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प्रश्न
The algebraic sum of the deviations of a set of n values from their mean is
पर्याय
0
n − 1
n
n + 1
उत्तर
If `bar( X) `be the mean of the n observations q `X_i,...,X_n`, then we have
`bar(X) = 1/nsum_(i=1)^n X_i`
`⇒ sum_(i=1)^n X_i = nbar(X)`
Let ` bar(X)` be the mean of n values `X_i,...,X_n` . So, we have
`bar(X) = 1/nsum_(i=1)^n X_i`
`⇒ sum_(i=1)^n X_i = nbar(X)`
The sum of the deviations of n values `X_i,...,X_n` from their mean`bar(X)` is
`(x_1 - bar(X) ) +(x_2 - bar(X) )+......+(x_n-bar(X))`
`= sum_(i=1)^n(x_i- bar(X))`
`= sum_(i=1)^n x_i- sum_(i=1)^n bar(X)`
`= nbar(X) - nbar(X)`
=0
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