<PRE><B>Question 443096</B>
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&lt;-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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