what is a zscore and how is it useful?

Question

anonymous · Answer

http://en.wikipedia.org/wiki/Standard_score - I haz no idea... read wiki

anonymous · Answer

Z-score formula for predicting bankruptcy  this is what wiki says

anonymous · Answer

other than that idk, haven't done them

amistre64 · Answer

the zscore is what we use to measure with when faced with 2 or more different relations.  suppose we have one normal distribution (A) with a mean of 80 and a standard deviation of 15.  and we would like to compare it to another normal distribution (B) that has a mean of 143 and a standard deviation of 52.

How could we compare the relative worth of values using a standard unit of measurement?  Enter in from stage left, the zscore.

The zscore is a measure of the "unit" normal distribution.  I say unit because everyone likes to compare things to a "unit" this or that; circle, hyperbola, vector, etc...

The unit normal distribution then has a mean of 0 and a standard distribution of 1.  This makes integrating the equation of the curve much more palatable.

amistre64 · Answer

So lets compare some made up values to see which is one is relatively closer to their mean.

(A) is 54,  and (B) is 102

in A, we have a mean and all its points spreading out from 80, so lets shift it all so that the mean is at 0:

   (54-80) - (80-80) = -26 - 0 = -26

same for (B); 

    (102-143) - (143-143) = -31 - 0 = = -31

amistre64 · Answer

But our distance from each mean is comparable to its standard deviation away from it; which is different in each case.

For (A) the sd is 15; and to get this to conform to the unit sd of "1"; we have to divide it by ... 15.  15/15 = 1

    so our zscore at (A) is:  -26/15 = -1.733

For (B) the sd is 52; and to get this to conform to the unit sd of "1"; we have to divide it by ... 52.  52/52 = 1

   and our zscore at (B) is:  -31/52 = -0.596

Now we are at a stage in the game where we can compare the 2 distribution on the same level, on equal playing fields.

the value for (A) is 1.733 zscores away from its mean

the value for (B) is 0.596 zscores away from its mean

And so the value for (B) is relatively closer to its mean than the value for (A).