Question

Statistics help, please
For a group of 200 candidates, the mean and standard deviation of scores were found to be 40 and 15. Later on, it was discovered that the scores 43 and 35 were misread as 34 and 53. Find the corrected mean and standard deviation

Answer 1

Just did it quickly to give an idea of the approach.
N=200
\(\mu\)'=40
\(\sigma\)'=15

\(\sum\)x=N(\(\sigma\)')-34-53+43+35
\(\mu\)=(\(\sum\)x)/N=39.955
\(\delta\)\(\mu\)=\(\mu\)-\(\mu\)'=39.955-40=-0.045

\(\sum\)(x-\(\mu')^2\)-\(\sum\)(x-\(\mu')^2\)
=\(\sum\)(x-(\(\mu+\delta\mu\)))^2+\)-\(\sum\)(x-\(\mu)^2\)
=\(\sum\)(2\(\mu \delta\)\(\mu -2x \delta\mu +\delta\mu^2)\)
=N\(\delta \mu^2\)


S'=\(\sum(x-\mu)^2\)
=\(\sum(x-\mu+\delta\mu)^2-N\delta\mu^2\)
=\(200*(15^2-0.045^2)\)
=44999.595
Finally, corrected sum
\(S=S'-(34-\mu)^2-(53-\mu)^2+(43-\mu)^2+(35-\mu)^2\)
\(=44999.595-(53-\mu)^2+(43-\mu)^2+(35-\mu)^2\)
\(=44827.785\)

\(\sigma=sqrt(S/N)=14.971\)

(please check derivation and numerical values)

Answer 2

@mathmate Thanks for the solution:
My question :I believe that there is a mistake at \(\sum(x-\mu')^2 -\sum(x-\mu')^2\)
It should be \(\sum(x-\mu')^2 -\sum(x-\mu)^2\)
but I don't get the third line and the last one which is \(=N\delta \mu^2\)

Answer 3

Yes, you're right. That was a typo.
If we expand the two expressions, and do the actual subtraction, we get
\(2\delta x\mu - 2x\delta x+(\delta x)^2\)
If we sum it over the N items, the expression becomes
\(2N\delta x\mu - 2\sum x\delta x+(\delta x)^2\)
which equals
\(2N\delta x\mu - 2N\mu\delta x+(\delta x)^2\) because \(\mu=(\sum x ) /N\)
the first two terms cancel and that leaves \((\delta x)^2\)

Answer 4

Sorry, the final result should read \(N(\delta x)^2\) since we summed a constant over N.