Question

what is the standard deviation of the data set given below
2 3 6 9 10
A: 10
B: square root of 10
C: square root of 12.5
D: square root of 3.5
E: 3.5
F: 12.5

Answer 1

Standard deviation:
\[\Large = \sqrt{\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar{x})^2} \]

Answer 2

can you tell me what to plug in where because i dont understand the formula

Answer 3

\(n\) is the sample size (the # of data points you have), which is 5
\(x_i\) is referring to each data point
\(\bar{x}\) is the mean of your data

so \(x_i - \bar{x}\) means to take a data point and subtract it from the mean
\[\sum_{i=1}^n (x_i-\bar{x})\] means take the 1st point and subtract it with the mean, then add that to (the 2nd point subtracted to the mean) and keep doing that until the fifth point.
\[\sum_{i=1}^n (x_i-\bar{x})^2\] is similar to what I describe above, except when you take the differences, you must square that result.

In your example:
The mean is \[ \frac{2+3+6+9+10}{5}=6\]
So the formula, once plugged in, should like like:
\[ \sqrt{\frac{1}{5-1}\left( (2-6)^2+(3-6)^2+(6-6)^2+(9-6)^2+(10-6)^2 \right)}\]

Answer 4

Thank you so much! it helped a ton!

Answer 5

awesome :)

Answer 6

that's is the sample standard deviation, I hope you don't need to population standard deviation.

I always thought the population was the default

Answer 7

hmm, I was always under the assumption that the sample one was the default. Population data are rare to come by. But I guess it's still a valid point. Hopefully your teacher told you what to assume. If it's a population standard deviation, you should use \(\Large \frac{1}{5}\) instead of \(\Large \frac{1}{5-1}\)

Answer 8

square root 12.5