00:01
So we have a distribution that is approximately normal, which means now we don't care how big the sample size is.
00:09
And we have the mean amount in being loaded into a car, coal, i believe it was, is 75 tons, and the standard deviation was 0 .8 tons.
00:22
And we have, what is the likelihood that you pick one car and it holds less than 74 .5 tons.
00:33
And because, again, this is a normal distribution, we can answer this question.
00:37
But we're non -normal, we wouldn't be able to answer it because the sample size is one.
00:42
So we need to convert that to a z value, and we know that we're going to take that 74 .5 minus the 75, and then divide it by 0 .8.
00:52
And that z value gives us, and that is a negative 0 .5, divided by 0 .8 and that is a z value of negative 0 .625.
01:06
Now if you're going to look it up in your table, you'll probably be using negative 0 .63.
01:10
And so when i look up negative 0 .63, that corresponds with 0 .2643.
01:16
Now if you use your software with normal cd -up, you'll get a little bit different answer, but it's still going to be pretty close to that.
01:23
Then part b asks, what if you use, take a sample size of 20 and that will be sufficient to cause my sampling distribution to be normal because my original distribution was normal.
01:35
And what's the likelihood of getting an x bar that is less than 74 .5? and think about what you're doing.
01:41
You're taking 20 cars at random and then adding together and finding out if that average is less than 74 .5.
01:49
And so we want to convert that to a z value and we have that 74 .5 minus 75.
01:56
So we have the numerators...