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Algebra 30 Online
OpenStudy (jessicawade):

what is the standard deviation rounded to the nearest 10th?

OpenStudy (jessicawade):

(52.1,45.5,51,48.8,43.6)

OpenStudy (jessicawade):

sorry guys im terrible with standard deviation lol

OpenStudy (lxelle):

((11668.06/5) - (48.2) square ) sqrt

OpenStudy (jessicawade):

my answer choices are 3.2, 6.9, 10.3 and 10.4

OpenStudy (lxelle):

3.2

OpenStudy (jessicawade):

ok but how? i want to learn how to find it

OpenStudy (jessicawade):

i also have one more i did by myself but im not sure if its right

OpenStudy (jessicawade):

56, 78, 123, 34, 67, 91, 20 my answers are 32.1 32.2 67.0 67.1 i got 32.2

OpenStudy (lxelle):

There's a formula for standard deviation

OpenStudy (jessicawade):

i know it has steps but i got confused

OpenStudy (lxelle):

Sqrt (sigma x square / n - mean square)

OpenStudy (lxelle):

As for your question refer to I grouped data formula for s.d.

OpenStudy (jessicawade):

OpenStudy (jessicawade):

thats right?

OpenStudy (lxelle):

Refer to the first.

OpenStudy (jessicawade):

ok

OpenStudy (lxelle):

First you square each of all your numbers, then plus up all together and decide by the number of data you have.

OpenStudy (lxelle):

Find your mean and square it. Mean is total data. Over number of data.

OpenStudy (lxelle):

Take the first step minus the mean square.

OpenStudy (jessicawade):

i got 32.18695 but i rounded it and got 32.2

OpenStudy (lxelle):

Lastly squareroot your answer.

OpenStudy (jessicawade):

because i got 67 as the mean

OpenStudy (lxelle):

Your answer is correct bingooo! :)

OpenStudy (jessicawade):

yay!! :) thanks

OpenStudy (lxelle):

No problem. :))

OpenStudy (jessicawade):

i have 1 more but i dont think it has to do with standard deviation

OpenStudy (jessicawade):

given the mean of 7.6 and a standard deviation of 3.1, within how many standard deviations of the mean do the values 10,9,11,5, and 3 fall?

OpenStudy (jessicawade):

i dont understand at all how to do this lol

OpenStudy (jessicawade):

the answers are 0,1,2,and 3

OpenStudy (jessicawade):

i was going to use the z-score but i dont think i can

OpenStudy (jessicawade):

|dw:1408723111394:dw|

OpenStudy (jessicawade):

@LXelle

OpenStudy (lxelle):

if the mean is 7.6 the 1 standard deviation from the mean is lower bound =7.6 - 3.1 and Upper bound = 7.6 + 3.1 so if a number is between 4.5 and 10.7 it is within 1 standard deviation of the mean. 2 standard deviations lower bound = 7.6 - 2 x 3.1 = 1.4 Upper bound = 7.6 + 2 x 3.1 13.8 so the boundary for 2 standard deviations are 1.4 to 4.5 then 10.7 to 13.8 so look at each number individually 10... between 4.5 and 10.7 so 1 SD 9 be

OpenStudy (jessicawade):

so its 1?

OpenStudy (lxelle):

9 between 4.5 and 10.7 so 1 SD Then keep going on. Yes you're right.

OpenStudy (jessicawade):

so i have 4.5 and 10.7 to take each number and see if they fall in the middle?

OpenStudy (jessicawade):

nvm because then 11 and 3 wouldnt fit and it would be 0

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