Well, I'm sorry it took you so long to find, but I am glad that you find it helpful. Do please have a look through some of my other videos in case they are of use to you.
Hello. It works exactly the same as the third example in the video: P(x>12). There is no difference for allowing "equals 6", since under the Normal distribution P(X=n)=0 for all n, i.e. the calculation for greater than is the same as for greater than or equals to. Hope that helps!
Hold on a second... I'm using a TI Voyager 200, made maybe around the same time as this calculator... NormalCdf is giving me very difference answers than your calculator.
Unfortunately it is not as straightforward as that. If you want p(x>a)=0.9 you are looking at a value to the left of the mean. The calculator only deals with the values to the right of the mean and then finding probabilities less than that. If you were answering the question you have posed, you would have to work out the answer to the question that I posed, then find the distance from the mean to the value the calculator gave (16.407 - 10 = 6.407), and then subtract this from the mean to get the answer to your question (10 - 6.407 = 3.593). I hope that makes sense.
Ignore this video, as there is easier functionality on the Casio CG 20 ...press menu 2 statistics, choose DIST, NORM then either Npd, Ncd or InvN. Each option prompts you to enter the required variables without the need to remember the formula format. This IS a much easier way to access the function.
exactly it's way better because it tells you where to put the bounds, SD and mean. you don't have to line them all up in order. but I guess that was the way Mr oyrs was more comfortable doing it
What do you mean by "standard normal"? If you mean X~N(0,1), simply use those values (0 and 1) in your expression for the mean and standard deviation. If you mean something else, please let me know and I will see if I can help.
It should work. Don't forget that when you enter those values into the expression that you swap the order - see the video. You might be entering 0 as the standard deviation, which cannot happen.
