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)
Answer:
x=15
Step-by-step explanation:
(10*9)/6=10*x/10
90/6=x
15=x
4
4*3 is 12 but 4 is not a multiple of 6
Answer:
x=2
Step-by-step explanation:
6250/2=3125
5^5=3125
x+3=5
x=2
The answer for x is:
x= 0.7-0.83 = -0.13
