Question

Let m1, r1 and v1 be the mean, the range and the variance of a group of numbers {x1,x2,x3...,x100} respectively. If m2, r2 and v2 are the mean, the range and the variance of the group of numbers {x1,x2,x3,...,x100,m1} respectively, which of the following must be true?
I. m1=m2
II. r1=r2
III. v1=v2

i know the first one is correct, how about the others?
@Callisto

Answer 1

@Callisto i have done this question before but i forgot :O

Answer 2

If \(m_1\) is the mean of the first set of data, then \(x_1\le m_1 \le x_{100}\), agree?!

Answer 3

why m1=x100 ??? :O

Answer 4

second one i agree ... [?]

Answer 5

It may not be the same as x100, but if x1=x2=...=x100, then m1=x100

Answer 6

So, you agree that \(x_1\le m_1 \le x_{100}\)??

Answer 7

yes

Answer 8

Then adding \(m_1\) to the first data set won't affect the range since m1 does not change the extreme values (the max and the min.), agree?

Answer 9

wait, let me organize o.0

Answer 10

yup, i agree

Answer 11

So, 2) is true.

Answer 12

what does 'variance' mean?

Answer 13

\(\sigma^2\), where \(\sigma\) = standard deviation.

Answer 14

oh i see o.0

Answer 15

Do you know if (3) is true?

Answer 16

the number of terms are not the same, so it's incorrect?

Answer 17

Hmm.. A better argument:
For the first set of data, we have
\(\sigma_1^2 = \frac{(x_1-m_1)^2+ .. + (x_{100}-m_1^2)}{100}\)
For the second set of data, we have
\(\sigma_2^2 = \frac{(x_1-m_1)^2+ .. + (x_{100}-m_1^2)+(m_1-m_1)^2}{101} = \frac{(x_1-m_1)^2+ .. + (x_{100}-m_1^2)}{101} <\sigma_1^2\)
So, it is not true

Answer 18

oh i forgot the standard deviation formula lol

Answer 19

Typo: It should be \((x_{100}-m_1)^2\) for all those three terms

Answer 20

umm let me organize this first o.0

Answer 21

Take your time

Answer 22

how do that two formula are out?

Answer 23

http://www.mathsisfun.com/data/standard-deviation-formulas.html

Answer 24

whooops. oh i should clear the concepts on SD tonight lol
thank you so muchhhhh <3

Answer 25

Welcome :)