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:
The wording "fewer than" is translated to the math symbol "less than". The variable n represents the number of students.
<em>the number of students</em> is fewer than 18
translates to ...
n < 18
__
The wording "at least" translates to the math symbol "greater than or equal to."
<em>the number of students</em> is at least 15
translates to ...
n ≥ 15
Answer:
31.4 is the answer
Step-by-step explanation:
Answer: 16, 10, 64
Step-by-step explanation:
64/4=16, so the 3rd number is 4x the first.
16-6=10. 6 less than the first number.
16+10+64=90.
15/x = 20/100
15*100 = 20*x
1500 = 20x
1500/20 = 20x/20
75 = x
15/75 = 20/100
the expected number of couples who met online is 75
