00:01
In this question, two parts are assembled and the standard deviation, sigma 1, that is equals to 0 .4 and sigma 2 is 0 .15.
00:09
In the a part, we have to calculate the average of clearance, we have to calculate here.
00:17
So here this x1 is such that n, mu, comma, sigma, square, and x2 is such that it is n, mu 2, this is mu 2, this is mu 1, okay, comma, sigma 2 square.
00:31
So now here x1 minus x2 this is n, this is mu 1 minus mu 2 comma sigma 1 square plus sigma 2 square.
00:42
So now it is x1 minus x2 it is n this is new 1 minus mu 2 comma this value is 0 .25.
00:50
Now in the a part we have to calculate average clearance.
00:54
So p of x1 minus x2 this is less than 0 .05.
00:59
So this value is 0 .003.
01:03
Okay.
01:04
So now if here this p z is less than 0 .005 minus mu 1 minus mu 2 and whole divided by 0 .5 so this value is jad of 0 .00006 okay solving this out so this value is minus 3 .43 by the z inverse table.
01:27
So from here p z is less than 0 .005 minus mu 1 minus mu 2 that is equals to 1 .716.
01:37
So from here, mu 1 minus mu 2, the value of this is equals to what, 1 .721.
01:45
What is the average clearance? average clearance here is equals to 1 .721.
01:54
This will be the answer for the a5...