Answer to Question #347244 in Statistics and Probability for Uya

Question

Suppose that the lifetimes of electric light bulbs follow an Exponential distribution with mean

1000 hours. A

random sample of 40 light bulbs is tested. The probability that the mean

lifetime is at least 900 hours is:

Answer 1

The Variance will be $Var(\bar{X})=\dfrac{\lambda^2}{n}.$

$\mu=1000\ h, n=40$

$$\lambda=\dfrac{1}{\mu}$$

$$\beta^2=Var(nX)$$

$$\beta=\sqrt{Var(X)}=\sqrt{\dfrac{\lambda^2}{n}}$$

$$=\sqrt{\dfrac{1}{40(1000)^2}}=1.58\times10^{-4}$$

Exponential distribution

$$P(X\ge900)=P(X>900)$$

$$=e^{-\beta(900)}=e^{-1.58\times10^{-4}(900)}=0.867448$$