Suppose the distribution of a set of scores has a mean of 47 and a standard deviation of 14. If 4 is added to each score, what will be the mean, standard deviation, and variance of the distribution of new scores?

Answers:

Explanation:

Consider the set {x1,x2,x3,...,xn}. The numbers 1,2,3,...,n represent the index of each x value. So we have n numbers we want to find the average.

The sum of those values is

S = x1+x2+x3+...+xn

When dividing that over n, we get the average A

A = S/n

A = (x1+x2+x3+...+xn)/n

Now consider adding 4 to each x value

We have this new set

{x1+4, x2+4,x3+4,...,xn+4}

The new sum T is

T = (x1+4)+(x2+4)+(x3+4)+...+(xn+4)

T = (x1+x2+x3+...+xn) + (4+4+4...+4)

T = S + 4n

Divide both sides by n

T/n = (S+4n)/n

T/n = (S/n) + (4n/n)

T/n = (S/n) + 4

T/n = 47+4

T/n = 51

The new average T/n is 4 larger compared to the old average S/n. So we just add 4 to the old average 47 to get 51 as the new average.

If we shift every value the same amount, then the standard deviation and variance will not change. This is because both of those items measure how spread out a distribution is. Shifting each item the same amount, keeps the same spread and shape of the distribution. If only affected some of the items, then the standard deviation and variance would change.

Note that,

variance = (standard deviation)^2

variance = (14)^2

variance = 196

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