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:
Step-by-step explanation:
i think b and c i know a and d are wrong but hope it helps
Answer:
easy :) answer is 12x-8y
Step-by-step explanation:
all you do is combine like terms
1. the 10x + 2x which is 12x
2. (this part gets tricky) you have a NEGITIVE 5 minus 3 so if you are subtracting by a negitive number you add (5+3=8) and keep the negitive (-8y)
Answer:
i know the answer its 13
Step-by-step explanation:
its 13
Answer: more than 15000 total enrollments) have some.... ... (more than 15,000 total enrollments) have some online offerings. Suppose you randomly pick 13 such institutions.
Step-by-step explanation: more than 15000 total enrollments) have some.... ... (more than 15,000 total enrollments) have some online offerings. Suppose you randomly pick 13 such institutions.
