Future Electronics makes compact disc players. Its research department found that the life of the laser beam device is normally distributed, with mean 4940 hours and a standard deviation of 480 hours.
(a) Find the probability that the laser beam device will wear out in 5000 hours or less.
(b) Future Electronics wants to place a guarantee on the players so that no more than 13% fail during the guarantee period. Because the laser pickup is the part most likely to wear out first, the guarantee period will be based on the life of the laser beam device. How many playing hours should the guarantee cover?
Solution :
Given that ,
mean = = 4940 hours
standard deviation = = 480 hours
a) P(x 5000)
= P[(x - ) / (5000 - 4940) / 480]
= P(z 0.125)
Using z table,
= 0.5497
b) Using standard normal table,
P(Z < z) = 13%
= P(Z < z ) = 0.13
= P(Z < -1.126 ) = 0.13
z = -1.126
Using z-score formula,
x = z * +
x = -1.126 * 480 + 4940
x = 4399.52
x = 4400 hours.
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