The following table gives the total caseload in the New York State courts form 2004 through 2009.
Year 2004 2005 2006 2007 2008 2009
Cases (in MM) 4.2 4.3 4.6 4.5 4.7 4.7
A) Find the mean and standard deviation of the data.
B) Consider the second table for the state of Pennsylvania
Year 2004 2005 2006 2007 2008 2009
Cases (in MM) 3.4 3.5 3.2 3.7 3.9 3.8
Find the mean and standard deviation of the second table
C) What does a comparison of the means say about the number of cases?
a) Find mean. To do this I made a table to include the cases as probability distributions.
Frequencies: 4.2 appears once, 4.3 appears once, 4.6 appears once, 4.5 appears once, 4.7 appears twice. 6 total items.
X 4.2 4.3 4.6 4.5 4.7
1/6 1/6 1/6 1/6 2/6
So then the mean is computed like this:
(4.2) * (1/6) + (4.3) * (1/6) + (4.6) * (1/6) + (4.5)* (1/6) + (4.7) * (2/6)
= 4.5
standard deviation :
set up like this ? (4.2)*(1/6-4.5)^2+(4.3)*(1/6-4.5)^2+(4.6)*(1/6-4.5)^2+(4.5)*(1/6-4.5)^2+(4.7)*(2/6-4.5)^2
Take the answer and square root it.(I think it is 20 but I maybe made a mistake or didn't round).
3 answers
Frequencies: 3.4 appears once, 3.5 appears once, 3.2 appears once, 3.7 appears once, 3.9 appears once, 3.8 appears once. 6 total items.
X 3.4 3.5 3.2 3.7 3.9 3.8
p(x=x) 1/6 1/6 1/6 1/6 1/6
(3.4) * (1/6) + (3.5) * (1/6)+(3.2) * (1/6) + (3.7) * (1/6)+ (3.9) * (1/6) + (3.8) * (1/6) = 3.6 = mean
standard deviation =
(3.4)*(1/6-3.6)^2+(3.5)*(1/6-3.5)^2+(3.2)*(1/6-3.6)^2+(3.7)*(1/6-3.6)^2+(3.9)*(1/6-3.6)^2+(3.8)*(1/6-3.6)
square the answer and i think it might be 16 but please correct me if i'm wrong.
mean = (4.2 + 4.3 + 4.6 + 4.5 + 4.7 + 4.7)/6 = 4.55
squares of differences:
(4.55-4.2)^2 = .1225
(4.55-4.3)^2 = .0625
(4.55-4.6)^2 = .0025
(4.55-4.5)^2 = .0025
(4.55-4.7)^2 = .0225
(4.55-4.7)^2 = .0225
sd = √(sum of above/6) = 19685..
I carried all decimals in my calculator, leave it up to you to round off if needed
repeat for Penns.