00:02
So in the question we have been given that it takes five days on an average to sell a car.
00:08
So first we'll define a variable x.
00:10
X is the waiting time time to sell a car.
00:21
And we know that the averaging the x is 1 upon 5.
00:27
It takes 1 upon it takes 5 days to make 1 save which implies that x is an exponential distribution with lambda.
00:36
Equal to 1 upon 5.
00:38
This will give us that the pdf or probability density function of x would come up to be 1 upon 5 4 e to the power minus of x upon 5.
00:54
Right.
00:56
So we'll be using this to solve our first part of the problem which is what is the probability that the wait time is greater than equal to 8 days.
01:06
Now the priority such that the wait time is later than 8 days would be integration of from 8 to infinity of p of x d x where p of x is the pdf of x and the pdf of x is 1 upon 5 e to the power minus x upon 5 d x which gives us that p of x greater than 8 comes out to be limits from 8 to infinity, e to the power minus x upon 5.
01:45
This gives us that p of x greater than 8 is e to the power minus 8 upon 5, which is approximately 0 .1019.
02:05
The second part of the problem, we have been asked that probability such that he will have to wait between 5 to 10 days, 6 to 10 days to make another cell...