00:01
Hey there, welcome to numerate.
00:03
So we are looking at the credit score of an average american and the average credit score is 682 with a standard deviation in 59.
00:12
So this is normally distributed this data here.
00:15
So we had to determine the credit score intervals for the specified amount of standard deviation.
00:23
So let's write our mean and standard deviation given to m.
00:29
Since this is dealing with the population, we're going to denoted as mule equals m.
00:37
Mule equals 682 and our standard deviation which is denoted as sigma sigma equals 59 so mean equals 682 and standard deviation equals 59 so let's jump into part a first so for a, determine the interval of the credit scores there are one standard deviation around the mean so basically the standard deviation shows how far apart the numbers are from the mean, right? so therefore, what we're going to do to find an interval is basically the mean, 682 plus and minus 59 because one standard deviation.
01:29
So basically you can denote it also as 59 times one because one standard deviation.
01:34
So let's calculate our intervals here.
01:36
So it's a lower interval when we subtract an upper interval when we add.
01:41
So with that, let's calculate it.
01:43
So we will have 682 minus 59, which equals 623, and we have 682 plus 59, which will equal 741.
02:04
Right, so that will be our interval here.
02:08
So let's go move on to b.
02:09
Speed is pretty similar, but we try to find two standard deviations away from the mean.
02:14
So basically the mean is 860, 682 plus and minus.
02:18
59 times 2.
02:22
Okay...