Here we have 12 refrigerators, seven of which have a defective compressor, and the other five do not. From the 12 we're going to randomly select six, and we're interested in the number of this election that have a defective compressor. So this is a hyper geometric random variable, the number that do you have a defective compressor because we have a population with two distinguishable groups that form of partition, and we're sampling six from that population without replacement. So the number X, which belongs to one of those groups, is a hyper geometric, random variable. So, as stated in the problem, we have a population of refrigerators, which equals 12. And in the group that we're interested in, that's the seven that have defective compressors that size of seven, and we're looking at six of them, so the selection sizes six or the sample size is six. So in party were asked for the probability that five of them of the sixth sampled have a defective compressor for hyper geometric random variable for the probability that five given a sample size of six, a group size of seven and a population size of 12. So before making this calculation. I'll just write out the general formula. So the probability mass function. So it's equal to amateurs X times and minus M over a small and minus X divided by population size. Choose the sample size so that corresponds to the following in our situation, and this should actually be choose. And this comes out to zero point 114 next for Part B were asked for the probability that at most, four have a defective compressor from our sample of six. So this is equal to this summation and notice that I'm summing Starting at X equals one so that we can't have X equals zero. Because if we're sampling six from a population of 12 then and seven of the 12 have a defective compressor, then at least one of the sampled refrigerators must have the defective compressor that's given by the following requirement for the number for X have the max of zero and n minus. N plus M is listener equal to X, which is Lessner equal to the minimum of small end. And, um so so that is, if X is a hyper geometric random variable, its value must be an integer that which satisfies this expression. Doing this calculation comes out to zero point 879 in Part C. We're looking for the probability that exceeds its mean value by more than one standard deviation. So we have to calculate both the mean value and the standard deviation for this random variable for hyper hyper geometric random variable. The expectation is given by this formula, which comes out to 3.5, and the variance is given by this formula. So we have 12 minus 6/12, minus one times six times 7/12 times one minus 7/12, which comes out to 0.795 which gives us a standard deviation, which is the square root of that value of 0.892 So we're looking for the probability that X is greater than the mean plus one standard deviation. So that's the probability that is greater than the mean, which is 3.5 plus the standard deviation 0.892 which is the probability that that's greater than 4.392 And since X can only take on your values, this is the probability that it is greater than five greater than or equal to five. And since our sample size is six, then this must be equal to the probability that X is equal to five, plus the probability that it is equal to six. This is actually the complement of what we calculated in Part B, and this is 0.1 to 1. And then, lastly, for Part D, were asked to assume that there are 400 refrigerators, 40 of which have defective compressors and were randomly selecting 15 of them. So whereas to approximate the probability that at most, five of that selection have defective compressors and we're as to provide, um, or convenient or less tedious way to calculate this than to use the hyper geometric distribution. So this is a situation where you can use the binomial distribution to approximate ah, hyper geometric. And this is admissible because we have a large number of population, so 400 and the possible number of successes is high, which is 40. That's the number of defective compressors among the 400 refrigerators. So if you have a large population and a large number of successes, which is M, you can use the binomial approximation. And to use the binomial approximation, this probability success is equal to am over the population size. So that's 40 over 400 in this situation or a 0.1. So to find the probability that at most five from our random selection have a faulty compressor. This is approximately the cumulative probability for a binomial based on a sample size of 15 and a probability success of 40 over 400 or 0.1. And this 15 is the size of our selection for Part D. So out of the 400 refrigerators, we're selecting a random sample of 15, and we're interesting the probability that less than five of these 15 have defective compressors, and this probability comes out to zero point 998