68 flyhigh airlines determined that the distance traveled...

6.8 FlyHigh Airlines determined that the distance traveled per aircraft per year is normally distributed, with a mean of 60 thousand miles and a standard deviation of 10 thousand miles. What proportion of aircraft can be expected to travel between 44 and 54 thousand miles in a year? What percentage of aircraft can be expected to travel either less than 25 or more than 70 thousand miles in a year? How many miles will be traveled by at least 70 percent of the aircraft? What are your answers to (a) through (c) if the standard deviation is 12 thousand miles?

Transcript

00:01 So we have a normal distribution for the number of miles flown, and we have the mean is 60 ,000 with a standard deviation of 10 ,000.

00:10 And we want to find in part a, what's the likelihood that the number of miles is between 44 ,000, and 54 ,000 if we get a, just choose a random plane.

00:24 And so we're going to take that 44 minus 60 divided by 10, and the 54 minus 60.

00:31 Divided by 10.

00:33 And that comes up to be this difference is 16 negative and divided by 10 is going to be negative 1 .6.

00:42 And that's supposed to be a z value now.

00:46 And this is negative 6 divided by 10.

00:49 So that's going to be negative 0 .6.

00:50 And if we find the area in our table, negative 0 .6, the area below is 0 .2743.

01:00 And then the area below negative 1 .6 is .0548.

01:06 And let's get what that difference is.

01:10 .2743 minus .0548 comes out being .295.

01:20 Now, i won't go back and change all the answers with the standard deviation change into 12.

01:26 You will do that here and then can find it exactly the same way.

01:31 So i'll continue, but i'm not going to do that with the part with changing the standard deviation.

01:37 So now what's the likelihood if they have less than 25 ,000 miles or more than 70 ,000 miles? so again, i'm going to use convert it to a z value, and that's 25 minus the mean divided by the standard deviation.

01:55 And likewise, 70 minus the mean divided by the standard deviation.

02:01 And so this one comes out to be the 25 minus the 60 comes out to be negative 3 .5 divided by 10.

02:10 That z value is less than negative 3 .5...

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