Question

What is the standard deviation of the data set given below?

5, 7, 9, 11, 13

Answer 1

Did you find the mean (or the average) of your data?

Answer 2

I think its 9

Answer 3

Hi @Gabe24Me ! Well, the group of numbers are 2 numbers apart. 5+2=7, 7+2=9, 9+2=11, ect.

Answer 4

Hop that helps a little bit.

Answer 5

A. square root of 10
B. square root of 8
C. 3
D. square root of 3
E. 8
F. 10

Answer 6

Yes, the mean is:
\(\large\color{black}{ \displaystyle \mu=9 }\)

Now, find \(\large\color{black}{ \displaystyle (x_i-\mu)^2 }\)

That is \(\Downarrow\)
Subtract the mean from each value in the list and square the difference

Answer 7

What is your explanation that the answer is a? @babylove18

Answer 8

Then add all of those differences squared.

Answer 9

ok thank you all for the help and A was correct

Answer 10

I mean:

\(\color{black}{ \displaystyle \frac{(5-9)^2+(7-9)^2+(9-9)^2+(11-9)^2+(13-9)^2}{5} }\)

then, the population varience is:
\(\color{black}{ \displaystyle \sigma_x{\tiny~}^2=\frac{(5-9)^2+(7-9)^2+(9-9)^2+(11-9)^2+(13-9)^2}{5} }\)

then the population standard deviation:
\(\color{black}{ \displaystyle \sigma_x=\sqrt{\frac{(5-9)^2+(7-9)^2+(9-9)^2+(11-9)^2+(13-9)^2}{5}} }\)

Answer 11

sample standard deviation is same, but you divide by "1 less then the amount of data in the list", and in this case by 5-1=4.

\(\color{black}{ \displaystyle s_x=\sqrt{\frac{(5-9)^2+(7-9)^2+(9-9)^2+(11-9)^2+(13-9)^2}{4}} }\)