00:01
For this problem, the speed of a file transfer from a 7 campus follows a normal distribution with an average of 60 kilobits per second and a standard deviation of 4 kilobits per second.
00:18
For the first question, we are to determine the probability that a file will transfer at a speed of 68 kilobits per second or more.
00:28
So i'm rating as the probability that x is greater than or equal to 68.
00:32
Now since this scenario follows a normal distribution, we're going to compute the z scores.
00:40
The formula is right here.
00:41
Z is equal to x minus the mean divided by the standard deviation.
00:50
Therefore this becomes the probability that z is greater than or equal to we have 68 minus the mean and we divided by the standard deviation 4.
01:09
That gives us the probability that z is greater than or equal to we have 68 minus the mean and we divide it.
01:12
Or equal to 2 .00.
01:20
Now i'm trying to illustrate this on a standard normal curve, which has 0 at the center.
01:32
1, 2, 3.
01:39
Now you're interested in the region to the right of 2 .00.
01:49
Now the cumulative probabilities under the z table correspond to the area to the left of z.
01:55
So if you want to find the area to the right of z, we need to subtract it from 1 because the total area under the curve is equal to 1.
02:05
Therefore this becomes 1 minus the normal value of 2 .00.
02:11
Now the normal value for 2 .00 is equal to 0 .973.
02:19
So we do the subtraction and we have the answer to be 0 .0227...