Question

Find the mean, standard deviation, and variance.

Answer 1

@rational

Answer 2

For mean, simply take the average

Answer 3

add up the values and divide by the number of values

Answer 4

I know that part...

Answer 5

5.10

Answer 6

yes thats the mean

Answer 7

okay. now how do I find the Standard deviation?

Answer 8

step1 : find the difference between mean and each observation

Answer 9

oh, so like 5.1-4.85 and 5.1-5.1 and so on?

Answer 10

observation mean-observation
4.85
5.10
5.50
4.75
4.50
5.00
6.00

Answer 11

yes fill that table

Answer 12

okay.

Answer 13

0.25
0
-0.4
0.35
0.6
0.1
-0.9
right?

Answer 14

square each of them

Answer 15

why?

Answer 16

@rational

Answer 17

standard deviation formula requires it haha

Answer 18

https://www.mathsisfun.com/data/images/standard-deviation-formula.gif

Answer 19

oh, can you still help me?

Answer 20

@rational

Answer 21

have you squared the results yet?

Answer 22

I thought I had to take the square root..

Answer 23

oops

Answer 24

I so read that wrong XD

Answer 25

the standard deviation is the sqrt of variance

variance is the average of the squared differences

Answer 26

anddd it disappeared.

Answer 27

do you click on the post button to post? sometimes it gets hidden behind the chat box.

i just tab-space to post

Answer 28

okay, so it's 0.0625, 0, 0.16, 0.1225, 0.36, 0.01, 0.81

Answer 29

I got it.

Answer 30

now take the average of those values .... and thats the variance

sqrt the variance to get standard deviation

Answer 31

average is 0.217857....

Answer 32

right?

Answer 33

if i had excel i could chk it rather quickly, but this is my sister computer so ill have to chk it the long way

Answer 34

okay..

Answer 35

and I just take the square root of that to find the standard deviation?

Answer 36

5.10 is the original mean ... now finding the diffs

http://www.wolframalpha.com/input/?i=mean%7B%284.85-5.10%29%5E2%2C+%285.10-5.10%29%5E2%2C+%285.50-5.10%29%5E2%2C+%284.75-5.10%29%5E2%2C+%284.50-5.10%29%5E2%2C+%285.00-5.10%29%5E2%2C+%286.00-5.10%29%5E2%7D

.2178... looks good to me

Answer 37

so variance of the population is .2178

if this had been a sample, we would have divided by n-1, instead of n

Answer 38

well it would be 0.2179, right?

Answer 39

the variance yes

the standard deviaion is he sqrt of variance

Answer 40

so 0.466797...

Answer 41

right?

Answer 42

correct, the wolf agrees :)

http://www.wolframalpha.com/input/?i=standard+deviation%7B4.85+5.10+5.50+4.75+4.50+5.00+6.00%7D&a=*C.standard+deviation-_*PopulationStandardDeviation-

Answer 43

okay, thanks so much!

Answer 44

youre welcome :)