If the median and mode of 4 observations is 4, 6 and the sum of squares of observations is 48. Then
standard deviation is

\sqrt{2}

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\sqrt{3}

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\sqrt{5}

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Solution

The correct option is C $\sqrt{3}$
Given that median = 4 and mode = 6. We know that
Mean - mode = 3 (Mean - Median)
Mean - 6 = 3 (Mean - 4)
$¯ x - 6 = 3 ¯ x - 12$
$¯ x = 3$
$σ^{2} = \frac{\sum x_{i}^{2}}{n} - {(\frac{\sum x^{i}}{n})}^{2} = \frac{48}{4} - (3)^{2} = 3$ $∴ σ = \sqrt{3}$

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If the median and mode of 4 observations is 4, 6 and the sum of squares of observations is 48. Then standard deviation is

The correct option is C √3Given that median = 4 and mode = 6. We know that Mean - mode = 3 (Mean - Median) Mean - 6 = 3 (Mean - 4) ¯x−6=3¯x−12 ¯x=3 σ2=∑x2in−(∑xin)2=484−(3)2=3 ∴σ=√3

If the median and mode of 4 observations is 4, 6 and the sum of squares of observations is 48. Then
standard deviation is

The correct option is C $\sqrt{3}$
Given that median = 4 and mode = 6. We know that
Mean - mode = 3 (Mean - Median)
Mean - 6 = 3 (Mean - 4)
$¯ x - 6 = 3 ¯ x - 12$
$¯ x = 3$
$σ^{2} = \frac{\sum x_{i}^{2}}{n} - {(\frac{\sum x^{i}}{n})}^{2} = \frac{48}{4} - (3)^{2} = 3$ $∴ σ = \sqrt{3}$