The mode of the following frequency distribut | vedclass

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(A) Given total frequency $N = 150$.Sum of frequencies: $6 + 8 + 17 + x + 42 + 30 + y + 8 = 150$$111 + x + y = 150 \implies x + y = 39$  --- $(1)$Sinc

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(A) Given total frequency $N = 150$.Sum of frequencies: $6 + 8 + 17 + x + 42 + 30 + y + 8 = 150$$111 + x + y = 150 \implies x + y = 39$  --- $(1)$Since the mode is $46$,the modal class is $40-50$.Here,$l = 40, f_1 = 42, f_0 = x, f_2 = 30, h = 10$.Mode formula: $	ext{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} ight) 	imes h$$46 = 40 + \left( \frac{42 - x}{2(42) - x - 30} ight) 	imes 10$$6 = \left( \frac{42 - x}{84 - 30 - x} ight) 	imes 10$$6 = \frac{10(42 - x)}{54 - x} \implies 3(54 - x) = 5(42 - x)$$162 - 3x = 210 - 5x$$2x = 48 \implies x = 24$Substituting $x = 24$ in $(1)$: $24 + y = 39 \implies y = 15$.Thus,$x = 24$ and $y = 15$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$6$	$8$	$17$	$x$	$42$	$30$	$y$	$8$

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$2$	$4$	$10$	$20$	$18$	$20$	$16$	$10$

The mode of the following frequency distribution is $46$ and the total frequency is $150$. Find the missing frequencies $x$ and $y$.

Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$

Frequency $6$ $8$ $17$ $x$ $42$ $30$ $y$ $8$

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(in $.55$)

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The mode of the following frequency distribution is $46$ and the total frequency is $150$. Find the missing frequencies $x$ and $y$. Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ Frequency $6$ $8$ $17$ $x$ $42$ $30$ $y$ $8$