Let O be the mean deviation of the first five | vedclass

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(B) The first five odd natural numbers are $1, 3, 5, 7, 9$. Their mean $\bar{x} = \frac{1+3+5+7+9}{5} = \frac{25}{5} = 5$.The mean deviation $O = \fra

Vedclass · Accepted Answer

$0.32$

Vedclass · Answer

(B) The first five odd natural numbers are $1, 3, 5, 7, 9$. Their mean $\bar{x} = \frac{1+3+5+7+9}{5} = \frac{25}{5} = 5$.The mean deviation $O = \frac{\sum |x_i - \bar{x}|}{5} = \frac{|1-5| + |3-5| + |5-5| + |7-5| + |9-5|}{5} = \frac{4+2+0+2+4}{5} = \frac{12}{5} = 2.4$.The first five prime numbers are $2, 3, 5, 7, 11$. Their mean $\bar{y} = \frac{2+3+5+7+11}{5} = \frac{28}{5} = 5.6$.The mean deviation $P = \frac{\sum |y_i - \bar{y}|}{5} = \frac{|2-5.6| + |3-5.6| + |5-5.6| + |7-5.6| + |11-5.6|}{5} = \frac{3.6 + 2.6 + 0.6 + 1.4 + 5.4}{5} = \frac{13.6}{5} = 2.72$.Therefore,$P - O = 2.72 - 2.4 = 0.32$.

Let $O$ be the mean deviation of the first five odd natural numbers about their mean and $P$ be the mean deviation of the first five prime numbers about their mean. Then $P-O=$

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