00:01
Hello students, according to the given question, x will follows exponential distribution with parameter lambda, f of x is equal to lambda e power minus lambda x for x greater than or equal to 0, lambda greater than 0.
00:14
For the first one, cdf is f of x is equal to probability of x less than or equal to x is equal to integral 0 to x f of x dx that is equal to integral 0 to x lambda e power minus lambda x dx that is equal to lambda into e power minus lambda x by lambda.
00:33
So, it to the limits 0 to x that is 1 minus e power minus lambda x.
00:39
So, here which is equal to the value here it is minus.
00:44
So, 1 minus e power minus lambda x.
00:47
In the second case probability of 0 less than or equal to x less than or equal to 5 is equal to probability of x less than or equal to 5 minus probability of x less than or equal to 0 that means 1 minus e power minus 5 x minus of 1 minus e power 0 that is equal to 1 minus e power minus 5 by 5 lambda.
01:11
In the third case expectation of x is equal to integral 0 to infinity x into f of x dx.
01:18
So, that is equal to integral 0 to infinity x into lambda e power minus lambda x dx that is equal to lambda integral x power 2 minus 1 into e power minus lambda x into dx.
01:32
So, by calculating we get expectation of x is 1 by lambda and expectation of x square similarly we get 2 by lambda square...