If -i - 19 xi - 5 - 9 and -i - 19 xi - 52 - 45- then the standard deviation of the 9 items x1-x2-x9 is

The correct option is A 2 S.D of xi 5 is σ =√(∑_i=1^9(xi 5)^29 [ ∑_i=1^9(xi 5)^29 ]^2) σ =√(5 1)=2 ...

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Standard XII

Mathematics

Geometric Mean

If ∑i = 19 ...

Question

If $9 \sum i = 1 (x_{i} - 5) = 9$ and $9 \sum i = 1 {(x_{i} - 5)}^{2} = 45$ , then the standard deviation of the $9$ items $x_{1}, x_{2}, . . . . ., x_{9}$ is

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Solution

The correct option is A $2$
S.D of $x i - 5$ is

$σ = \sqrt{\frac{\sum_{i = 1}^{9} (x i - 5)^{2}}{9} - {[\frac{\sum_{i = 1}^{9} (x i - 5)^{2}}{9}]}^{2}}$

$σ = \sqrt{5 - 1} = 2$

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If 9∑i=1(xi−5)=9 and 9∑i=1(xi−5)2=45, then the standard deviation of the 9 items x1,x2,.....,x9 is

The correct option is A 2S.D of xi−5 isσ= ⎷∑9i=1(xi−5)29−[∑9i=1(xi−5)29]2σ=√5−1=2

If $9 \sum i = 1 (x_{i} - 5) = 9$ and $9 \sum i = 1 {(x_{i} - 5)}^{2} = 45$ , then the standard deviation of the $9$ items $x_{1}, x_{2}, . . . . ., x_{9}$ is

The correct option is A $2$
S.D of $x i - 5$ is

$σ = \sqrt{\frac{\sum_{i = 1}^{9} (x i - 5)^{2}}{9} - {[\frac{\sum_{i = 1}^{9} (x i - 5)^{2}}{9}]}^{2}}$

$σ = \sqrt{5 - 1} = 2$