# Normal Distribution burnout… of lightbulbs.

Thank you for looking through this problem, much appreciated! I tried to work out the answer for a, but I got .2946 when the actual answer is .3085... How do I start this? By the way, I just want to take the time to genuinely thank all the people that answer questions here, you are life savers, and I give partial credit to the grade I get on my next exam to you.

"Suppose that the length of life in hours of a light bulb manufactured by company A is N(800, 14400) and the length of life in hours of a light bulb manufactured by company B is N(850, 2500). One bulb is randomly from each company and is burned until "death".

a.) Find the probability that the length of life of the bulb from company A exceeds the length of life of the bulb from company B by at least 15 hours.

b.) Find the probability that at least one of the bulbs "lives" for at least 920 hours."

Let $W=X-Y$. Then $W$ has normal distribution, mean $-50$, and variance $14400+2500=16900$. So the standard deviation of $W$ is $130$.
We want $Pr(W\ge 15)$. This is the probability that $\frac{W-(-50)}{130}\ge \frac{15-(-50)}{130}$, that is, the probability that $Z\ge 0.5$, where $Z$ is standard normal. Tables or software now give the answer. It is approximately $0.3085$.
Remarks: For the probability that at least one bulb lasts $920$ hours, it is easiest to first find the probability of the complementary event, that both have life $\lt 920$. If $X$ is the lifetime of the first bulb, and $Y$ is the lifetime of the second, the probability that they both have life $\lt 920$ is $\Pr(X\lt 920)\Pr(Y\lt 920)$.