Okay, so we're given a normal distribution with a mean of 10 and a standard deviation of two. And what we're trying to find is the probability that the mean x value is at most 11 Over two different days. And so what I'm gonna do is cut this up into day one and day two and find the probabilities on day one and day two, and then Go from there. So on day one we have five people filling out the form, so and is equal to five. And what I'm gonna do is use something called a Z score, which basically just transforms our normal distribution into the standard normal distribution and transforms our Maximum value, or 11 into whatever that value would be on the standard distribution table, so that we can just look at that table to find our probabilities. So the Z score is the mean x value minus are you value divided by the standard deviation divided by the square root of the number of trials. And and so if I do this for 11, we would have 11 minus 10 divided by our standard deviation, which is too divided by squared five since we have five trials and this is equal to one over two, divided by the square to five. And now you can just simplify this two square to 5/2 and we can plug this into our probability. So we say the probability of Z being less than or equal to one, sorry, Square to 5/2. This is equal to our probability of our X men being less than equal to 11. So now we can just use the standard normal distribution table to find our probability. Um And so the last thing we're gonna want to do is Actually divide are square to five by two. So we have a value to look at for the standard distribution table. Standard normal distribution table, so this is squared of 5/2 is equal to one point 12 about. So if you look this up on the standard distribution table, we'd see that The probability of Z being less than or equal to 1.12 is eight 6.8686. And now what we want to do is find the probability on day two. So for day two we have N is equal to six and we're trying to find the same probability. So we're trying to find the probability of C sorry, of X being less than or equal to 11. So we can do the same thing. We can set our Z R. Z score equal to I mean X value minus you over the standard deviation divided by the squared of end. And if we do this for 11, again, going to see the only thing that changes is the number of trials or this squared of n. And so last time we had one over two divided by the square of five. This time we have 1/2 divided by the square of six. So that's just the squared of six over to And if we plug this into a calculator, we'll see that The square root of six divided by two is equal to around one 22 So now we can find the probability of RZ being less than or equal to 1.22, Which is it should be a little bit bigger than the other one. And it actually comes out to be .88 88 And now the last thing that we do is that we take these two probabilities .8888 10.8686. And since we're assuming that are probabilities are independent of each other, the two days are independent events. Where you can just say that the probability of, I'm gonna say of both days having a mean value of less than or equal to 11 is equal to the probability of the first day, multiplied by the probability of the second day. And we can do this because they're independent events. So This is equal 2.8. So the first day was point 8686, and the second day it was .88 88 And if we plug this into a calculator, You can see that this comes out to about .772.