If mean and standard deviation of 5 observations x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance of 6 observations x1, x2, x3, x4, x5 and

Engineering

Mathematics

Measure of Central Tendency

Measure of Dispersion

Question

If mean and standard deviation of 5 observations x₁, x₂, x₃, x₄, x₅ are 10 and 3, respectively, then the variance of 6 observations x₁, x₂, x₃, x₄, x₅ and – 50 is equal to

586.5

582.5

507.5

509.5

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$\sum x = 50$

${(3)}^{2} = \frac{1}{5} ({ex}^{2} - \frac{{(ex)}^{2}}{5})$

$9 = \frac{1}{5} (\sum x^{2} - \frac{2500}{5})$

$∴ \sum x^{2} = 545$

New variable $= \frac{1}{6} (3045 - \frac{0}{6}) = 507 .5$