SOLUTION: We did not find results for: the life expectancy of a particular brand of light bulb is normally distributed with a mean of 1500 hours and a standard deviation of 75 hours. What pe
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Question 443096: We did not find results for: the life expectancy of a particular brand of light bulb is normally distributed with a mean of 1500 hours and a standard deviation of 75 hours. What percentage of the light bulbs would be expected to last less than 1390 hours?.
Suppose the manufacturer would like to warranty the life of the light bulbs. They do not want anymore than 3% of the bulbs returned under the warranty. What time period should be covered?
A large office complex has ordered a case of 120 light bulbs. What is the probability the mean life of these light bulbs is between 1495 and 1505 hours?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
We did not find results for: the life expectancy of a particular brand of light bulb is normally distributed with a mean of 1500 hours and a standard deviation of 75 hours. What percentage of the light bulbs would be expected to last less than 1390 hours?.
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z(1390)= (1390-1500)/75 = -1.467
P(z<-1.457) = 0.0712
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Suppose the manufacturer would like to warranty the life of the light bulbs. They do not want anymore than 3% of the bulbs returned under the warranty. What time period should be covered?
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Find the z-value with a left tail of 3%
invNorm(0.03) = -1.8808
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Find the corresponding x-value:
x = zs+u
x = -1.8808*75+1500 = 1358.94
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A large office complex has ordered a case of 120 light bulbs. What is the probability the mean life of these light bulbs is between 1495 and 1505 hours?
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Find the z-values of 1495 and 1500.
Find the probability that z lies between those z-values.
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Let me know if you need further help.
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Cheers,
Stan H.
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