The life of a semiconductor laser at a constant power is normally distributed wi

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

Q.4 Mean = 10 years

Standard Deviation = 1 year

(a) Pr ( t > 12; 10 ; 1) = 1 - Pr ( t <= 12; 10 ; 1)

value of Z = (12 -10)/1 = 2

so  Pr ( t > 12; 10 ; 1) = 1 - (2)

where is the cumulative normal distribution function

so Pr ( t > 12; 10 ; 1) = 1 - (2) = 1 - 0.9772 = 0.0228

(ii) Let say the time is T years

Pr ( t > T; 10 ; 1) = 0.95

so  Pr ( t > T; 10 ; 1) =1 - Pr ( t <T ; 10; 1) = 0.05

so (z) = 0.05

Z = -1.645

so (T - 10)/1 = -1.645

T = 10 - 1.645 = 8.355 Year

(iii) for a normal laser life range = 7 years to 13 years

so we have to calculate the probability of an abnormal laser

1 - Pr ( 7 < t < 13 ; 10; 1) = 1- [Pr(t <13; 10; 1) - Pr(t <7; 10; 1)]

= 1 - [ 0.9987 - 0.0013] = 0.0026

(iv) the sample here is

8, 11, 6,13, 7, 10 , 9 , 15

by making it in ascending order

6,7,8,9,10,11,13,15

median = (9 + 10)/2 = 9.5

IQR = Q3 - Q1

R= 25/100 x (8 + 1) = 2.25

Q1 = 7 + (8-7) * 0.25 = 7.25

R = 75/100 x (8 + 1) = 6.75

Q3 = 11 + (13-11) * 0.75 = 12.5

IQR = 12.5 - 7.25 = 5.25

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.