u = mean = 875
s = standard deviation = 80
Thus,
z = (x - u) / s = 14.0625
Thus, using a table/technology, the right tailed area of this is
P(z > 14.0625 ) = 0 [ANSWER]
YES, IT IS VERY UNUSUAL, that its probability is very close to 0.
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b)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 5000
u = mean = 10000
s = standard deviation = 1500
Thus,
z = (x - u) / s = -3.333333333
Thus, using a table/technology, the left tailed area of this is
P(z < -3.333333333 ) = 0.00042906 [ANSWER]
yes, this is unusual, as P < 0.05. [ANSWER]
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c)
As the middle area is
Middle Area = P(x1<x<x2) = 0.99
Then the left tailed area of the left endpoint is
P(x<x1) = (1-P(x1<x<x2))/2 = 0.005
Thus, the z score corresponding to the left endpoint, by table/technology, is
z1 = -2.575829304
By symmetry,
z2 = 2.575829304
As
u = mean = 10000
s = standard deviation = 1500
Then
x1 = u + z1*s = 6136.256045
x2 = u + z2*s = 13863.74396
Hence, between 6136.256 and 13863.744 hours [ANSWER]
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d)
No. Even a 2000 hour incandescent bulb is rare as in part A, how much more for meven the lower bound on c, 6136 hours? There is very very small chance.
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