When a die is rolled 42 times, it is observed | vedclass

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(A) The frequency distribution is given by $f_i = 2i$ for $x_i = i$ where $i \in \{1, 2, 3, 4, 5, 6\}$.$x_i$$1$$2$$3$$4$$5$$6$$f_i$$2$$4$$6$$8$$10$$12

Vedclass · Accepted Answer

$\frac{80}{63}$

Vedclass · Answer

(A) The frequency distribution is given by $f_i = 2i$ for $x_i = i$ where $i \in \{1, 2, 3, 4, 5, 6\}$.$x_i$$1$$2$$3$$4$$5$$6$$f_i$$2$$4$$6$$8$$10$$12$Total frequency $N = \sum f_i = 2(1+2+3+4+5+6) = 2(21) = 42$.Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{2(1^2+2^2+3^2+4^2+5^2+6^2)}{42} = \frac{2(91)}{42} = \frac{182}{42} = \frac{13}{3}$.Mean deviation from mean $MD = \frac{\sum f_i |x_i - \bar{x}|}{N}$.$MD = \frac{2|1-\frac{13}{3}| + 4|2-\frac{13}{3}| + 6|3-\frac{13}{3}| + 8|4-\frac{13}{3}| + 10|5-\frac{13}{3}| + 12|6-\frac{13}{3}|}{42}$.$MD = \frac{2(\frac{10}{3}) + 4(\frac{7}{3}) + 6(\frac{4}{3}) + 8(\frac{1}{3}) + 10(\frac{2}{3}) + 12(\frac{5}{3})}{42}$.$MD = \frac{20 + 28 + 24 + 8 + 20 + 60}{3 	imes 42} = \frac{160}{126} = \frac{80}{63}$.

$x_i$	$6$	$12$	$18$	$24$	$30$	$36$	$42$
$f_i$	$4$	$7$	$9$	$18$	$15$	$10$	$5$

When a die is rolled $42$ times, it is observed that a face having the digit $i$ occurred $2i$ times. The mean deviation from the mean of this discrete frequency distribution is

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