Question

What is the standard deviation of the data set?
15 32 28 26 23 17 27

Please explain!

Answer 1

That has something to do with mean right?

Answer 2

So I guess you should find that.

Answer 3

idk that's why i need help! oh i forgot to include answer choices! my bad hold on a sec..

Answer 4

a. about 5.7

b. about 6.1

c. about 15.0

d. about 37.3
ANY IDEAS?????

Answer 5

Uhm, I cannot help you at this time. Sorry.
I was looking at this earlier.
www.mathmotivation.com/symbolic/standard-deviation.html
Maybe that might help.

Answer 6

Hmm it is not as simple as thought. It has a lot of calculation so brace yourself...

Answer 7

okay, can someone walk me through it please cuz i am super confused...

Answer 8

First, find the mean. Second, find the distance of each point from the mean. Then, square each of those numbers. Then, average the results, and square root that average. Voila, standard deviation.

Answer 9

is the answer either a or b??? I don't know which one????!!!

Answer 10

Once you have done the calculation you will know which one :)

Answer 11

ahhh! idk how to do the calculation tho!! please help me @nbouscal !!! And EVeryone else?!?!! i really appreciate your help pplz :)

Answer 12

What do you mean you don't know how? I gave you step-by-step instructions very clearly.

Answer 13

Given the data:
15 32 28 26 23 17 27

1) Calculate the mean/average,
15+32+28+26+23+17+27= 168
168/7 = 24

2) Find the difference of each point to the mean,
15 - 24= -9
32-24 = 8
28-24= 4
26-24=2
23-24=-1
17-24 -7
27 - 24 = 3

3) Then square them all and add them

-9^2 + 8^2 +4^2 +2^2 + -1^2 + -7^2 + 3^2
= 81+64+16+4+1+49+9
=224

4) Divide by the number of data values,
224/7= 32

5) Finally, Square root it
sqrt(32)= 5.65 approx. 5.7

Answer 14

Like i don't get how to find the mean, how to square those numbers, how to average the results and square root the average, etc...

Answer 15

I find it very hard to believe you don't know how to square numbers.

Answer 16

A calculator with statistical functionality is usually permissible. Using such a calculator gives sigma as 5.6568

Answer 17

thanks @tiaph for the very detailed explanation! i totally understand it now! :) I was sooo confused earlier hahaa

Answer 18

@nbouscal sorry i just got super confused and really needed an explanation like the one @tiaph gave which helped me a ton

Answer 19

Yes, I'm sure tiaph's answer was very helpful, especially the part that had the answer in it.

Answer 20

<3 that sarcasm in it. Well i know most of the people that come here, come for the answers. I deliver but i try my best to show as much as the working in the simplest way possible.. so pple make an effort to read through everything and come off with a bit of understanding. Sorry if you feel offended by me spoonfeeding the answers to people.

Answer 21

Not offended, just something that I would prefer to avoid. It's pretty clear when certain people are not here to make an effort themselves, but rather just to get the answer to their homework. For those people specifically, I prefer not to provide answers.

Answer 22

@nbouscal no offense and all (and love the sarcasm btw seriously cuz sarcasm is cool) lol but i had a feeling that was the answer already but I just needed to be shown the steps and all to get to the answer so I could LEARN how to do it... That;s why i asked for an explanation!!! So i could LEARN!!!! Thank you btw for the fantastic explanation @tiaph :) props to you! :)

Answer 23

\[\sigma^2=\langle x^2 \rangle-\langle x \rangle^2\]
\[\langle x \rangle=\sum\limits_{n=1} x_iP(n)\]
\[\langle x^2 \rangle=\sum\limits_{n=1}x_i^2P(n)\]

Answer 24

You had a feeling that was the answer? How exactly did you have a feeling that the result of a relatively involved computation was 5.7 as opposed to 6.1? Math is not about feelings. Math is about computation, a computation that is relatively simple albeit long, and a computation that you elected to have someone do for you instead of doing yourself.

Answer 25

Yea i understand what you mean. I know some people are blatantly just asking for answers in the face. Bu i know iheartfood is not someone like that, cause i previously helped with some questions of his/her.

Answer 26

okay there @nbouscal jeez no need to get so heated and mad about this lol thanks @tiaph i really do appreciate your help! and i am a her :) I want to learn how to get the answer and stuff :) @nbouscal chill out cuz i know ppl just want the answers but I want to know how to get to the answer and LEARN!!! :)

Answer 27

i*n

Answer 28

and sorry if that sounds blunt lol i mean to sound polite not rude lol sorry if it comes out that way! haha

Answer 29

I'm neither heated nor mad, nor do I need to chill out. I'm explaining my point of view calmly and clearly. You tried to tell me that you "don't get how to square numbers." That's just a bad joke.

Answer 30

oh haaha i meant i don't know how to square the numbers since i don't know how to get to the number haaha i know how to square numbers (in my head... simple ones tho...) but just didn't know how to do so since i did not know how to get to the numbers i needed to square... sorry if i offended you @nbouscal i just really want help.. that's why i am on here...

Answer 31

Again, I'm not offended. You keep attributing emotions to me that I don't have. Standard deviation is a simple computation, the most advanced step is taking the mean of a set, which is something usually learned long before an introduction to standard deviations. I find it very hard to believe that you couldn't have done the computations just fine on your own.

Answer 32

@tiaph one should divide by n-1=7-1=6
to get the standard deviation
\[
4 \sqrt{\frac{7}{3}}=6.1101

\]

Answer 33

@eliassaab That is only true if you are looking for a sample standard deviation. If we assume that this is a population standard deviation, then @tiaph calculated correctly.