employee weekly salary anja $245 raz $300 natalie $325 mic $465 paul $100 what is the variance for the data…

employee weekly salary anja $245 raz $300 natalie $325 mic $465 paul $100 what is the variance for the data? variance $s^{2}=\frac{(x_{1}-overline{x})^{2}+(x_{2}-overline{x})^{2}+cdots+(x_{n}-overline{x})^{2}}{n - 1}$ 118.35 132.32 14006 17507.5

employee weekly salary anja $245 raz $300 natalie $325 mic $465 paul $100 what is the variance for the data? variance $s^{2}=\frac{(x_{1}-overline{x})^{2}+(x_{2}-overline{x})^{2}+cdots+(x_{n}-overline{x})^{2}}{n - 1}$ 118.35 132.32 14006 17507.5

Answer

Explanation:

Step1: Calculate the mean

The data set is (245,300,325,465,100). The mean (\bar{x}=\frac{245 + 300+325+465+100}{5}=\frac{1435}{5}=287).

Step2: Calculate the squared - differences

((245 - 287)^2=( - 42)^2 = 1764), ((300 - 287)^2=13^2 = 169), ((325 - 287)^2 = 38^2=1444), ((465 - 287)^2=178^2 = 31684), ((100 - 287)^2=( - 187)^2=34969).

Step3: Calculate the sum of squared - differences

(1764+169+1444+31684+34969 = 70030).

Step4: Calculate the variance

Using the formula (s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}), with (n = 5), we have (s^{2}=\frac{70030}{4}=17507.5).

Answer:

17507.5