Calculating Changes in Standard Deviation without Given Summation Values

tzx9633

Homework Statement


There are 15 numbers on a list, and the mean is 25. The smallest number on the list is changed from 12.9 to 1.29.

a. Is it possible to determine by how much the mean changes? If so, by how much does it change?

b. Is it possible to determine the value of the mean after the change? If so, what is the value?

c. Is it possible to determine by how much the median changes? If so, by how much does it change?

d. Is it possible to determine by how much the standard deviation changes? If so, by how much does it change?

I have problem with part d , i don't have the ans .

Homework Equations

The Attempt at a Solution



So , we need to compute the new ∑x and ∑(x^2 ) , the new ∑x = old ∑x + 1.29 -12.9

the new ∑(x^2 ) = old ∑(x^2 ) - (12.9^2) + (1.29^2) ...
However , the old https://www.physicsforums.com/file:///C:/Users/User/AppData/Local/Temp/msohtmlclip1/01/clip_image002.pnghttps://www.physicsforums.com/file:///C:/Users/User/AppData/Local/Temp/msohtmlclip1/01/clip_image004.png∑x and ∑(x^2 )are not provided , so we can't calculate the changes



Correct me if i am wrong
[/B]
 
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You can find the old sum of values. Check what you did in (a).

You cannot find the new standard deviation as absolute number, but can you express the new standard deviation as function of the old one?
 
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mfb said:
You can find the old sum of values. Check what you did in (a).

You cannot find the new standard deviation as absolute number, but can you express the new standard deviation as function of the old one?
ok , i am able to get old ∑x , but , i can't get ∑(x^2) , how to express the new standard deviation as function of the old one ??
 
tzx9633 said:

Homework Statement


There are 15 numbers on a list, and the mean is 25. The smallest number on the list is changed from 12.9 to 1.29.

a. Is it possible to determine by how much the mean changes? If so, by how much does it change?

b. Is it possible to determine the value of the mean after the change? If so, what is the value?

c. Is it possible to determine by how much the median changes? If so, by how much does it change?

d. Is it possible to determine by how much the standard deviation changes? If so, by how much does it change?

I have problem with part d , i don't have the ans .

Homework Equations

The Attempt at a Solution



So , we need to compute the new ∑x and ∑(x^2 ) , the new ∑x = old ∑x + 1.29 -12.9

the new ∑(x^2 ) = old ∑(x^2 ) - (12.9^2) + (1.29^2) ...
However , the old https://www.physicsforums.com/file:///C:/Users/User/AppData/Local/Temp/msohtmlclip1/01/clip_image002.pnghttps://www.physicsforums.com/file:///C:/Users/User/AppData/Local/Temp/msohtmlclip1/01/clip_image004.png∑x and ∑(x^2 )are not provided , so we can't calculate the changes



Correct me if i am wrong [/B]

You should have no trouble doing (a) and (b). The answer to (c) is quite straightforward---just think some more about what is meant by "median". Finally, you can determine the change in variance, if you use the correct formulas. Again, write things down and think about what you know and don't know, or about what you can find out from the given problem data.
 
tzx9633 said:
ok , i am able to get old ∑x , but , i can't get ∑(x^2) , how to express the new standard deviation as function of the old one ??
Just call the old ∑(x^2) = c, a new variable I introduced.
Now express the new sum of squares as function of c, and then express both old and new standard deviation as function of c.
If you can reverse the first equation (express c as function of the standard deviation), you can plug it in the other one and get the new standard deviation as function of the old one.
 
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mfb said:
Just call the old ∑(x^2) = c, a new variable I introduced.
Now express the new sum of squares as function of c, and then express both old and new standard deviation as function of c.
If you can reverse the first equation (express c as function of the standard deviation), you can plug it in the other one and get the new standard deviation as function of the old one.
Here my ans , it's rather weird ... How to get the new standard deviation as function of the old one ?
 

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Something went wrong with 12.9^2 (or elsewhere in that line).
You just followed one part of my suggestions.
 
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mfb said:
Something went wrong with 12.9^2 (or elsewhere in that line).
You just followed one part of my suggestions.
which part is wrong ? can you point out ? I have checked thru the working , can't find which part is wrong
 
How do you get 8600 by squaring 12.9 and then subtracting something?
 
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