standard deviation and mean
Message posted by Jamie Arsenault (via 198.164.200.233) on September 24, 2001 at 5:51 PM (ET)

I can't figure out this problem:
Age students faculty
0-2 23 30
3-5 33 47
6-8 63 36
9-11 68 30
12-14 19 8
15-17 10 0
18-20 1 0
21-23 0 1

Find the two means, the two standard deviations and the two mean absolute deviations.
Please help, thanks

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Re: standard deviation and mean
Message posted by Darius (via 200.23.217.10) on September 26, 2001 at 3:21 PM (ET)

With N = Sum(frecuencies)

Class_Mark(i) = Lim_Sup(i)-Lim_Inf(i)
SFC=Sum[frecuency(i)*Class_Mark(i)]
SFC2=Sum[frecuency(i)*Class_Mark(i)^2]
Mean for grouped data = SFC/N
DS for grouped data = Sqrt{(N*SFC2-SFC^2)/[N*(N-1)]}

So: Mean1=7.79; DS1=3.79; Mean2=5.875;DS2=3.65