1.

Find the mean of the following data . Range of first n natural numbers , range of negative integers from - n to -1 (where -n lt -1) , range of first n positive even integers and range of first n positive odd integers .

Answer»

`(3)/(2) ( n-1)`
`(3n-2)/(2)`
`(3)/(2) (n-2)`
`(4n-3)/(2)`

SOLUTION :Range of the first 'n' NATURAL numbers =n - 1.
Range of the last n negative INTEGERS = `-1 - (-n) = n-1`.
Range of the first 'n' positive even integers = 2n - 2 .
Range of first 'n' positive odd integers = (2n+ 1) - (1) = 2n -2 .
`therefore` Mean of the given data is
`((n- 1) + ( n - 1) + (2n -2) + (2n - 2))/(4)`
`= (6n-6)/(4) = (3n-3)/(2)`
`= (3)/(2) (n-1)`.


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