Acertain brand of flood lamps has a lifetime that is normally distributed with a mean of 3,750 hours and a standard deviation of 300 hours. sketch a. what is the probability that a lamp will last for more than 4,000 hours? b. what is the probability that a lamp will last less than 3,000 hours? c. what lifetime should the manufacturer advertise for these lamps in order that only 4% of the lamps will burn out before the advertised lifetime?

Given that mean=3750 hours and standard deviation is 300:
Then:
a. The probability that a lamp will last for more than 4,000 hours?
P(x>4000)=1-P(x<4000)
but
P(x<4000)=P(z<Z)
where:
z=(x-μ)/σ
z=(4000-3750)/300
z=0.833333
thus
P(x<4000)=P(z<0.8333)=0.7967
thus
P(x>4000)=1-0.7967=0.2033

b.What is the probability that a lamp will last less than 3,000 hours?
P(x<3000)=P(z<Z)
Z=(3000-3750)/300
z=-2.5
thus
P(x<3000)=P(z<-2.5)=0.0062

c. .What lifetime should the manufacturer advertise for these lamps in order that only 4% of the lamps will burn out before the advertised lifetime?
the life time will be found as follows:
let the value be x
the value of z corresponding to 0.04 is z=-2.65
thus
using the formula for z-score:
-2.65=(x-3750)/300
solving for x we get:
-750=x-3750
x=-750+3750
x=3000

Yes, that is the correct formula. to solve this problem given the number of years, plug in the number of years for t and plug the whole thing into the