# Thread: Calculating correct mean and standard deviation

1. ## 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.

2. 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?

3. Thanks It'll make the standard deviation smaller, right?

4. Originally Posted by blackdragon190
Thanks It'll make the standard deviation smaller, right?
No,
the standard deviation will remain the same because it's calculated using the "differences" between the samples and the mean
(the deviations from the mean remain the same).

$\displaystyle\huge\sigma=\sqrt{\frac{[(a+3.5)-(\mu+3.5)]^2+[(b+3.5-(\mu+3.5)]^2+....}{n}}=\sqrt{\frac{(a-\mu)^2+(b-\mu)^2+...}{n}}$

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