00:01
Here we have a data set which is nice because it's organized already from least to greatest.
00:06
So it's from 1 to 10 values and what we're gonna do is find the mean median and standard deviation so the mean x bar of the sample.
00:17
I'm gonna assume this is a sample from a larger population so x bar is equal to the sum of all the x values so add all these up and divide by n how many there are and the standard deviation s is somebody taking the square root of the sum of each x value minus the sample mean squared divided by n minus 1.
00:43
So we're gonna again assume this is a sample so we're divided by m minus 1 if it's the populate relation standard deviation, which would be sigma.
00:50
The only difference is that it's the square root of the sum of squares of x which ss sub x is the same as that numerator divided by n.
00:59
So it's the same thing just a little different just to change the denominator.
01:04
But again, we're gonna assume it's a sample.
01:05
So we're gonna use this one and then the median is equal to what's the middle value? and so since there are 10, there's no exact middle value so we're gonna take essentially the average or the mean of these middle two values of the fifth and six values all right, so let's go ahead and do the mean and standard deviation, so first we're gonna do is sum all the x values together one hundred thirteen point three two divided by ten you get eleven point three three two if we do so, we'd probably surround the mean then to be eleven point three all right.
01:53
Now we're gonna do this as the sums of squares of x again, that's taking x minus x bar so this new here here we have this column is each x value minus the mean x bar value so five point seven eight minus this mean value squared you get this six point seven one minus eleven point three three two squared you get this value and so on so forth i'm up and this my friends right here this piece.
02:22
This is the the sum of squares of x 71 so we get the square root of that number three one three point three will say divided by nine because n is 10 10 is 9 and that is gonna give us our standard deviation of our sample five point eight nine nine.
02:42
So we'll just round that to five point.
02:48
All right so now we're gonna find the median and while we find the median, we're gonna go ahead and find some some other things, but it's gonna be the the five number summary so we're gonna find the median we're gonna find the first quartile q1 we're gonna find the minimum value which we can already see that's five point seven eight then we're gonna find the third quartile and then we're gonna find the max value which we already know.
03:25
This is called the five number summary.
03:26
We're not specifically asked for it, but just just for completeness we'll do that so the median like we said is gonna be the exact middle value.
03:36
So let me draw that in this anymore the exact middle value we don't have one.
03:46
So it's gonna be the middle here.
03:48
So this is going to be where the median is located the first and third quartile, it's essentially the medians of each of these halves after you we cut them we cut we use the mean to cut cut the data set in half so q1 is going to be the median of this upper half.
04:07
So this is q1.
04:09
It's this exact middle value and this lower half from those six to ten the exact middle value is fourteen point seven one so this is q3 all right, then we're gonna do that and then we're gonna get the inner quartile range iqr which is the difference q3 minus q1 all right, let's go do that get all right, there's that here's our here's that five number summary iqr the difference of q3 and q1 to get 7 .87 now what we want to do is identify outliers.
05:01
That's what this piece is for that's what all this stuff is for...