the weekly salaries of a sample of employees at the local bank are given in the table below.\n| employee |…

the weekly salaries of a sample of employees at the local bank are given in the table below.\n| employee | weekly salary |\n|--|--| \n| anja | $245 |\n| raz | $300 |\n| natalie | $325 |\n| mic | $465 |\n| paul | $100 |\nwhat is the variance for the data?\nvariance: $s^{2}=\frac{(x_{1}-overline{x})^{2}+(x_{2}-overline{x})^{2}+cdots+(x_{n}-overline{x})^{2}}{n - 1}$
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 ((x_i-\bar{x})^2) for each (x_i)
For (x_1 = 245), ((245 - 287)^2=(-42)^2 = 1764). For (x_2 = 300), ((300 - 287)^2=(13)^2 = 169). For (x_3 = 325), ((325 - 287)^2=(38)^2 = 1444). For (x_4 = 465), ((465 - 287)^2=(178)^2 = 31684). For (x_5 = 100), ((100 - 287)^2=(-187)^2 = 34969).
Step3: Sum up ((x_i-\bar{x})^2) values
(\sum_{i = 1}^{5}(x_i-\bar{x})^2=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}{5-1}=\frac{70030}{4}=17507.5).
Answer:
(17507.5)