SOLUTION: 4. 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 sc
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Question 1083698: 4. 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 mathmate(429) (Show Source): You can put this solution on YOUR website!
Question:
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?
Solution.
All questions are based on the normal distribution with parameters N(0,1), i.e. mean, mu = 0, standard deviation, sigma = 1.
All data should be normalized to the distribution N(0,1) to use the normal distribution table.
Normalized value is
Z=(X-mean)/sigma=(X-1400)/200
then P(X<=Z) can be found from the probability table.
(for example, http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf)
(i)
Z(1600 hours)=(1600-1400)/200=1
Z(1900 hours)=(1900-1400)/200=2.5
P(1600<= X <= 1900)=Z(1900)-Z(1600)=P(2.5)-P(1)=0.9938-0.8413=0.1524
(ii)
Assume N hours will give an expected defective percentage of 0.02, then
P(X P(Z=2.054) from table.
Therefore
(N-1400)/200=2.054 => N=1400+200*2.054=1810.75 hours
(iii)
X=1850, Z=(1850-1400)/200=2.25
P(x<=X)=P(Z<2.25)=0.9878
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