00:01
So for part a here, we have that x, the number of children in the sample who suffer from the condition is going to be distributed effectively as a binomial with 1 ,797 trials, and the probability of, quote, unquote, success, or the probability of kleinfelter syndrome is 0 .0014.
00:23
So for the probability that more than two of the male births have klein -felter syndrome, so probability that x is strictly greater than 2, we can calculate as 1 minus the probability that x is less than or equal to 2.
00:38
So we can do that as 1 minus be.
00:44
Now the probability that x equals 2, we would do as 1 ,797 choose 2, or in my software here, that's binomial 1797 comma 2, times probability of success, 0 .0014 to the power of 2 times probability of failure, 1 minus 0 .0014, to the power of 1797 minus 2.
01:07
And then we would subtract off the same for when we have value of 1.
01:13
So 1797 choose 1, 0 .0014 to the power of 1 times that are, of 1797 minus 1.
01:24
And then lastly, we do, technically it's 1797 choose 0, which is just going to be 1.
01:31
And we have 0 .0014 to the power of 0, which is also just going to be 1.
01:35
And then we have 1 minus 0 .0014 to the bar of 1797.
01:39
So we get that our probability is going to be roughly 0 .460 to 3 decimal places, 0 .4603.
01:48
Then for part b, using the poisson approximation, we have that the lambda for our poisson distribution...