Question

What are the mean, variance, and standard deviation of these values? Round to the nearest tenth.

92, 97, 53, 90, 95, 98

Answer 1

@phi

Answer 2

the number are few so it shouldnt be to bad to freehand it

Answer 3

mean is: add em up, and divide by how many there are

Answer 4

and once you have variance, sd = sqrt(var0

Answer 5

sqrt(var) ... keep undoing the shift and that 0 slips in

Answer 6

Okay so add them up and divide it by 6?

Answer 7

yep, theres 6 data points

Answer 8

I did and i got 87.5 now i need to find my variance and standard devation

Answer 9

variance is a process

create a new set of data to work with by subtracting the mean from the given data points

Answer 10

well my variance not my SD

Answer 11

92-mean, 97-mean, 53-mean, ...

Answer 12

or mean-53 ... if thats simpler to do by hand. The difference is important here and not the signage

Answer 13

is my variance 4 or 245.6 ?

Answer 14

dunno, im not creating a set, and i havent seen yours yet

Answer 15

Well idk what it is neither

Answer 16

start by working thru the process

subtract the mean (or find the difference) from the data points given

Answer 17

Okay let me try.

Answer 18

Iv subtracted , divided , multiply and add and still didnt get either of those answers . ?

Answer 19

Well im basically stuck between 2 answers because i can't figure out what my variance is ?

Answer 20

{92, 97, 53, 90, 95, 98}
-87.5..........................
----------------------
{4.5, 9.5, 34.5, 2.5, 7.5, 10.5}

square the numbers in this new set

{4.5, 9.5, 34.5, 2.5, 7.5, 10.5} ^2

{4.5^2, 9.5^2, 34.5^2, 2.5^2, 7.5^2, 10.5^2}

and add em all up again .... 1473.5

and divide by 6 to find the average of the 6 numbers: 245.583 or so

Answer 21

By looking at that. Thank you very much this helped me a lot. I came across as this as my answer mean = 87.5
variance = 245.6
standard deviation = 15.7

Answer 22

those look fine to me. assuming this is a population and not a sample of course.

Answer 23

I hope this is my right answer lol. Thank you so much for your help

Answer 24

good luck :)

Answer 25

process is:\[mean=\frac{sum}{\#}\]
\[var=\frac{sum~\{(a-mean)^2,(b-mean)^2,(c-mean)^2,...\}}{\#}\]