A: 1/5( 1+2+3+4+5)=3.

b: 1/5( 10 1+ 102+ 103+ 104+ 105)= 103。

c: 1/5(3+6+9+ 12+ 15)= 9。

Difference:

A: 1/5[( 1-3)? +(2-3)? +(3-3)? +(4-3)? +(5-3)? ]=2

b: 1/5[( 10 1- 103)？ +( 102- 103)? +( 103- 103)? +( 104- 103)? +( 105- 103)? ]=2

C: 1/5[(3-9)？ +(6-9)? +(9-9)? +( 12-9)? +( 15-9)? ]= 18

Standard deviation: the square root of variance

A: Root number 2

B: root number 2.

C: triple radical number 2

2. the data stability of group a and group b is the same, and the data of group b is higher than that of group a on average.

3. Among the data of group A and group C, the data of group A is stable, and the data of group B is higher than that of group A on average.

4.( 1) Average value

Average value of ∵ x 1, x2, ... xn is a.

∴ X 1+X2+X3+X4+……+Xn=an

∴x 1-3+x2-3+x3-3+x4-3+……+xn-3 = an-3n = n(a-3)

∴ x 1, x2, x3...xn average a-3.

discrepancy

∫x 1, x2, ... xn variance is b.

∴[(X 1-a)？ +(X2-a)？ +......+(Xn-a)？ ]/n=b

∴[(X 1-3-a)？ +(X2-3a)？ +……+(Xn-3-a)？ ]/n=b

Maybe you will ask why it's still B, hehe, it's like this. The variance depends on the stability of the data. Although this set of data X 1, X2......Xn are each reduced by 3, the overall stability is not broken, is it? In fact, just look at the variance of the previous question 1, 2, 3, 4, 5 and10/,102, 103, 104, 105. Are they the same? This is equivalent to adding 100 to the second set of data, and the variance has not changed. )

Standard deviation

The standard deviation is the square root of variance. Since the variance is all b, its square root is still C.

(2) Average value

Average value of ∵ x 1, x2, ... xn is a.

∴x 1+x2+....+xn=na

∴2x 1+2x2+2x3+....+2xn=2+(x 1+x2+....+xn)=2na

The average level of ∴ is 2a.

Difference:

(In fact, this problem is the same as the first big problem, which is equivalent to the comparison of group C data 3, 6, 9, 12, 15 and group A data 1, 2, 3, 4, 5. Group c data is multiplied by 3 on group a data. So the variance is the square of the multiplier, which is 3? , which is 9. )

Standard deviation:

The standard deviation is 3.

(3) Average value

Average value of ∵x 1, x2, ... xn is a.

∴x 1+x2+x3+....+xn=na

∴ 1/2x 1-2+ 1/2x2-2+...+ 1/2xn-2 = 1/2(x 1+x2+....+xn)-2n = 1/2an-2n =( 1/2a-2)n

∴ Average index: 1/2a-2

discrepancy

∫x 1, x2, ... xn variance is b.

∴[(X 1-a)？ +(X2-a)？ +......+(Xn-a)？ ]/n=b

∴[( 1/2x 1-2-a)？ +( 1/2x2-2-a)？ +....+( 1/2xn-2-a)？ ]/n= 1/4b

Well, first of all, addition and subtraction don't work on it, as I said before. The key is multiplication and division. In this problem, multiply by 1/2, so the final variance is multiplied by 1/4. )

Standard deviation

The standard deviation is the square root of variance 1/2c.

(4) This group can be digitized into 2+ 1646, 4+ 1646, 6+ 1646, 8+ 1646, 10+ 1646.

In this way, the problem is much simpler and becomes a deformation of 2, 4, 6, 8, 10.

So just work it out according to the law you just found.

In fact, it is enough to find the variance of 2,4,6,8, 10, because the added 1646 doesn't work.

The variance is 1/5[(2-6)? +(4-6)? +....+( 10-6)? ]=8

I hope I can help you!