Answer:
0.7026
Step-by-step explanation:
Let X denote the lifetime of light bulb. Given [tex] X \sim Exp(\lambda) [/tex] where the mean is [tex] E(X) = 340 =\frac{1}{\lambda} \implies \lambda = \frac{1}{340} = 0.00294[/tex].
Recall that,
[tex] \displaystyle P(X>x) = 1 - \int_0^x e^{-\lambda x} = 1 - (1 - e^{-\lambda x} = e^{-\lambda x} [/tex]
[tex]\displaystyle P(X>220 | X>100) = \frac{P(X>220,X>100)}{P(X>100)} = \frac{e^{-\lambda \times 220}}{e^{-\lambda \times 100}} = \frac{0.5236}{0.7452} = 0.7026[/tex]
