So we're going to have to distributions in this problem. One is the distribution for individual nails and looking at their heights, and we're assuming the main is 68 inches, with a standard deviation of three inches. So Approximately here is 71 1 standard deviation away, and the next one is going to have groups of size nine And that sampling distribution and then group of size nine, and then you find the main. So our sampling distribution is going to have a smaller standard deviation, still is going to be centered At 68", but now it's going to have a standard deviation Of three divided by the square root of nine, which is one. So our first question asked, what's the likelihood that you sample one person and you get that one person to be between 67 And 69 tall. And we can see that way, that means we would need to convert this to a Z value, so I have 67 minus 68 divided by the standard deviation. Now it's a Z value And I have 69 -68 Over three. And so we can see in both cases that we get uh negative one third, can I get positive one third? So we're symmetrical and I'm going to utilize my normal CDF for this And second and distribution normal CDF, and I'm going to use the lower value is negative .3 Repeating, and the upper is .3 repeating. And leave the mean at zero and a standard deviation at one. And when I do that, I find out that that probability comes out to be about 26%. Part B says now we're looking at this distribution and what's the likelihood of having the mean be between 67 and 69. And when we convert that into Z values again we take the number minus the mean, divided by the standard deviation, which is that one? Yeah. And this month groups this is the 69 minus this, 68 Do I, too? By that standard deviation up here, which is one. And so I get my Z values to be negative 1-1. Yeah, and again, I know that's about 68% but I'm going to use my normal CDF button again, normal cbf use my lower as negative one. My upper is one. And leave the mean at zero and standard deviation at one And we get .68, 2 rounds to seven. Now, part C. Yes, this answer is much larger, we expect that. But why? Basically? Because this standard deviation of this second problem is much smaller because this is the standard deviation that I use For groups of size nine. So we expect for those same numerical values to have a much higher percentage of the X bars being between these two numbers than individual values. So I didn't write all that down. But hopefully you see that this standard deviation smaller, therefore We would expect more values to be in between 67 and 69.