00:01
Hi, so we are given from the question at software support center.
00:07
There is a specialist answering the phones to help solve the customer problems.
00:13
So in a part we have to determine the fraction of time each specialist is busy while given average rate is 60 per day and average time sorry exponentially distributed with an average of 40 millimeters.
00:35
So basically average rate we have been given that means you are given lambda is equals to 60 per hour.
00:43
So this is in per hour.
00:45
So this said be equals to 2 .5 per hour as the average service.
00:55
Now as per the question mean of population mean should be equals to 1 in 40 minutes.
01:05
That means 1 .5 per hour number of service and as we are given equals to 3.
01:16
So let it is started with a part the fraction of time each service is busy fraction of time each service is busy that will become equals to lambda by s into mu.
01:31
Let's plug in the values lambda is 2 .5 divided by 3 into mu is 1 .5.
01:40
So this will comes out was to 0 .56.
01:45
So this is the required answer for a part.
01:47
Now, let's have a look to be part.
01:49
So we have to use the following two formulas to find out the service.
01:55
Let's say that p not to for that the average wasting time that should be w of q lambda given mu given a standard deviation given by using lambda equals to 2 .5 music was to 1 .5 s is equals to 3.
02:16
You will get p not is equals to 0 .173 and wasting time wq is equals to 0 .15 hours.
02:29
So clearly for standard deviation equals to 0 lambda by mu times of that mean host to the power s by s factorial.
02:45
So this should be comes 2 .5 by 1 .5 rest to the power.
02:50
So for s to be 0 this will become 0 by 0 factorial.
02:55
That means it will be goes to 1 similarly when standard deviation will be equals to 1 in that case lambda will be 2 .5 divided by 1 .5 as mu rest to the power s is now 1 by 1 factorial will comes out 1 .6667.
03:13
Similarly, at s is equals to 2 the value will become 2 .5 by 1 .5.
03:18
This to the power at 2 by 2 factorial that will be goes to 1 .3889.
03:26
Similarly, for s is equals to 3 you want to going to get a 2 .5 by 1 .5 rest to the power 3 by 3 factorial will comes out 0 .7716.
03:38
So by using this submission of lambda by mu that hold to the power s by s factorial only up to 2 this will be goes to 1 plus 1 .6667 plus 1 .3889 only we are taking up to 2...