A continuous random variable x has the following probability...

A continuous random variable X has the following probability density function F(x) = cx^3, 0<x<2 and 0 otherwise (a) Find the value c such that f(x) is indeed a density function. (b) Write out the cumulative distribution function of X. (c) P(1 < X < 3) =? (d) Write out the mean and variance of X. (e) Let Y be another continuous random variable such that when 0 < X < 2, and 0 otherwise. Calculate the mean of Y.

Transcript

00:01 The function given here as f of x is equal to c x q where x is in between 0 and 2.

00:10 So for solution of first part we know that we know that integration of 0 to 2 for f of x d x is equal to 1 and now if f of x d x is equal to 1 then we will find as c 0 to 2 x q d x will be equal to 1 and therefore c when we simplify our value here this would be execute plus 1 that is 3 plus 1 4 x to the power of 4 divided by 4 divided by 4 and now our limit 0 to 2 is equal to 1 so when we simplify with our limits we will get here c by 4 and 8 i'm sorry 2 to the power of 4 4 is equal to 1.

01:05 So 2 to the power of 4 is 16.

01:07 C by 4 for the multiplied by 16 is equal to 1.

01:12 So this will simplify as 4.

01:15 So 4 c is equal to 1.

01:17 Therefore, c is equal to 1 apart 4.

01:20 So here's the solution of 1.

01:23 This is the answer for a part.

01:27 Next part is b.

01:29 So in b part, we'll find a cumulative.

01:34 Cumulative.

01:37 This is the answer.

01:37 Distribution of x so there as f of x is given 0 to x for f of x d x so we will put down here 1 by 4 integration of 0 to x for x q d x that is 1 by 4 again x to the power of 4 and the limit 0 to x this will be equal to 1 by 4 multiplied by 1 by 4 and x to the power of 4.

02:18 So therefore we will get here 1 upon 16 multiplied by x to the power of 4 when x is in between 0 to so our final answer for f of x i'm sorry this is capital x is equal to x to the power of 4 divided by so now this is our solution for b part of the question answer for b.

02:43 Then the next part is part c so in the c part here we'll find a probability for getting x in between 1 and 3 so x take value of x will take take take values value in 0 comma 2 interval therefore probability of getting 1 x 3 will be equal to probability of getting 1 x less than x less than 2 and that is simplified as 1 to 2 integration f of x d x so as earlier we have simplified this value we will do the same here so that is 1 by integration of 1 to 2 for 1 upon of x which is 1 by 4 x q d x we will put down here our previously therefore this is 1 by 4 and x to the power of 4 divided by 4 next we have to put here our limits which are 1 and 2 then this is 1 upon 16 within the bracket we have 2 to the part of 4 minus 1 so that is 1 upon 16 16 minus 1 will become 15 by 16...

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