Question

the data set on the right represents the population. which formula should be used to calculate the variance? a teacher recorded all of his students grades on the final exam as: 62, 77, 78, 80, 82, 82, 83, 84, 85, 87, 89, 95. consider the formulas: a: $s^{2}=\frac{(x_{1}-overline{x})^{2}+(x_{2}-overline{x})^{2}+cdots+(x_{n}-overline{x})^{2}}{n - 1}$, b: $s=sqrt{\frac{(x_{1}-overline{x})^{2}+(x_{2}-overline{x})^{2}+cdots+(x_{n}-overline{x})^{2}}{n - 1}}$, c: $sigma^{2}=\frac{(x_{1}-mu)^{2}+(x_{2}-mu)^{2}+cdots+(x_{n}-mu)^{2}}{n}$, d: $sigma=sqrt{\frac{(x_{1}-mu)^{2}+(x_{2}-mu)^{2}+cdots+(x_{n}-mu)^{2}}{n}}$. what is the variance?

Explanation:

Step1: Calculate the population mean $\mu$

$n = 12$, $\sum_{i = 1}^{12}x_i=62 + 77+78 + 80+82+82+83+84+85+87+89+95 = 974$, $\mu=\frac{\sum_{i = 1}^{12}x_i}{n}=\frac{974}{12}\approx81.17$

Step2: Calculate the sum of squared - deviations

$(62 - 81.17)^2+(77 - 81.17)^2+(78 - 81.17)^2+(80 - 81.17)^2+(82 - 81.17)^2+(82 - 81.17)^2+(83 - 81.17)^2+(84 - 81.17)^2+(85 - 81.17)^2+(87 - 81.17)^2+(89 - 81.17)^2+(95 - 81.17)^2$
$=(-19.17)^2+(-4.17)^2+(-3.17)^2+(-1.17)^2+(0.83)^2+(0.83)^2+(1.83)^2+(2.83)^2+(3.83)^2+(5.83)^2+(7.83)^2+(13.83)^2$
$=367.49+17.39+10.05+1.37+0.69+0.69+3.35+8.01+14.66+34.00+61.31+191.21$
$=709.22$

Step3: Calculate the variance using the population - variance formula

$\sigma^{2}=\frac{\sum_{i = 1}^{N}(x_i-\mu)^2}{N}$, where $N = 12$
$\sigma^{2}=\frac{709.22}{12}\approx59.10$

Answer:

$59.10$