Answer to Question #152849 in Statistics and Probability for samantha

Question

An Urban council has installed 2000 lamps with mercury bulbs in the streets of town area. The lifetimes of these bulbs are normally distributed with a mean of 1200 burning hours and having a standard deviation of 200 hours.
(i) How many of these bulbs can be expected to fail.
(a) In the first 800 burning hours?
b) Between 900 and 1750 burning hours?
(c) 10 % of the bulbs would fail?
(d)150 bulbs are still in good condition?

Answer 1

(i) a) X=800

z=$\frac{X-\mu}{\sigma}=\frac{800-1200}{200}=-2$

P(x<z)=0.228

Number of bulbs expected to fail:

N=2000*0.228=456

b) X₁=900

z₁=$\frac{X_1-\mu}{\sigma}=\frac{900-1200}{200}=-1.5$

X₂=1750

z₂=$\frac{X_2-\mu}{\sigma}=\frac{1750-1200}{200}=2.75$

P(z₁<x<z₂)=0.930

N=2000*0.930=1860

c) P(x<z)=0.1

z=-1.282

10 % of the bulbs would fail in:

X=$z*\sigma+\mu$ = -1.282*200+1200 = 944 hours

d) P(z<x)=150/2000=0.075

z=1.44

150 bulbs are still in good condition in:

X=$z*\sigma+\mu$ = 1.44*200+1200 = 1488 hours