All right. So a poem is we have Owen, we're doing a simple random sample, which means we're taking 350 randomly. A simple kind of sample of 300 the engines and probably one of each engine. In a population running is 10.999 and exit poll number number running. And we're supposed to show a couple things. First is supposed to show that X is ah binomial, Let's get first. So for extra by, no meal needs to satisfy five properties. First is that we need to have in identical trials. This is true because we're doing the same experiment every time describing engine, turning it on, seeing what happens then. Second is that they're only two possible outcomes. This is again true because every engine will either run or not run. Then we need to have that the n Well, I do. Yes, you do. In in trials are independent. This is true, because if I turned in Injun on whether it runs or not, will have no impact on whether or not the next one runs or not, we need to have a probability of success equal a pee and then probability of failure he's equal to. And I wrote that, sloppy. I'm sorry, right? That better is equal one minus B, which is the case because we have other successes in this case, 0.99 then failure every 0.1. And in. The last thing is that X is equal to the total number of success is. And that is also true because we are letting ex count the total number of interests that are running after an hour so that we have shown that it is by no meal. So ex is by no meal. All right now for B here's was to find mu of X and signal Vex all supposed to interpret our results. All right, well, this is actually pretty straightforward. Mu of X is equal to, and by normal distribution is in times P, which is 350 times 0.999 which is 350 3 49.65 And that just means that if I do some I do for any simple random sample, the average or expected well, I will use expected value. The average number of running engines will be 349.65 In other words, this is Well, this is This is what is trying to say earlier. This is the expected value. So if I were to say, how many do you expect to run? Given a simple experiment, you would say 349.65 because that is the average average is tthe e expected value. And so we've interpreted. Let's see, it's that one. And then we need to do staring deviation of X that is equal to the square root of the variance of X, which is equal to the square root of N. P. Q is equal to this word of 350. His 0.999 has 0.1 is equal to 0.591 And what does that mean? That means that for any supporting a sample, the number of running engines will be within 0.591 of you have X, which was equal Thio. What do we say? It was equal to 3 49.65 In other words, if we think of the binomial CDF, I think the binomial CDF um, the average is going to be here. This is gonna and then right here. But it's too full, Stephen. So if I draw this kind of draw an example here, if I do the experience over and over again, I would expect that because I know the standard deviations 0.591 I respect that 60. That's 68% of the experiments that I worked hard to keep doing this simple random sample over and over again. I would expect that about 60% them would result in the average being well that the number of running engines being let's see of one sir deviation away from the mean is 10.591 So if I do this experiment over and over again, or take 350 random ones and I check how many ran, I would expect that the averages 349.65 and at every experiment that 60% of experiments would lie within 349.65 minus one standard deviation, which was or 2.59 2 343 149.65 plus 0.59 which would be 349 point. Those six 2 300 50 point 24. So, yes, if I did, like conceited 10 experiments. If I did 10 experiments, I would expect that of those 10 like roughly like 6 to 7 of them, or 60% of those 10 so 6 to 7 of them would result in the number of engines running, being somewhere between 304 9.6 to 350.24 In other words, I would expect 60% of time for all of them to work, since the whole number between these two numbers is 3 50 But that is like going way. The odd or trying to answer the simple answer is this furnace to standard, simple random sample. The number of running engines will be within 0.591 of New Vex. All right, and now the last one and was see fine. Probably the X is less cynical. 3 48 It's the probability and X is less than equal to 348. Well, most, um, a little note here. Most calculators, like you're using a T I 84. We'll give you the probability that X is less than or equal something So if we're using, uh, most car caters to answer this problem, we would just do the binomial CDF of. We have 350 trials, probably of successes 0.999 and we want to go up to 348. And that will get me. I used my online calculator 0.0 49 ish. So pretty unlikely another way to get the same answer. But if you were to do it by not using the sea of calculator and calculate the terms individually, you could do the probability of X. A single to 348 is equal to the probability earlobes one minus the probability that X is greater than 348 which is equal to one minus probability. That X is equal to 349 minus probability that X that's equal to 3 50 and I would be a simple way to do it as well. But if you calculate those two things and subtracting from one, you would get this answer. So the question was that if we have two of them fail out of this simple random sample experiment, um, and we have to fail What does that say about the advertised success rate of point 999? Well, 0.49 implies for the probability, all right, that little friendly probably that X less than or equal to 348. Being equal 2.49 implies that is very unlikely. Two engines fail, so the advertised of 99.9% is most. That's a bad most, isn't it most likely inaccurate? Because if we have to fail, that's incredibly unlikely to happen. So it would suggest that this advertised rate is false and we are done.