I think you need to go back and review just working with x's and z-scores based on populations and then re-attempt doing it with samples.
That standard deviation is small because the numbers you're working with to begin with are small. That is not relevant. Standard deviations are also not "to the right or left" on the graph. A typical graph will have three standard deviations both to the right and to the left. A standard deviation is not one point on the graph, nor an area. I don't have any idea how you used that to get 5720.
Do you know how to use a z chart? You have not even brought that into this that I can see. Do you know how to find the z score for a population based on the population mean and standard deviation? It helps to practice that first and get a feel for it.
Doing it for a sample is basically the same, except that you're now working with an x-bar instead of just an x, and you have to find the standard deviation for the sample mean. What you have is for the sample, not the sample mean. That equation is:
When working with a sample, you won't even be using the .11.
What is on the graph is the mean in the center, and then standard deviations out to both the right and left. You need to compare the x you're looking for to where your mean is to find out if it's to the right or the left of the mean. The line is the ct. numbers. That is, 1.52 goes in the center.
You also need to relate these two different values to each other: uncut and cut. You're wanting 1 ct after it's cut, which is 2/3 of the uncut. Since everything else is given as uncut figures, what would the original uncut number be if 1 ct is 2/3 of it?
Once you have all that, go do some reading on this thread and see what of this you can figure out:
https://www.askmehelpdesk.com/math-s...le-392066.html
Whatever you come up with, it helps if you show your calculations so we know what you're doing, where you might be going wrong, and therefore how we can correct you.