Question

I need a data set containing 9 numbers where the standard deviation is 1?

Answer 1

1,2,3,4,5,6,7,8,9

maybe?

Answer 2

you need a variance of 1 as well

variance is the sum of the squares divided by n

Answer 3

so the sum of the squares has to be 9 as well

Answer 4

\[\sum_{1}^9x^2=9\]

Answer 5

9 different numbers? or just a data set of 9 elements?

Answer 6

There must be 9 numbers in the set where the standard deviation would be 1

Answer 7

well, the sum of x^2 from 1 to 9 is 285; so id say divide that by 285

Answer 8

\[\frac{1}{\sqrt{285}},\frac{2}{\sqrt{285}},\frac{3}{\sqrt{285}},\frac{4}{\sqrt{285}},\frac{5}{\sqrt{285}},\frac{6}{\sqrt{285}},\frac{7}{\sqrt{285}},\frac{8}{\sqrt{285}},\frac{9}{\sqrt{285}}\]

Answer 9

still off

Answer 10

X (X-M)2 (1 or -1)2 1 M=

1+1+1+1+1+1+1+1+1=9

(1/9) x 9 = 1

√1 = 1

X= number in data set M=mean of data set #s

Answer 11

1 has no deviation

Answer 12

what is it? from which course?

Answer 13

business math

Answer 14

a set of 1s has no deviation; no variance

Answer 15

0,1,1,1,1,-1,-1,-1,-1 is fine tho

Answer 16

ugh, sample or population?

Answer 17

http://www.wolframalpha.com/input/?i=variance+%7B1%2C1%2C1%2C1%2C1%2C1%2C1%2C1%2C1%7D&a=*C.variance-_*PopulationVariance-

Answer 18

the question just says List a data set in which the standard deviation is 1. it's for a discussion forum. Neither sample or population is stated

Answer 19

well, sample deviation would be the 0,1,1,1,1,-1,-1,-1,-1

or you could pull out some z scores off the normal distribution curve

Answer 20

http://www.wolframalpha.com/input/?i=variance+%7B0%2C1%2C1%2C1%2C1%2C-1%2C-1%2C-1%2C-1%7D

Answer 21

Is there any way for population to have a standard deviation of 1 with 9 elements?

Answer 22

yes, but its tricker

Answer 23

assuming a mean of zero, you would need 9 values that sum their squares to 9

Answer 24

\[\sum_{-4}^{4}x^2 = 60\]

Answer 25

How about this? http://www.wolframalpha.com/input/?i=variance+%7B-1%2C-1%2C-1%2C-1%2C0%2C1%2C1%2C1%2C1%7D

Answer 26

so each term is 1/sqrt(60)

Answer 27

which is off since that would equate to 1 not 9

Answer 28

Why wouldn't -1,-1,-1,-1,0,1,1,1,1 work?

Answer 29

k sqrt(9/60), for k=-4 to 4

would work

Answer 30

Has a sample standard deviation of 1

Answer 31

that wont work for a population variance since 8/9 is not 1

Answer 32

Are we using population or sample?

Answer 33

either one is what they say, but they are now curious about pop var

Answer 34

@vinnv226 neither was stated, but my guess is population.

Answer 35

-4 sqrt(3/20)
-3 sqrt(3/20)
-2 sqrt(3/20)
-1 sqrt(3/20)
0 sqrt(3/20)
1 sqrt(3/20)
2 sqrt(3/20)
3 sqrt(3/20)
4 sqrt(3/20)

Answer 36

My statistics need some work, so I'll leave this to you guys

Answer 37

http://www.wolframalpha.com/input/?i=sum+%28sqrt%283%29x%2Fsqrt%2820%29%29%5E2%2C+x%3D-4+to+4

Answer 38

http://www.wolframalpha.com/input/?i=variance%7B-4sqrt%283%2F20%29%2C+-3+sqrt%283%2F20%29%2C+-2+sqrt%283%2F20%29%2C+-sqrt%283%2F20%29%2C+0%2Csqrt%283%2F20%29%2C+2+sqrt%283%2F20%29%2C+3+sqrt%283%2F20%29%2C+4+sqrt%283%2F20%29%7D&a=*C.variance-_*PopulationVariance-

Answer 39

ok, and to find a set of 9 elements that have a standard deviation of 0, wouldn't all 9 elements be the same number?

Answer 40

yes

Answer 41

if noting varies ... then there is no deviation

Answer 42

Thank you

Answer 43

youre welcome