00:01
Okay, so we've got a variable x, which is the amount of time in minutes that a student spends on one chapter of homework.
00:09
And the population, it's normally distributed with the population, well, it doesn't actually say it's normally distributed.
00:17
It just says the mean of x is given by 100, and the standard deviation of x is given by 100 as well.
00:27
It then says we take a sample size of n equals 64 and asks us to answer some questions.
00:34
And so it wants us to describe x bar for this sample size of 64 and state its distribution.
00:42
So this is a sample of size greater than 30 drawn from a population where we know the standard deviation.
00:53
And we'll assume, you know, these 64 students are independent and that means that by the central limit, theorem, this x bar, i'm going to denote it with a subscript 64, just to remind me what the sample size is, but you don't have to.
01:06
Follows a normal distribution with mean equal to the population mean for a single one, which is mu x or 100.
01:15
And then standard deviation given by the population standard deviation divided by the square root of the sample size.
01:24
And so its variance is just the square of that.
01:27
And so the distribution is a number.
01:29
Normal distribution with mean 100 and standard deviation 100 over 8.
01:49
And it wants us to describe it in words.
01:51
So this is the average time spent on chapter 1 by a student in this sample drawn from the population.
02:14
Okay, question 7 says a calculate probability, see that the that this x -bar was between 102 and 110.
02:24
So that's obviously going to be the same that the probability that x -bar was just less than 110, all values less than 110, minus the probability that x -bar was less than 102.
02:37
And then we can standardise these, since it follows a normal distribution, as described up here, we can say this is the same as z being less than equal to 110 minus 100 divided by the standard deviation, 100 over 8.
02:51
And then the exact same thing for the second one.
02:55
This one is the same as the son of normal variable z being less than equal to 102 minus 100 divided by 100 over 8.
03:10
And this first one if you plug it into your calculator gives you 0 .80 and this second one gives you 0 .16.
03:22
You can use your z tables to find out what these two probabilities are.
03:28
It turns out they are 0 .781 and 0 .5636 respectively.
03:37
And this gives you 0 .2 to 5 to 3 decimal places, which is what it asks for.
03:44
Question 8 then says, calculate the interval for the middle 45 % for this x bar.
03:53
So we basically want values min and max such that the probability that x lies between them is 45%.
04:11
Now, if we do the same for z, so if we're looking at the standard normal distribution, we can say we want z max and zmin, which bound the central 45%, then because it's central, it means that we must have 100 minus 45 divided by 2 % either side, which is 27 .5 % either side.
04:48
And so this zed max value is simply the value below which 72 .5 % of the z values lie, right? all of these values add up to 72 .5%.
05:05
And so we know that the probability that the standard normal variable z is less than this z max is equal to 0 .725.
05:19
And so you can find from your tables that such a z max is given by 0 .59...