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: 1/4
Step-by-step explanation: Anything to a negative power is the same as the reciprocol to the absolute value of the power
The eccentricity of an eclipse is found by using the
formula, eccentricity = c/a
Where,
c represents the distance between the center and a focus.
a represent the distance between that focus and a vertex
The numerical value of the eccentricity of an eclipse ranges
between 0 and 1