# Calculating Changes in Standard Deviation without Given Summation Values

• tzx9633
I'm sorry I can't do your homework for you, and I'm also not really sure if you are just trying to cheat or actually learn. Please show your work and explain your reasoning, and I can try to guide you in the right direction.
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 .

## 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 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?

tzx9633
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 .

## 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.

tzx9633
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.

tzx9633
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?

## 1. What is Standard Deviation?

Standard Deviation is a measure of how spread out a set of data is from its average value. It measures the amount of variation or dispersion from the average or mean of a data set.

## 2. How is Standard Deviation calculated?

To calculate Standard Deviation, you need to follow these steps:1. Calculate the mean of the data set.2. For each data point, subtract the mean and square the result.3. Sum up all the squared differences.4. Divide the sum by the total number of data points.5. Take the square root of the result to get the Standard Deviation.

## 3. What does it mean when Standard Deviation changes?

When Standard Deviation changes, it indicates that the data points are either more spread out or less spread out from the mean. A higher Standard Deviation means more variation in the data, while a lower Standard Deviation means less variation.

## 4. How does Standard Deviation affect the shape of a distribution?

The Standard Deviation directly affects the shape of a distribution. A higher Standard Deviation results in a wider and more spread out distribution, while a lower Standard Deviation results in a narrower and more concentrated distribution.

## 5. Can Standard Deviation be negative?

No, Standard Deviation cannot be negative. It is always a positive value or zero. A value of zero indicates that all the data points are the same, while a positive value indicates the amount of variation from the mean.

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