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Exponential distribution $p\left ( x\right )=\lambda e^{-\lambda x}$ where mean and variance is $\frac{1}{\lambda },\frac{1}{\lambda }$

Now,

$84)$here mean=$2$

 $p\left ( Y\geq t \right )=$$\frac{1}{2}.e^{-\frac{1}{2}.x}$

So, ans will be $(C)$


 

$85)$ In Exponential distribution $E\left ( x \right )=\mu$

Now, $E\left ( x_{1} \right )=\mu$

$E\left ( x_{2} \right )=\mu$

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then $E\left ( x_{n} \right )=E\left ( \frac{x_{1}+x_{2}+.....+x_{n}}{n} \right )$

Here

$E\left ( Y \right )=E\left ( \frac{\left ( X-2 \right )\cap \left ( X> 2 \right )}{X> 2} \right )$

Now putting $X=2$

$E\left ( 0 \right )=\mu =2$

Ans $D)$

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Please Explain  2nd part ?
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