The arithmetic mean of the observations 10, 8 | vedclass

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(D) Given observations are $10, 8, 5, a, b$.Arithmetic Mean,$\bar{x} = \frac{10+8+5+a+b}{5} = 6$$23 + a + b = 30 \implies a + b = 7$ ... $(i)$Variance

Vedclass · Accepted Answer

$12$

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(D) Given observations are $10, 8, 5, a, b$.Arithmetic Mean,$\bar{x} = \frac{10+8+5+a+b}{5} = 6$$23 + a + b = 30 \implies a + b = 7$ ... $(i)$Variance $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2 = 6.8$$\frac{10^2 + 8^2 + 5^2 + a^2 + b^2}{5} - (6)^2 = 6.8$$\frac{100 + 64 + 25 + a^2 + b^2}{5} - 36 = 6.8$$\frac{189 + a^2 + b^2}{5} = 42.8$$189 + a^2 + b^2 = 214$$a^2 + b^2 = 25$ ... $(ii)$We know that $(a + b)^2 = a^2 + b^2 + 2ab$$(7)^2 = 25 + 2ab$$49 = 25 + 2ab$$2ab = 24 \implies ab = 12$

Classes	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequencies	$5$	$8$	$15$	$16$	$6$

The arithmetic mean of the observations $10, 8, 5, a, b$ is $6$ and their variance is $6.8$. Then $ab$ is equal to:

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