Best Estimate from mean and mean squared deviation

Hi,

I don't have a clue to the correct way of attempting this question:

Eight measurements of the volume of a block of iron had a mean value of 26.52cm3 and a mean square deviation 0.025cm6.

Fifteen measurements of a block of aluminium gave the corresponding values 8.72cm3 and 0.058cm6.

If the density of iron is 7.88g/cm3 and of aluminium is 2.70g/cm3, what is the best estimate of the total mass of the two blocks and of the precision of the method of measurement?

Re: Best Estimate from mean and mean squared deviation

Quote:

Originally Posted by

**zeeshahmad** Hi,

I don't have a clue to the correct way of attempting this question:

Eight measurements of the volume of a block of iron had a mean value of 26.52cm3 and a mean square deviation 0.025cm6.

Fifteen measurements of a block of aluminium gave the corresponding values 8.72cm3 and 0.058cm6.

If the density of iron is 7.88g/cm3 and of aluminium is 2.70g/cm3, what is the best estimate of the total mass of the two blocks and of the precision of the method of measurement?

Hi zeeshahmad! :)

The formula for the total mass is:

$\displaystyle m_{tot} = \rho_{iron} V_{iron} + \rho_{alum} V_{alum}$

Just fill in the numbers...

When you multiply by a constant, the mean deviation goes up by the same factor.

Since you have a mean *squared *deviation, that one goes up by the factor squared.

When you add to measurements, their respective squared deviations should be added.

Re: Best Estimate from mean and mean squared deviation

Hi,

This method of answering doesn't make use of "Eight measurements.." and "Fifteen measurements" though..

Re: Best Estimate from mean and mean squared deviation

Quote:

Originally Posted by

**zeeshahmad** Hi,

This method of answering doesn't make use of "Eight measurements.." and "Fifteen measurements" though..

Correct.

Those numbers have already been processed, giving a *mean *square deviation.

Apparently the blocks have been measured that many times to get the precisions into a desirable range.

Re: Best Estimate from mean and mean squared deviation

Oh wait, it may be that they want you to calculate the proper standard deviations of those measurements yourself.

Since they gave you a *mean *squared deviation, that would be the sum of the squared deviations, divided by the number of measurements.

However, the proper squared standard deviation, is the sum of the squared deviations, divided by the number of measurements* ***minus one**.

It would mean that you have to multiply the mean squared deviation by n/(n-1).

That is, in the case of the iron:$\displaystyle \text{squared standard deviation} = \text{mean squared deviation} \times {8 \over 8 - 1}$

Re: Best Estimate from mean and mean squared deviation

Thank you very much for your help!

I've done it quicker then expected.

Having done the question means I don't have to stay up late worrying.

:)