Question

If `bar"X"` is the mean of n observations x₁, x₂, x₃,..., x_n then the mean of `x_1/"a", x_2/"a", x_3/"a",...,x_"n"/"a" "is" bar"X"/"a"`, where a is an non-zero number.

i.e., if each observation is divided by a non-zero number, then the mean is also divided by it.

Answer 1

We have

`bar"X" = (1)/"n"(sum_(i = 1)^"n" x_i)`    ...(i)

Let `bar"X"` be the item of `x_1/a, x_2/a,...,x_"n"/a`. Then

`bar"X" = (1)/"n"(x_1/a + x_2/a + ... + x_"n"/"n")`

= `(1)/"n"((x_1 + x_2 + ... + x_"n")/(a))`

= `(1)/a((x_1 + x_2 + ... + x_"n")/"n")`

= `(1)/a[(1)/"n"(sum_(i = 1)^"n" x_i)]`

= `(1)/a(bar"X")`,    ...[Using (i)]

= `bar"X"/a`.