Calculus II

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Expert Answers

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a)

For function `f` to be probability density function the following must hold

`int_(-oo)^oof(x)dx=1`

this is because probability of whole sample space must be 1. Hence we have

`int_(-oo)^oof(x)dx=int_0^ooaxe^(-x^2)dx=|(u=x^2),(du=2xdx),(u_1=0),(u_2=oo)|=`

`a/2int_0^ooe^(-u)du=a/2(-e^(-t))|_0^oo=a/2(0+1)=a/2`

If we now return to our formula we get

`a/2=1`

`a=2` <--Your solution

b)

In this case `P(0leqXleqt)` is actually expression for cumulative distribution function which we can calculate in following maner

`P(0leqXleqt)=int_0^tf(x)dx=`

`int_0^t2xe^(-x^2)dx=`

But we have already calculated similar integral in the part a) so we know that it is equal to

`-e^(-x^2)|_0^t=-e^(-t^2)+e^0=1-e^(-t^2)`

Thus your solution is

`P(0leqXleqt)=1-e^(-t^2)`

In the image below blue is probability density function `f` and red is cumulative distribution function `P(0leqXleqt).`  

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