Supposing, for the sake of illustration, that the mean is 31.2 and the std. dev. is 1.9.
This probability can be calculated by finding z-scores and their corresponding areas under the std. normal curve.
34 in - 31.2 in
The area under this curve to the left of z = -------------------- = 1.47 (for 34 in)
1.9
32 in - 31.2 in
and that to the left of 32 in is z = ---------------------- = 0.421
1.9
Know how to use a table of z-scores to find these two areas? If not, let me know and I'll go over that with you.
My TI-83 calculator provided the following result:
normalcdf(32, 34, 31.2, 1.9) = 0.267 (answer to this sample problem)
The first one is 15 mornings
The value of is 10 I think
Answer:
So to graph using a table, the first step is...well... make a table. so you have a column of x and y. under the x column write down a bunch of values you would like to test. I am going to test -2 through 2.
x | y
-2|
-1|
0 |
1 |
2 |
Next, we have to plug each of those values into the equation y=2x to see what the corresponding y value is.
when x=-2 y=2(-2)=-4, when x=-1 y=2(-1)=-2, and so on...
x | y
-2|-4
-1|-2
0 |0
1 |2
2 |4
Next, graph those points on a coordinate plane. Each row on the table is a point on the graph, (x,y). Hope that helps!
