Question

find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 1.46​

Answer 1

$\begin{lgathered}\begin{tabular}{| c | c | c | c | c | c | c| c|}\cline{1-8} \bf{No. of accident (x)} & \sf{0} & \sf{1} & \sf{2} & \sf{3} & \sf{4} & \sf{5}& \sf{Total}\\ \cline{1-8}\bf{Frequency (f)} & \sf{46} & \sf{?} & \sf{?} & \sf{25} & \sf{10}& \sf{5}& \sf{200} \\ \cline{1-8} \end{tabular}\end{lgathered}$

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☯ Let the missing frequencies be $\bf f_i\;and\;f_2$

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$\boxed{\begin{array}{cccc}\bf x_i&\bf f_i &\bf f_ix_i \\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad\qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 0&\sf 46&\sf 0\\\\\sf 1 &\sf f_1&\sf f_1\\\\\sf 2 &\sf f_2&\sf 2f_2\\\\\sf 3&\sf 25&\sf 75\\\\\sf 4&\sf 10&\sf 40\\\\\sf 5&\sf 5&\sf 25\\\frac{\qquad\qquad\qquad \qquad \qquad}{}&\frac{\qquad \qquad\qquad \qquad \qquad}{}&\frac{\qquad\qquad \qquad \qquad \qquad\qquad}{}\\\sf & \sf N = 86 + f_1 + f_2& \sf \sum f_i x_i = 140 + f_1 + 2f_2 \end{array}}$

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We have: N = 200

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Therefore,

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$\dashrightarrow\sf 200 = 86 + f_1 + f_2$

$\dashrightarrow\sf f_1 + f_2 = 200 - 86$

$\dashrightarrow\sf f_1 + f_2 = 114\qquad\qquad\bigg\lgroup\bf eq\;(1)\bigg\rgroup$

Also,

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Mean = 1.46

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$\dashrightarrow\sf 1.46 = \dfrac{ \sum f_i x_i}{N}$

$\dashrightarrow\sf 1.46 = \dfrac{ 140 + f_1 + 2f_2}{200}$

$\dashrightarrow\sf 292 = 140 + f_1 + 2f_2$

$\dashrightarrow\sf f_1 + 2f_2 = 152\qquad\qquad\bigg\lgroup\bf eq\;(2)\bigg\rgroup$

☯ Now, Solving eq (1) and (2), we get

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$\dashrightarrow{\boxed{\frak{\pink{f_i = 76\;and\;f_2 = 38}}}}\;\bigstar$