Asked by lijm
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).
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).
Answers
Answered by
lijm
For the second table:
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.
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.
Answered by
lijm
For part c, it looks like the mean is larger for NY so I think NY has more court cases than Pennsylvania.
Answered by
Reiny
New York:
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.
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.