The life expectancy of a typical lightbulb is normally distributed with a mean of 2,000 hours and a standard deviation of 27 hours. What is the probability that a lightbulb will last between 1,975 and 2,050 hours?

Question

ybarrap · Answer

$$
\large
P(1,975 \le X \le 2,050)=\Phi(\cfrac{2,050-2,000}{27})-\Phi(\cfrac{1,975-2,000}{27})
$$
Where $\Phi(x)$ is the CDF for the Normal Distribution.

ybarrap · Answer

*$\Phi(x)$ is the CDF of the $\color{red}{Standard}$ Normal Distribution.

ybarrap · Answer

You can use this calculator, just plug in x for the endpoints - http://www.wolframalpha.com/input/?i=standard+normal+cdf+calculator

ybarrap · Answer

http://en.wikipedia.org/wiki/Normal_distribution#Cumulative_distribution

anonymous · Answer

I have no idea what to even plug in for the calculator. At all.

anonymous · Answer

@helpme1.2

anonymous · Answer

a. 0.17619 

b. 0.32381 

c. 0.79165 

d. 0.96784

ybarrap · Answer

ybarrap · Answer

Where $X$ above represents the random variable: hours the lightbulb survives.

anonymous · Answer

why is it .968-.1772 shouldn't it be a right there answer?

ybarrap · Answer

what do you mean?

anonymous · Answer

nothing, Thank you for your help. I started out with choosing the answer D. I just wanted to make sure it was right. Im taking an exam and I have to get a good grade on it.