00:01
Hello students, given f of x is equal to k into x where 0 less than or equal to x less than 1, k x square where 1 less than or equal to x less than 2, 0, other, twice, party, we know that given k is equal to 6 divided by 17, then integration minus infinity to infinity f of x dx, this is equal to integral 0 to 1 k x that is 6 divided by 17 into x into dx plus 1 to integration 1 to 2, 6 divided by 17 x square dx, the last one is 0.
01:02
So this is equal to 3 by 17 into x square with interval 0 to 1 plus 2 by 17 into x cube with 1 integral 1 to 2, this is equal to 3 by 17 plus 14 divided by 17 that is equal to 1.
01:35
F of x is a valid pdf, then part b, x is when 0 less than or equal to x less than 1, then f of x is equal to integration 0 to x, 6 divided by 17 x dx that is equal to 3 by 17 x square with 0 to x interval which is equal to 3 by 17 x square.
02:10
When 1 less than x less than or equal to x less than or equal to 2, f of x equal to integration 0 to 1, 6 divided by 17 x dx plus integration 1 to x, 6 divided by 17 x square dx that is equal to 3 by 17 x square with 0 to 1 plus 2 by 17 x cube with 1 to x, this is equal to 3 by 17 plus 2 by 17 into x cube minus 1 which is equal to 2 by 17 x cube plus 1 by 17 that can be written as 2 into x cube plus 1 divided by 17.
03:28
Therefore, cumulative probability density function f of x is equal to 0 minus infinity less than x less than 0, 3 by 17 x square 0 less than x less than 1 and 2 x cube plus 1 divided by 17 if 1 less than x less than or equal to x less than 2 and 1 if x greater than or equal to 2...