Answer
given,
installed electric lamp = 2000
average life = 1000 hrs
standard deviation = 200 hrs
a)
let after p hors 10% bulb fails ,
P(X<p) = 0.10
[tex]P(X< \dfrac{x-\mu}{\sigma})= 0.10[/tex]
from Normal distribution table,
P(Z<-1.28) = 0.10
on comparing
[tex]\dfrac{x-\mu}{\sigma} = -1.28[/tex]
[tex]\dfrac{x-1000}{200} = -1.28[/tex]
x = -1.28 x 200 + 1000
x = 744 hours
b) P( 900 < X < 1300) =
=[tex]P(\dfrac{900 -1000}{200} < Z <\dfrac{1300-1000}{200})[/tex]
= P(-0.5 < Z < 1 .5)
= P(Z<1.5) - P(Z<-0.5)
= 0.93319 - 030854
= 0.62465
lamps may be expected to fail between 900 and 1300
= 0.62465*2000
= 1249.3
= 1249 (approx)
