00:01
We're given a random variable x, representing the amount of gasoline and gallons purchased by a randomly selected customer at a gas station.
00:11
We're told that the mean and standard deviation of x are 11 .5 and 4 .0.
00:17
In part a, we're asked to find, in a sample of 50 randomly selected customers, the approximate probability that the sample mean amount purchased is at least 12 gallons.
00:29
So we have that x bar is the sample average amount of gas purchased by 50 customers.
01:09
Now we know by the central limit theorem, since 50 is greater than 30, this applies, we have that x bar is approximately normal.
01:33
With expected value of x bar being the same as the average of the population, which we're given is 11 .5, and the standard deviation.
01:50
Of x bar is the standard deviation of the population sigma over the square root of the number of samples n and we're given that sigma is 4 .0 and we have that n is 50 and so we have 4 over root 50 and therefore the probability that x bar is going to be at least 12 this this is the same approximately as 1 minus phi of 12 minus the mean 10 .5, sorry, 11 .5, i mean.
03:10
In standard deviation, 4 .0 over root 50.
03:19
This is approximately 1 minus phi of .88.
03:31
And using a calculator or a table, we find that this is equal to approximately .1 94 .1894.
03:46
Next, in part b, we're given a sample of 50 randomly selected customers and we're asked to find the approximate probability that the total amount of gasoline purchased is at least 600 gallons.
04:00
So now we're dealing with the total amount of gasoline purchased...