00:01
For this problem we're told that an american worker spends an average of 528 minutes a day on the job with a standard deviation of 44 minutes.
00:09
It says that when we sample 64 workers their mean time spent on the job or average time spent on the job is 531 minutes.
00:17
There's a few different questions here so we'll start with the first one which i'll just call a which asks what is n.
00:24
Now n is always the sample size so since it says we sample uh 64 workers its n is equal to 64.
00:33
So the problem just came out and told us that.
00:35
B asks what is mu.
00:37
Mu is the population mean so we were told that a worker in the u .s.
00:43
Spends on average 528 minutes a day on the job so mu is 528.
00:49
It is not the 531 because the 531 is the mean for the sample but 528 is just for the population because it's workers in general.
01:00
C asks for the standard deviation of sample means.
01:04
So the standard deviation of the sample mean is equal to the population standard deviation divided by the square root of the sample size n.
01:12
So we were told the population standard deviation is 44 and we just found our sample size which is 64.
01:18
So if we plug this into our calculator 44 divided by the square root of 64 we get 5 .5 as the standard deviation of sample means.
01:31
D asks what is x bar.
01:34
X bar is the sample mean so we were told that when we take that sample of 64 workers their mean time is 531.
01:43
So that means 531 is x bar our sample mean.
01:49
Lastly then we have e which asks for the probability that the sample mean is more than 531 minutes.
01:58
So we're going to basically use this standard deviation that we found here in order to solve for this.
02:07
So what we'll do is we'll find a z score for 531...