00:01
Hello students, according to the given question by observing the wear out of a machine part is normally distributed with 90 % of failure occurring between 200 and 270 hours.
00:17
So, in the a bit we have to calculate the 90 % of failures lie symmetrically between 200 and 270.
00:29
So, integral 200 to 235 f of t d t will be equal to integral 235 to 270 f of t d t which will be equal to 0 .45.
00:54
So here, new will be equal to 235 hours.
00:58
So the standard normal variant will be probability of 235 minus mu by sigma less than t minus mu by sigma less than 270 minus mu by sigma so which will be equal to 0 .45.
01:23
So it becomes probability of 235 minus 230 minus 2 .30 is divided by sigma less than z less than 270 minus 235 is divided by sigma so which is equal to 0 .45 so we get probability of 0 less than z less than 35 is divided by sigma which is equal to 0 .45 so from the standard normal tables, phi of z will be equal to 0 .45, that implies z will be equal to 1 .645.
02:12
Therefore, 35 is divided by sigma will be equal to 1 .645.
02:20
So sigma will be equal to 21 .28 hours.
02:26
Now in the b bit we have to calculate r of t.
02:33
So r of t will be equal to integral t to infinity f of t d t...