Question

help need clarification on the topic of Calculating Variance. I have one step of the process I'm stuck on screenshots below

Answer 1

working on part 1 and the #'s i wrote are for Class A. Im wondering step 2 its confusing i did the whole subtracting mean but the squared difference part is confusing is square difference just the square root and the # for ex. 9 the squared difference is 81? or 3?

Answer 2

if this helps any heres the question

Answer 3

@InExileWeTrust

Answer 4

Let \(x_1, x_2, ..., x_n\) be the n data values.

Mean: \( \bar{x} = \dfrac{x_1 + x_2 + ... + x_n}{n}\)

Variance = \(\dfrac{(x_1 - \bar{x})^2 + (x_2 - \bar{x})^2 + ... + (x_n - \bar{x})^2}{n}\)

Answer 5

If the mean is \(\mu\) then the squared difference is \((x-\mu)^2\).

Answer 6

I got a different mean for class A ,

Answer 7

ok if i understand right the numbers i have there are right then right? now all i have to do is ^2 each?

Answer 8

i just downed openoffice today ... excited to use its excel clone

Answer 9

Hmmm... I got a Class A average of 79 as well. So, this is how variance would look worked out:
(80 - 79)^2 = (1)^2 = 1
(85 - 79)^2 = (6)^2 = 36
(87 - 79)^2 = (8)^2 = 64
...
(62 - 79)^2 = (-17)^2 = 289

Then add up all those results
1 + 36 + 64 + ... +289 = 4558

and divide by 15
4558/15 = 303.866666.......

Answer 10

it works like a remember it did :)

Answer 11

Thanks :D i think i understand I'm going to ttry it again and i will come back here if i don't get it