Question

find the median of data 10,20,5,4,20,5,8 and 10​

Answer 1

Given: The average of 5 number is 496. If 2 of them are 117 and 140.

Need to find: The average of remaining three numbers.

❒ Let the five numbers be $\sf x_{1}, x_{2}, x_{3}, x_4 \;and\; x_5.$

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$\underline{\bf{\dag} \:\mathfrak{As\; we \; know \; that \; :}}$

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$\star\;\boxed{\sf{\pink{Average = \dfrac{ Sum\; of\; observation}{Total \; number \; of \; observation}}}}$

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Also,

The average of 5 number is 496.

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Therefore,

$:\implies\sf 496 = \dfrac{ x_1 + x_2 + x_3 + x_4 + x_5}{5} \\\\\\:\implies\sf Numbers = 496 \times 5 \\\\\\:\implies{\underline{\boxed{\frak{\purple{ 2480}}}}}\;\bigstar$

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$\qquad\quad\boxed{\bf{\mid{\overline{\underline{\bigstar\: According\: to \; the\; Question \: :}}}}\mid}$

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The average of 5 number is 496. And, If two of them are 117 and 140.

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Therefore,

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$\sf x_1$ = 117

$\sf x_2$ = 140

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Now,

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$:\implies\sf 117 + 140 + x_3 + x_4 + x_5 = 2480 \\\\\\:\implies\sf x_3 + x_4 + x_5 = 2480 - 257 \\\\\\:\implies{\underline{\boxed{\frak{\pink{ x_3 + x_4 + x_5 = 2223}}}}}\;\bigstar$

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By using same formula, finding the average of remaining three numbers.

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Hence,

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$:\implies\sf \dfrac{x_3 + x_4 + x_5}{3} \\\\\\:\implies\sf \cancel\dfrac{ 2223}{3} \\\\\\:\implies{\underline{\boxed{\frak{\purple{741}}}}}\;\bigstar$

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$\therefore{\underline{\sf{Hence, \: the\; average\; of \: remaining \; three \; numbers \;is \; \bf{741}.}}}$