00:01
Okay, so we have a random variable x with mean mu and variance sigma squared.
00:06
We want to use shebyshev's inequality to find the smallest k in the natural numbers, so positive integers, for which this probability is at least 0 .95 and at least 0 .99.
00:18
So firstly, before we start making a move on these questions, we can simplify this and use shebysheves inequality, and then we can just apply what we get to both parts a and b.
00:28
So i'm going to rearrange this.
00:30
This is the same as the probability of minus k sigma being less than equal to x minus mu, less than equal to k sigma.
00:42
So all i did was subtract mu from all sides of these inequalities.
00:47
And now this is the same as the probability that the absolute value of x minus mu is less than or equal to k sigma.
00:59
Okay, but shibbyshev's inequality tells us.
01:01
Something about the absolute value of x minus mu being greater than some value k sigma here.
01:08
So we can rewrite this as one minus the probability that x minus mu is greater than k sigma.
01:19
And now shabby shev's inequality tells us that this is less than or equal to 1 over k squared.
01:25
Since we have a minus, we have that this whole thing is greater than or equal to minus 1 over k squared.
01:31
So this is greater than or equal to 1 minus 1 over k squared.
01:37
So these steps are just simple use of probabilities.
01:41
And then this step from here to here is shebyshev's inequality.
01:45
And now we can look at part a and part b.
01:49
So part a, we want this probability to be at least 0 .95...