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Question
If `bar"X"` is the mean of n observations x1, x2, x3,..., xn 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.
Solution
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`.
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