Find Correct standard deviation

Question

For 100 chemistry tests, the mean score was 40 and the standard deviation was 10

However two tests were wrongly scored as 30 and 70 in place of 3 and 27 respectively. Find the correct standard deviation.

Answer 1

Luc, it turns out, you CAN adjust the standard deviation accurately, using a VERY useful algebraic identity.

As Dan said, we don’t have all the x_i or the x_i - mean, but here’s the nice identity:

sum( (x_i - mean )²) = sum(x_i²) - (1/n)( sum(x_i) )²

in other words,

(sum of the squares) minus (1/n)*(square of the sum)

I’ll show you how it helps.

We will start with the original numbers, before the error was detected:

STD = 10

Sum(x_i) = 4000

using the new identity,

std = sqrt(sum(x_i²) - (1/n)( sum(x_i) ))²

From this we can solve for sum(x_i²)

std²= sum(x_i²) - (1/n)(( sum(x_i) )²)

sum(x_i²) = std² + (1/n)( (sum(x_i))² )

100 + 4000²/100 = 160100

Now we can change those numbers to make the new standard deviation

The new sum(x_i ) is the old sum minus the erroneous scores (30 and 70) plus the correct values (3 and 27) = 

4000 -30 -70 + 3 + 27 = 3930

The new sum(x_i²) Is the old sum minus the squares of the erroneous scores (900 and 4900) plus the squares of the correct scores (9 and 729) = 160100 -5800 + 738 = 154300+738 = 155038

Now the new standard deviation is

std = sqrt(sum(x_i²) - (1/n)( (sum(x_i))²)

= sqrt( 155038 - 3930²/100)

= sqrt( 155038 - 15444900/100)

= sqrt( 155038 - 154449)

= sqrt( 589 ) = roughly 24.25

I think the most important lesson of this exercise is that identity

sum( (x_i - mean )²) = sum(x_i²) - (1/n)( sum(x_i) )²

Answer 2

Because there isn't 100 scores I'll stick with avg; then 3 and 27. The avg is 40*98+3+27=3950/100=39.5.

standard=1/amount((score-avg)^2 amt times (in this case 100)).