Normal distribution question:
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Normal distribution question:

[From: ] [author: ] [Date: 11-12-20] [Hit: ]
..a lot more than 2% would be claiming warranty Ill tell ya that much!)-Pr(X a = 26.......
please help!!! statistics final is tomorrow :(

the life of electric bulbs has a normal distribution with a mean of 35 months and a standard deviation of 4 months. What should the warranty period be if the company that manufactures these bulbs does not want to replace more than 2% of the bulbs?

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z = (x - 35)/4

We want a z-score giving 0.98 since 98% will lie beneath it (meaning the light bulb lasting normally, the above 2% did not last).

This corresponds to a z-score of 2.05

2.05 = (x - 35) / 4
x = 43.2 months

This makes sense since it is above the mean...

(If you use 0.02, which is 2% you will find that this corresponds to a z-score of -2.05 giving x = 26.8 months which doesn't make sense since this is below the mean...a lot more than 2% would be claiming warranty I'll tell ya that much!)

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Pr(X < a) = 2% ----> a = 26.8 months (calculator or normal table)
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