00:01
So for this problem, the first thing that i'm going to do here is create a table of the dice outcome, then the payout that results.
00:11
So we know that the different possible outcomes on die, or pardon me, on the dice or on the die, however you want to say it, would be 1, 2, 3, 4, 5, or 6.
00:22
We know that we win nothing if the number of spots is odd, so that would be zero payout if we have 1, 3, or 5.
00:31
And we're told it's $1 for a $2 or a $4 and $10 for a 6.
00:37
Now we know that the probability of each outcome is going to be equal to 1 over 6 because we're assuming at least it's a fair die.
00:47
So to find the expected value now for part a, to find the expected value of the payout, so that's e of x, what we do is we take the sum of the product of each probability with the corresponding payout.
01:03
So i'll ignore all of the ones where the payout is zero, since zero times anything is still zero.
01:09
But we'd have payout of one with probability 1 over 6, then payout of 1 with a probability of 1 over 6.
01:17
Then we have a payout of 10 with a probability of 1 over 6.
01:21
So we have an expected value of 2 .0.
01:25
Then to find the standard deviation, the first thing that i'll do is find the expected value of x squared.
01:31
So here we're squaring each one of the payouts, which i'll note we can quite easily see, well, the payouts were 1 and then 1, first of all.
01:41
So 1 squared is still 1.
01:43
We don't need to change those.
01:45
But for that 10, we will need to square that.
01:48
We can see that the expected value of x squared is equal to 17 .0.
01:53
Now, the standard deviation is equal to the square root of the expected value of x squared minus the square of the expected value of x so plugging in all of our values here this is going to be equal to the square root of 17 .0 minus two squared so we find that the standard deviation is equal to approximately 3 .606 i'll round that to three decimal places 3 .606 then for part b if we play twice then we know that the different possible outcomes will be a little bit more complicated because well we could have one and one actually one second here just create a entire new outcome table the different possible outcomes well we could have one and one have one in two one in three one and four one and five one and six then go on continuing on downwards so actually the way that'll set up the table just to make this kind of efficient to do is we have i1 result i 2 result erase what i have so far so we would start out with just one with all of its possible companions so to speak so we can have one and one two three four five or six then pattern is going to continue on downwards we have two or let's see here have two and one two and one two 2 and 2, 2 and 3, 2 and 4, 2 and 5, 2 and 6, and so on.
04:02
And then to, i'll pause the recording and fill out the rest of the table here.
04:07
All right, so here we can see i've gone through and calculated what would be the result for each one of the possible pairings.
04:15
Now, what i'll do is create a table of just the individual unique payout values and their corresponding probabilities.
04:26
So the unique payout values are 0, 1, 2, 10, 11, 20.
04:34
And the different possible individual payout values would be, or pardon me, the different probabilities can find by counting up the number of times that payout occurs and dividing by the total number of observations, which is 36.
04:48
So we'd have 9 over 36, for instance, for the first, and then one moment here.
04:54
I'm just going to do a quick calculation off screen.
05:01
One moment.
05:03
Okay, so figuring out the different counts and their corresponding probabilities, we'll find that the different possible payout or probabilities for each payout are 0 .25, 0 .33, 0 .11, 0 .11, 0 .11, and 0 .03 .03.
05:21
So now we can go through and apply the same procedure as we did before for finding the expected value in the standard deviation...