00:03
We're given the random variable x, which represents height and y, which represents weight for an american male.
00:10
And we're told that x and y have a bivariate normal distribution, that the mean and standard deviation of the heights are 70 and 3 inches, and the mean and standard deviation of the weights are 170 pounds and 20 pounds, and that the correlation coefficient is row equals 0 .9.
00:30
In part a, we're asked to determine the distribution of y, given, that x is equal to 68 so the weight distribution for a 5 foot 8 inch american male well we have that the conditional variable y given x equals x is normal since y and x are bivariate with a mean which is the mean of y mu 2 plus the correlation code coefficient row times the standard deviation of y, sigma 2, times x minus sigma or mu 1 over sigma 1 and a variance which again by a formula from this section which is 1 minus the correlation coefficient row squared times standard deviation or the variance of y.
02:09
So in part a we want to just substitute into these expressions and therefore it follows that the mean is going to be equal to what we're given that the average weight is 170, so 170 plus we're given that row is 0 .9.
02:29
We're given that the standard deviation of y is 20 times, and then we're given that we're looking at 68 inch persons, so 68 minus the mean 70, over the standard deviation, which we're given is 3.
02:52
And this is a proxy.
02:54
Approximately 158.
02:57
Because we're looking at weight, this is pounds.
03:10
That's the mean, and the variance is going to be 1 minus, we're given this is 0 .9 squared, times the variance of y, we're given the standard deviation of y is 20, so this is times 20 squared, and this works out to be 76.
03:39
And therefore, it follows that the standard deviation of y is the square root of this, which is calculated to be approximately 8 .72 pounds.
03:58
Therefore, it follows that the weight distribution of american males with a height of 5 feet 8 inches, well, this is going to be normal with a mean of 158 pounds, and a standard deviation of 8 .72 pounds.
04:51
Next, in part b, we're asked to determine the distribution of y, given the x is equal to 70.
04:58
In other words, the weight distribution for a 5 -10 -inch american males...