00:01
Given exponential distribution with mean theta equal to 3 per minute.
00:17
So density function f of x is equal to 1 divided by theta multiplied with e to the power negative x divided by theta substituting the theta value 1 divided by 3 e to the power negative x divided by 3 where x is greater than 0 less than infinity.
00:40
Part a, p of x greater than 5 that is equal to integral 5 to infinity 1 divided by 3 e to the power negative x divided by 3 dx on integrating 1 divided by 3 e to the power negative x divided by 3 divided by negative 1 divided by 3 with the limits 5 and infinity.
01:06
This can be rewritten as negative 3 divided by 3 e to the power negative x divided by 3 with 5 and infinity as limit.
01:19
So negative 1 applying the limits e to the power negative infinity minus e to the power negative 5 divided by 3.
01:28
So negative 1 multiplied with negative e to the power negative 5 divided by 3 because e to the power negative infinity is 0 and hence e to the power negative 5 divided by 3 is our answer which is equal to 0 .1889.
01:52
Part b, p of x greater than 10 plus 5 given that x greater than 5 can be written as p of x greater than 10.
02:03
So p of x greater than 10 is equal to integral 10 to infinity f of x dx.
02:14
So integral 10 to infinity 1 divided by 3 e to the power negative x divided by 3 dx.
02:24
On integrating we will be getting 1 divided by 3 multiplied with e to the power negative x divided by 3 divided by 1 divided by 3 with the limits 10 and infinity...