Find out standard deviation of the monthly income of 5 person, as stated below:

Explanation

`sigma=sqrt((sumdx'^(2))/(N)-((sumdx')/(N))^(2))xxC`
`=sqrt((68)/(5)-((0)/(5))^(2))xx100`
`=sqrt(13.6)xx100`
`=3.6878xx100=368.78`
Standard Deviation`(sigma)`=368.78.