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प्रश्न
If `barx` is the mean of x1, x2, ..., xn, then for a ≠ 0, the mean of `ax_1, ax_2, ..., ax_n, x_1/a, x_2/a, ..., x_n/a` is ______.
विकल्प
`(a + 1/a)barx`
`(a + 1/a) barx/2`
`(a + 1/a)barx/n`
`((a + 1/a)barx)/(2n)`
उत्तर
If `barx` is the mean of x1, x2, ..., xn, then for a ≠ 0, the mean of `ax_1, ax_2, ..., ax_n, x_1/a, x_2/a, ..., x_n/a` is `underlinebb((a + 1/a) barx/2)`.
Explanation:
Given: `barx` is the mean of x1, x2, ..., xn.
Then, `barx = (x_1 + x_2 + ... + x_n)/n`
Let the mean of the data set `ax_1, ax_2, .... ax_n, x_1/a, x_2/a , ... , x_n/a` is `bary`.
So, `bary = (ax_1 + ax_2 + ... + ax_n + x_1/a + x_2/a + ... + x_n/a)/(2n)`
`bary = (a(x_1 + x_2 + ... + x_n) + 1/a (x_1 + x_2 + ... + x_n))/(2n)`
`bary = ((a + 1/a)(x_1 + x_2 + .... + x_n))/(2n)`
From equation (I):
`bary = ((a + 1/a)barx)/2`
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