SOLUTION: 1.1 A batch of 5000 electric lamps has a mean life of 1000 hours and a standard deviation of 75 hours. Assuming a normal distribution: 1.1.1 how many lamps will fail before 90

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Question 1116915: 1.1 A batch of 5000 electric lamps has a mean life of 1000 hours and a standard deviation of 75
hours. Assuming a normal distribution:
1.1.1 how many lamps will fail before 900 hours?
1.1.2 how many lamps will fail between 950 and 1000 hours?
1.1.3 and given the same mean life, what would the standard deviation have to be to ensure
that not more than 20% of lamps fail before 916 hours?
1.2 Components are placed into bins containing 100. After inspection of a large number of bins the
average number of defective parts was found to be 10 with a standard deviation of 3. Assuming
that the same production conditions continue, except that bins containing 300 were used:
1.2.1 what would be the average number of defective components per larger bin? (5)
1.2.2 what would be the standard deviation of the number of defectives per larger bin?
1.2.3 how many components must each bin hold so that the standard deviation of the number
of defective components is equal to 1% of the total number of components in the bin?
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
z fpr 900 is (900-1000)/75 or -100/75=-1.333
probability of z < = -1.33 is 0.0912
multiply that by 5000 to get 456 lamps.
between 950 and 1000 hours is a z between -2/3 and 0.
That is a p-value of 0.2475,and multiplied by 500 gives 1237.53 round to 1238.
the z-value for 20% ile is -0.84 from the table
-0.84*sd=(916-1000)
-0.84*sd=-84
sd=100
The sd increases as the first power of the scale factor.
scale it up 3
defective parts are 30 with sd of 9. This continues with a sd of 3% of the total number of components in the bin.