Question

Compute the standard deviation for the set of sample data

90, 81, 73, 73, 81

Answer 1

doing this by hand is annoying

Answer 2

to do so you first have to find the average of the set

Answer 3

and recall that sample data has to /(n-1)

Answer 4

something to do with the degree of freedom

Answer 5

the average is 79.6

Answer 6

2(81) + 2(73) + 90
---------------- = mean
4

Answer 7

now take the square of each number subtracted from the mean

Answer 8

u lose a degree of freedom coz u estimate population mean by taking the mean of the sample.

Answer 9

2(81+73) + 90

2(154)+ 90= 398


99.5 is the mean if i did it right
-----
4|398
36
38
36
20

Answer 10

so

(90-79.6)^2+(81-79.6)^2+...

Answer 11

amistre64, the mean is 79.6, she is correct

Answer 12

it wouldnt make sense for the mean to be more than any of the numbers in the set...

Answer 13

i see that :) means i typoed or mismathed lol

Answer 14

im thinking of the n-1 in a different spot .. arent i

Answer 15

yup

Answer 16

perhaps?

\[\sqrt{\frac{\sum{(x-\bar x)^2}}{n-1}}\]

Answer 17

thats the equation

Answer 18

so i got 207.14

Answer 19

you should have gotten 199.2

Answer 20

my ti83 pops out Sx = 7.057 to check with

Answer 21

ya, thats what i got in excel too

Answer 22

from the 199.2, you divide by n-1 where n is the number of numbers in the series

Answer 23

then square root

Answer 24

I would use the formula

\[s=\frac{n\sum x^2-(\sum x)^2}{n(n-1)}\]

Answer 25

it's quicker

Answer 26

i still wonder why; since sd is the average distance from the mean, why we dont take the mean of the absolute values of x - mean ... i fhtat makes sense

Answer 27

square root of that

Answer 28

its actually longer to do by hand...

Answer 29

\[s=\sqrt{\frac{n\sum x^2-(\sum x)^2}{n(n-1)}}\]

Answer 30

quicker than using the definition by hand

Answer 31

not for me, guess im weird

Answer 32

you are then ;)