00:01
Here, x is normally distributed with mu is equal to 700 and sigma square is equal to 100 raised to the park 2.
00:11
Here x represents number of kilometers traveled per airplane per year.
00:17
Now, we have given mean and a standard deviation and using this we have to answer the following question.
00:24
So, the first question says that what percentage of planes is expected to travel between 4 ,415 and 700 ,000 kilometers in a year.
00:39
So we have to find the probability of 415 less than x is less than 700.
00:46
So for this we can write it as p of 415 lays than x is lased than 700 is equals to probability of x is less than 700 minus p of x is less than 400 minus p of x is less than 400 so this is equals to p of z is less than 700 minus 700 divided by the mu divided by the sigma that is 100 minus p of z is less than 415 minus 700 divided by 100.
01:24
So this is equals to p of z is less than 0 minus p of z is less than minus 2 .8.
01:34
This is equals to 0 .5 minus 0 .002186.
01:44
So this is equal to 0 .4978.
01:47
So it implies that 49 .49 .78 % of planes is expected to travel between 415 to 700.
02:16
Thousand kilometers in a year.
02:25
This is the answer for the first part.
02:28
Now in the second part we have to find the probability that what percentage of this place is expected to travel more than 650 .56 ,000 per kilometer that is p of 650 .56 is less than x.
02:48
So this we have to find so consider p of x is greater than 650 .56...