Question

complete the table to determine the standard deviation for the density of the aluminum foil.
trial $x_i - x_{mean}$ $(x_i - x_{mean})^2$
1 0.28783 0.0828461
2 0.49903 0.249030
3 -0.840351 0.706189
4 0.08676 0.00752729
5 -0.03324 0.00110489
standard deviation for the density of the aluminum foil.
calculate the sum of the squares.
1.04669

Explanation:

Step1: Recall standard - deviation formula

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$, where $n$ is the number of trials, $\sum_{i = 1}^{n}(x_i-\bar{x})^2$ is the sum of the squared differences, and $\bar{x}$ is the mean. Here $n = 5$ and $\sum_{i=1}^{5}(x_i - \bar{x})^2=1.04669$.

Step2: Calculate the standard deviation

Substitute $n = 5$ into the formula: $s=\sqrt{\frac{1.04669}{5 - 1}}=\sqrt{\frac{1.04669}{4}}=\sqrt{0.2616725}\approx0.5115$.

Answer:

$0.5115$