Question

What is the standard deviation of the following data? If necessary, round your answer to two decimal places.
6, 7, 10, 11, 11, 13, 16, 18, 25


5.48

5.58

5.68

5.78

Answer 1

\[\begin{matrix}{} j &|& 6 &7 & 10 & 11&13&16&18&25 \\ \hline N(j)&|& 1 &1& 1 & 2 & 1 & 1&1&1 \end{matrix}\]

Answer 2

\[N=\sum\limits_{j=0} ^{\infty} N(j)\]

Answer 3

5.58

Answer 4

\[\langle j\rangle=\sum\limits_{j=0}^{\infty} jP(j)=\sum\limits_{j=0}^{\infty} j\frac{N(j)}{N}\]\[=\frac{1}{N}\sum\limits_{j=0}^{\infty} j{N(j)}\]

Answer 5

is my answer correct?

Answer 6

i dont know i haven't got that far yet, what did you do?

Answer 7

\[\langle j^2 \rangle =\sum\limits_{j=0}^{\infty}j^2P(j)=\frac{1}{N}\sum\limits_{j=0}^{\infty}j^2N(j)\]

Answer 8

\[\sigma^2={\langle(\Delta j)^2\rangle} \]

\[\sigma=\sqrt{\langle(\Delta j)^2\rangle} =\sqrt{\langle j^2 \rangle -\langle j \rangle^2}\]

Answer 9

for N i am getting 9,

Answer 10

ok

Answer 11

Data: 6,7,10,11[2],13,16,18,25, n= 9
Σx= 117 , Σx²= 1801 , µ= 13 , σn-1=5.91608, σn=5.577734

Answer 12

?

Answer 13

for the average im am getting 13

Answer 14

average squared i am getting 169

the square of the average im am getting 1447/9~164.111

Answer 15

ok

Answer 16

so the difference (ie the standard deviation) is 4.888,....
damn i messed up

Answer 17

your
Σx= 117,
µ= 13
are correct,
check your
Σx²= 1801 i am getting 1477

Answer 18

I think i am doing this wrong somehow
however you had
σn=5.577734
which matches with b. so that is probably it