The standard deviation of first 50 even natural number is ` lamda ` then value of ` [lamda] ` is ( where [.] is gretest interger )…

Answer

28

Explanation

`SD = sqrt((sumx_(i)^(2))/(n)-((sumx_(i))/(n))^(2)) = sqrt((2^(2) + 4^(2) + 6^(2)+........+100^(2))/(50) - ((2+100)/(2))^(2))`
`= sqrt((4(1^(2) + 2^(2)+3^(3) +.......+50^(2)))/(50) - (51)^(2)) = sqrt((4 xx 50 xx 51 xx101)/(50 xx 6) - (51)^(2))`
`SD = sqrt(51((202)/(3) - 51)) = sqrt(51((202-153)/(3))) = sqrt(17 xx49) = sqrt(833) = lambda, [lambda] = [28.86] = 28`