Question

Find the mean and variance of the following data 52,34,64,49,55,26,38,63

Answer 1

Answer:

Mean = 47,625

Variance, ignoring the commas = 157

Explanation:

The mean is quite simple: is the mid value of an amount (m') of numbers.

Examples:

Mean of 2, 4:  2 + 4/2 = 3

Mean of 5, 6 and 8: 1+6+8/3 = 15/3 = 5

Generalizing: M' = An + Bn-1 + Cn-2 +... + Y1/ n

So, for the enunciated, we have:

M = 52 + 34 + 64 + 49 + 55 + 26 + 38 + 63/8

M = 47,625

Variance is quite harder.

The mathematics definition is also quite harder:

∑n, i=1 (Xi - X')²/n

Traduction: the sum of the difference of the square of the dispersion measures by the arithmetic mean of them, divided by the number of observed terms is the square of the standard deviation, that is, the variance.

Don't worry, you don't need to understand that. I'll do the question, then you will see in practice what that means.

First, we need the mean of the observerd terms. We already have done that.

M = 47,625.

Next, we will do the the difference of the square of the dispersion measures by the arithmetic mean of them. Which are:

A.52 - 47,625 = (4,375)²

B.34 - 47,625 = (-13,625)² - Because of a value like that, you know see why

C.64 - 47,625 = (16,375)²

D.49 - 47,625 = (1,375)²

E.55 - 47,625 = (7,375)²

F.26 - 47,625 = (-21,625)² - we need to square everything: because a negati

G.38 - 47,625 = (-9,625)² - ve result doest not help in the statisc.

K.63 - 47,625 = (15,375)²

I am using a calculator, but really with that, I'll aproximately the values. If you need a accurate measure, just square them all the way.

A = 16     D  = 1     G = 100

B = 169    E = 49    K = 225

C = 256   F = 441

The last part is the sum of the difference of the square of the dispersion measures by the arithmetic mean of them, dividing them by the number of observations.

16 + 169 + 256 + 1 + 49 + 441 + 100 + 225/ 8 = 157,125

8 is the number of observations.

___________________________

But why we do the variance? The variance serves to us for know how much the mean is precise.

So, the lower is the variance, the better is the mean.

Take a look on this example:

Imagine three persons, A, B and C.

They are in pub, talking about how much they receive. A gets, per month, 1000US$, b 2000 and C 3000US$.

So, their mean are 3000US$.

Now imagine that Bill Gates joins the group. And Bill, last month, got 100 milion US$.

Now the mean will be almost 25 milions of dollars.  So, she is VERY far from the reality of the major part of the group. And is in there that cames the variance. Looking to the variance, we will see that there is a big discrepancy for each observed term.