SOLUTION: A light bulb manufacturer produces bulbs which have a mean life of 1400 hours and a standard deviation of 200 hours. Assuming a normal distribution.
(i) What proportion of scores
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(i) What proportion of scores
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Question 1083766: A light bulb manufacturer produces bulbs which have a mean life of 1400 hours and a standard deviation of 200 hours. Assuming a normal distribution.
(i) What proportion of scores would be expected to have a life of between 1600 and 1900 hours?
(ii) What number of hours of life can the manufacturer guarantee so that there are no more than 2%
rejects?
(iii) What percentage of light bulbs would have a life of more than 1850 hours? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
a. z is between (1600-1400)/200 or +1 and +2.5, that probability is 0.1524
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b. no more than 2% rejects: z-value for 0.02 probability or 2nd percentile is -2.055
that * sd=-411 from mean
That is 989 hours. Note, the bulbs that are in the upper percentile are fine.
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c. z > =(1850-1400)/200=450/200=2.25
That probability is 0.0122
