So I have a question in front of me which reads: 'If component lifetime is exponentially distributed with parameter,obtain an expression for the proportion of components whose lifetime exceeds the mean value by more than 1 standard deviation.'

I need some guidance in approaching this question, any tips on solving it would be great. Thinking it through, it seems as though I may need to use:

the standard variance formula, but before I do that I need to find the mean of the exponential distribution, and only then find the variance.

Am I on the right path ?

Any help is appreciated.


If T (random variable) is the lifetime of a component then $P(T>t)=\lambda {{e}^{-\lambda t}}$. The mean lifetime is
$$\mu =E(T)=\int_{0}^{\infty }{t{{e}^{-\lambda t}}dt}$$

with variance
$${{\sigma }^{2}}=E({{(T-\mu )}^{2}})=\int_{0}^{\infty }{{{(t-\mu )}^{2}}{{e}^{-\lambda t}}dt}$$

It is well worth calculating these yourself rather than looking them up. Your question now asks you to calculate

$$P(T>\mu +\sigma) $$


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