standard deviation

[abzzzz]

Hey guys, quick question.
When finding the SD, do you divide the range by 4 or 6?

i heard it was by 4, but i'm getting the wrong answer. If i were given numbers, i could plug it in list1 on my calc, but i have to read a graph.

just wondering how you guys work it out?

[Natters]

range/4 is only an estimation to determine the standard deviation im pretty sure
like, you know that 68/95/99.7 rule yeah
if it was range/4 to find the exact value then 100% of the data would lie between two standard deviations from the mean (=4 deviations) so 95%=100% bumbow.
you gotta do this big table thing and find like a billion different values to do it by hand, and i only know how to use the TI-89, what calc you using?

[abzzzz]

i use the ti-89 too aha, it's simple finding it through that, but by hand, bit confusing.

[Natters]

it's not that bad, just time consuming as fck
basically, you just find the mean, write all your x values in a vertical line, then the x-mean values to the right of that line, then your (x-mean)^2 values in another line to the right of that
add the last line together, divide it by the sample size-1 and then square root that number and you have your answer

if any of you math gods out there know a way to do it faster,,, i want

[MJRomeo81]

it's not that bad, just time consuming as fck
basically, you just find the mean, write all your x values in a vertical line, then the x-mean values to the right of that line, then your (x-mean)^2 values in another line to the right of that
add the last line together, divide it by the sample size-1 and then square root that number and you have your answer

Then you missed a decimal place somewhere, and VCAA says "to one decimal place" just to piss you off.

Nah look do it through the calc. There is so much time in the exam. People keep raving about how you need to work fast. Please. If you've mastered one module (or possibly more) there's no need to be worried about time. Plug in the values in the calc, and let it do the dirty work. While you're at it, you get the mean, median, whole 5 number summary, etc. Usually VCAA like to ask for a few values, not just standard deviation.

[abzzzz]

from 2010 exam 1 i think q3, how would you find the SD? what would you plug into the calc

[acrimony]

Ayri just use the 68–95–99.7% rule

[Danielhogz]

IN the exam last year you had to estimate using 68-95-99.7 rule. To do this you had to divide the range by 6 as essentially 99.7 or approx 100% of data lies within 6 standard deviations

[Hodgeyhodgey]

With Q3 on the 2010 MC exam, the simplest way of estimating the SD without using any formulas would be to look at the graph. You know that 68% lie within 1 SD of the mean right? Or even 34% either side?

Well, if you look at the graph and you use the median 180 (rather than calculating the mean, as it's approx. symmetrical so it shouldn't matter too much) all you need to do it count 34% either side of 180 which will give you ~178 and ~182 and there you have it, a SD of 2!

[Keki]

Ahhh so that's how you work that question! I was confused by that

[nemolala]

you might her helpful for certain topics http://www.youtube.com/user/vcefurthermaths#p/u/7/w3FAiRoiLAA

[iroflmfao]

divide by 6 if it is normal distribution

[chelseaFC]

I have been told its range divided by 4 not 6?

[Wezanator123]

as mentioned above dividing by 4 is only an estimation

[spikey]

With Q3 on the 2010 MC exam, the simplest way of estimating the SD without using any formulas would be to look at the graph. You know that 68% lie within 1 SD of the mean right? Or even 34% either side?

Well, if you look at the graph and you use the median 180 (rather than calculating the mean, as it's approx. symmetrical so it shouldn't matter too much) all you need to do it count 34% either side of 180 which will give you ~178 and ~182 and there you have it, a SD of 2!

how come it is a SD of 2 from 178 - 182?