00:02
Hello student, in the given question let x is the total time spent in bookstores until n finds the book.
00:27
Therefore, here we given that the time spent on each book, each bookstore is exponentially distributed with rate is equal to lambda equals to 1 upon mu.
00:44
Now, here we can assume that the number of bookstore, number of bookstores visited, bookstores visited follows geometric distribution with parameter p and let y be the number of bookstores visited, number of bookstores visited, visited.
01:13
So, here expectation of y is equal to 1 upon p and variance of y is equal to 1 minus p upon p square.
01:31
Now, conditional on the number of bookstore visited, the total time spent on bookstore follows gamma distribution, gamma distribution with shape parameter k is equal to y and rate parameter lambda.
01:48
Therefore, the mean and variance of mean and variance of total time spent, total time spent in bookstore are, in bookstore are expectation of x is equals to expectation of, expectation of x given y which is equals to expectation of y upon lambda which is equals to expectation of y upon lambda.
02:35
Here we taking only expectation of y because lambda is constant and expectation of constant is constant which is equals to 1 upon p of lambda.
02:47
So, here we taking expectation of y value is equals to 1 upon p.
02:51
So, 1 upon p into lambda which is equals to mu upon p into 1 minus mu.
03:01
This is the value for expectation of x is equals to mu upon p into 1 minus mu.
03:12
Now, variance of x is equals to expectation of variance of x given y plus variance of expectation of x given y which is equals to expectation of y upon lambda square plus variance of y upon lambda which is equals to expectation of y upon lambda square plus variance of y upon lambda square...