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The lifetime of a component of a certain type is a random variable whose probability density function is exponentially distributed with parameter $2$. For a randomly picked component of this type, the probability that its lifetime exceeds the expected lifetime (rounded to $2$ decimal places) is ____________.
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$\frac{1}{e}$
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For this question, we don’t even need the value of $\lambda$ to get the answer.

It is good to remember this integration for solving questions on exponential distribution quickly.

$\int_{0}^{a} \lambda e^{-\lambda x} = 1 - e^{-\lambda a}$

So, with this,

$P (X \leq a) = 1 - e^{-\lambda a}$,

$P(X > a) = 1 – P(X \leq a) = e^{-\lambda a}$

$P (a \leq X \leq b) = P(X \leq b) – P(X \leq a) = (1 - e^{-\lambda b}) – (1 - e^{-\lambda a}) = e^{-\lambda a} – e^{-\lambda b}$

So, for this question the answer will be $P(X > E(X)) = P(X > \frac{1}{\lambda})$, which will be $e^{-\lambda *\frac{1}{\lambda}} = e^{-1} = 0.37$

Since the answer does not depend upon the value of $\lambda$, it means that it holds true for any exponential distribution.
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Not very sure of how good the free resources are out there in this topic. @ankitgupta.1729 Sir, do you have any good recommendations?

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  1. Harvard Stat 110 [Video Lectures]
  2. Sheldon M. Ross [book]
  3. Introduction to Probability by Dimitri P. Bertsekas and John N. Tsitsiklis [MIT Lecture Notes] 

I found these resources useful not only for random variables or expectations but for whole probability as a subject.

Instructor of Harvard Stat 110 course is Joseph K. Blitzstein who also published a book, so you can explore that also if you have time, since, I have not read that book, so can’t comment about it.    

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thanks a lot @ankitgupta.1729

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30 votes
30 votes

For exponential distribution:

${\displaystyle f(x;\lambda )={\begin{cases}\lambda e^{-(\lambda x)}&x\geq 0,\\0&x<0.\end{cases}}}$

$\text{Expected value} = \text{mean}= \frac{1}{\lambda }$

Given $\lambda =2$ , so $\text{mean} = \frac{1}{2} = 0.5$

Probability of lifetime(X) greater than expected lifetime is, 

$P(X>\text{mean})=P(X > 0.5)$

$\qquad = \int_{0.5}^{\infty }\lambda e^{-\lambda x} dx =  \int_{0.5}^{\infty }2 e^{-2 x} dx$

$\qquad = \frac{2}{2}e^{-2\times \frac{1}{2}} =e ^{-1} =0.3678$

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