The life of a semiconductor laser at a constant power is normally distributed wi
ID: 3234810 • Letter: T
Question
The life of a semiconductor laser at a constant power is normally distributed with a mean of 10 years and a standard deviation is 1 year. i) What is the probability that a laser fails after 12 years? ii) What is the life in years that 95% of the laser, exceed? iii) The life of a normal laser is between 7 and 13 years but that of an abnormal laser is out of the range. What is the probability that a laser is abnormal? iv) If five lasers are used in a product and they are assumed to fail independently, what is the probability that all five are operating after 12 years. Suppose that 8 11 6 13 7 10 9 15 are the life times of 8 lasers tested. What are the median and inter quantile (IQR)?Explanation / Answer
Solution:
The life of a semi conductor laser at a constant power is normally distributes with,
Mean () = 10 years = 87600 hours
S.D () = 1 year = 8760
(i) The probability that laser fails before 12 years?
12 years = 105120 hours
P[ X < 105120] = P[X-87600/8760 < 105120-87600/8760]
= P[Z < 2]
= 0.97725
(ii) What is the life in years that 95% of the lasers exceed?
P[X > x] = 0.95
=> P[X-87600/8760 > x -87600/8760] = 0.95
=> P[z > x -87600/8760] = 0.95
=> 1 - P[z x -87600/8760] = 0.95
=> P[z x -87600/8760] = 0.05
=> x -87600/8760 = -1.645
=> x = 87600 - (1.645)(8760)
=> x = 73189.8
(iii) The life of a normal laser is between 7 and 13 years, but that of an abnormal laser is out of the range, what is the probability that a laser is abnormal ?
P[61320 < X < 113880] = P[61320 -87600/8760 < X-87600/8760 < 113880 -87600/8760]
= P[-3 < Z < 3]
= P [ Z < 3 ]P [Z < 3 ]
= 0.9987 - 0.0013
= 0.9974
(iv) If five lasers are used in a product and they are assumed to fail independently, what is the probability that all five are operating after 12 years
Given there laser, that is, n = 3
The probability that it operating after 87600 hours is,
P[X > 105120] = 1 - P[X < 87600]
= 1 - P[X-87600/8760 < 87600-87600/8760]
= 1- P[z < 0]
= 1 - 0.5 = 0.5
The probability that all three still operating after 87600 hours = (0.5)^3 = 0.125
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