Question

Let X be an exponential random variable with parameter λ=2 . Find the values of the following. Use 'e' for the base of the natural logarithm (e.g., enter e^(-3) for e−3 ).
a) E[(3X+1)2]=
b) P(1≤X≤2)=

Answer 1

Answer:

a

$E[(3X+1)^2]= 8.5$

b

$P(1 < X < 2)=0.1170$

Step-by-step explanation:

From the question we are told that

   The parameter of  X  is  $\lambda = 2$

Generally the expected value of  X is  

    $E(X) = \frac{1}{\lambda }$

     $E(X) = \frac{1}{2}$

=>  $E(X) = 0.50$

Generally we have that

     $E(X^2) = E(X)^2 + E(X)^2$

=>  $E(X^2) = [\frac{1}{2}] ^2 + [\frac{1}{2} ]^2$

=>  $E(X^2) = 0.5$

Generally

    $E[(3X+1)^2]= E(9x^2 + 1 + 6x)$

=>  $E[(3X+1)^2]= 9E[X^2] + 1 + 6 E[X])$

=>    $E[(3X+1)^2]= 9* 0.5 + 1 + 6 * 0.5$

=>   $E[(3X+1)^2]= 8.5$

Generally  

   $P(1 < X < 2)= P(X < 2) - P(X < 1)$

Here  $P(X < 2 ) = e^{- 2 * \lambda }$

=>      $P(X < 2 ) = e^{- 2 * 2 }$

=>       $P(X < 2 ) = e^{- 4}$

and  

    $P(X < 1 ) = e^{- 1 * \lambda }$

    $P(X < 1 ) = e^{- 1 * 2 }$

    $P(X < 1 ) = e^{-2 }$

So

   $P(1 < X < 2)= e^{-2 } - e^{- 4}$

  $P(1 < X < 2)=0.1170$