PQ- Show that the two formulae for the standard deviation of ungrouped data -1-n- x i - X 2 and -1- n - x i - X 2 are equivalent-wehere X -1- n - x i-

Δ = √(1/n∑ (x_i X̅)^2) δ^1 = √(1/n)∑^2_i x^ 2 (x_i x̅)^2 = x^2_i + x^ 2 2x_i x̅ ∑ 2x_i x̅ = 2 x̅∑ x_i = 2nx^x 2 1/n∑ (x_i x̅)^2 = ∑ (x^2_i + x^ 2 2x ...

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Standard XII

Mathematics

Standard Deviation

PQ. Show that...

Question

PQ. Show that the two formulae for the standard deviation of ungrouped data $δ = \sqrt{\frac{1}{n} \sum (x_{i} - ¯ X)^{2}}$ and $δ = \sqrt{\frac{1}{n} \sum (x_{i} - ¯ X)^{2}}$ are equivalent, wehere $¯ X = \frac{1}{n} \sum x_{i}$ .

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Solution

$δ = \sqrt{\frac{1}{n} \sum (x_{i} - ¯ X)^{2}}$

$δ^{1} = \sqrt{\frac{1}{n}} \sum_{i}^{2} - x^{- 2}$

$(x_{i} - ¯ x)^{2} = x_{i}^{2} + x^{- 2} - 2 x_{i} ¯ x$

$\sum 2 x_{i} ¯ x = 2 ¯ x \sum x_{i} = 2 n x^{x - 2}$

$\frac{1}{n} \sum (x_{i} - ¯ x)^{2} = \frac{\sum (x_{i}^{2} + x^{- 2} - 2 x_{i} ¯ x)}{n}$

$= \frac{\sum x_{i}^{2} + \sum_{x}^{- 2} - \sum 2 x_{i} ¯ x}{n}$

$= \frac{1}{n} \sum x_{i}^{2} + \frac{n x^{- 2} - 2 n x^{- 2}}{n}$

$= \frac{1}{n} \sum x_{i}^{2} - x^{- 2}$

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PQ. Show that the two formulae for the standard deviation of ungrouped data δ=√1n∑(xi−¯X)2 and δ=√1n∑(xi−¯X)2 are equivalent, wehere ¯X=1n∑xi.

δ=√1n∑(xi−¯X)2 δ1=√1n∑2i−x−2 (xi−¯x)2=x2i+x−2−2xi¯x ∑2xi¯x=2¯x∑xi=2nxx−2 1n∑(xi−¯x)2=∑(x2i+x−2−2xi¯x)n =∑x2i+∑−2x−∑2xi¯xn =1n∑x2i+nx−2−2nx−2n =1n∑x2i−x−2

PQ. Show that the two formulae for the standard deviation of ungrouped data $δ = \sqrt{\frac{1}{n} \sum (x_{i} - ¯ X)^{2}}$ and $δ = \sqrt{\frac{1}{n} \sum (x_{i} - ¯ X)^{2}}$ are equivalent, wehere $¯ X = \frac{1}{n} \sum x_{i}$ .