Answer: σ = 0.0154
Step-by-step explanation:
1. Calculate the mean for the recorded data.
(0.48 + 0.51 + 0.47 + 0.49 + 0.49 + 0.50 + 0.52 + 0.48 + 0.49 + 0.51)/10 =
4.94 / 10 = 0.49
The mean is represented with the greek letter μ, so μ = 0.49
2. For each element in the serie, subtract the mean (0.49) and square the result
(0.48 - 0.49)^2 = (-0.01)^2 = 0.0001
(0.51 - 0.49)^2 = (0.02)^2 = 0.0004
(0.47 - 0.49)^2 = (-0.02)^2 = 0.0004
(0.49 - 0.49)^2 = 0
(0.49 - 0.49)^2 = 0
(0.50 - 0.49)^2 = (0.01)^2 = 0.0001
(0.52 - 0.49)^2 = (0.03)^2 = 0.0009
(0.48 - 0.49)^2 = (-0.01)^2 = 0.0001
(0.49 - 0.49)^2 = 0
(0.51 - 0.49)^2 = (0.02)^2 = 0.0004
So we get these results: 0.0001, 0.0004, 0.0004, 0, 0, 0.0001, 0.0009, 0.0001, 0, 0.0004
3. Now, we calculate the mean of these squared differences
(0.0001 + 0.0004 + 0.0004 + 0 + 0 + 0.0001 + 0.0009 + 0.0001 + 0 + 0.0004) / 10 =
(0.0024) / 10 = 0.00024
4. Now to get the standard deviation, we need to get the root of this last value (0.00024).
√(0.00024) = 0.0154
5. So that will be our standard deviation value:
σ = 0.0154
DONE!
![x^ \frac{m}{n}= \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%20%5Cfrac%7Bm%7D%7Bn%7D%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%20%20)
![5^ \frac{2}{5}= \sqrt[5]{5^2}= \sqrt[5]{25}](https://tex.z-dn.net/?f=5%5E%20%5Cfrac%7B2%7D%7B5%7D%3D%20%5Csqrt%5B5%5D%7B5%5E2%7D%3D%20%5Csqrt%5B5%5D%7B25%7D%20%20)