Variance and standard deviation

Any idea how to solve 12.1

Scremin34Egl said:

Any idea how to solve 12.1

View attachment 3219

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Don't let the "a" scare you - it is just a number

1) find the mean of the four numbers

2) find the mean of the squares of the four numbers

3) Variance = the mean of the squares minus the square of the mean

4) For the final part, evaluate Variance when a=5, and take the square root.

DrPhil said:

Don't let the "a" scare you - it is just a number

1) find the mean of the four numbers

2) find the mean of the squares of the four numbers

3) Variance = the mean of the squares minus the square of the mean

4) For the final part, evaluate Variance when a=5, and take the square root.

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Thanks, solved it although it was quite long. Is there a formula I could have used?

Scremin34Egl said:

Thanks, solved it although it was quite long. Is there a formula I could have used?

Click to expand...

The only other option would be first to calculate the mean, then subtract the mean from each number and average the squares of the deviations. The two formulas are

\(\displaystyle \displaystyle V = \dfrac{1}{n}\sum{x_i^2} - \left( \dfrac{1}{n}\sum{x_i}\right)^2 \)

\(\displaystyle \displaystyle V = \dfrac{1}{n}\sum \left(x_i - \dfrac{1}{n}\sum{x_i}\right)^2 \)

As a programmer, I generally prefer the first formula because it needs only a single pass through the data, forming the two sums simultaneously. On the other hand, if floating-point precision of the sums is compromised by underflow, then the second form may be required.