A certain brand of flood lamps has a lifetime that is normally distributed with a mean of 3,750 hours and a standard deviation o
f 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: <span>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-</span>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
<span>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. </span><span>.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</span>