# Calculating correct mean and standard deviation

• Aug 14th 2010, 06:47 AM
blackdragon190
Calculating correct mean and standard deviation
This question is asking me to calculate the correct mean and standard deviations for data that was over-measured by 3.5cm. The problem is that I only have the sample mean and standard deviation, both calculated from the incorrect measurements (which it does not give). What do I do...? There were 25 measurements taken. I have a feeling that I should calculate the mean and standard deviation using 25 values of "3.5cm" and then subtract the results from the original incorrect values, but this is a total guess.
• Aug 14th 2010, 07:43 AM
Quote:

Originally Posted by blackdragon190
This question is asking me to calculate the correct mean and standard deviations for data that was over-measured by 3.5cm. The problem is that I only have the sample mean and standard deviation, both calculated from the incorrect measurements (which it does not give). What do I do...? There were 25 measurements taken. I have a feeling that I should calculate the mean and standard deviation using 25 values of "3.5cm" and then subtract the results from the original incorrect values, but this is a total guess.

If the all the data samples were overmeasured by 3.5cm, then the sample mean is 3.5cm too high.

$\displaystyle\huge\frac{a+3.5+b+3.5+c+3.5+......}{ 25}=\frac{a+b+c+....}{25}+\frac{25(3.5)}{25}$

where a, b, c.... are the true sample values.

Will this make any difference to the sample standard deviation?
• Aug 16th 2010, 11:51 PM
blackdragon190
Thanks :) It'll make the standard deviation smaller, right?
• Aug 17th 2010, 01:44 AM
$\displaystyle\huge\sigma=\sqrt{\frac{[(a+3.5)-(\mu+3.5)]^2+[(b+3.5-(\mu+3.5)]^2+....}{n}}=\sqrt{\frac{(a-\mu)^2+(b-\mu)^2+...}{n}}$