If for some x in R^{+} cup {0},the frequency | vedclass

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(B) The total number of students is $N = 20$.Sum of frequencies $\Sigma f_i = (x+1)^2 + (2x-5) + (x^2-3x) + x = 20$.Expanding the terms: $(x^2+2x+1) +

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$2.8$

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(B) The total number of students is $N = 20$.Sum of frequencies $\Sigma f_i = (x+1)^2 + (2x-5) + (x^2-3x) + x = 20$.Expanding the terms: $(x^2+2x+1) + 2x - 5 + x^2 - 3x + x = 20$.$2x^2 + 2x - 4 = 20$ $\Rightarrow 2x^2 + 2x - 24 = 0$ $\Rightarrow x^2 + x - 12 = 0$.Factoring the quadratic: $(x+4)(x-3) = 0$.Since $x \in R^{+} \cup \{0\}$,we have $x = 3$.Now,calculate the frequencies for $x=3$:Marks $2$: $(3+1)^2 = 16$.Marks $3$: $2(3)-5 = 1$.Marks $5$: $3^2-3(3) = 0$.Marks $7$: $3$.Check sum: $16+1+0+3 = 20$. Correct.Mean $\bar{x} = \frac{\Sigma f_i x_i}{N} = \frac{2(16) + 3(1) + 5(0) + 7(3)}{20} = \frac{32 + 3 + 0 + 21}{20} = \frac{56}{20} = 2.8$.

Marks:	$2$	$3$	$5$	$7$
Frequency:	$(x+1)^2$	$2x-5$	$x^2-3x$	$x$

Value	$1$	$2$	$3$	$4$
Frequency	$5$	$4$	$6$	$f$

If for some $x \in R^{+} \cup \{0\}$,the frequency distribution of the marks obtained by $20$ students in a test is given by the table below,then find the mean of the marks.
Marks: $2$ $3$ $5$ $7$
Frequency: $(x+1)^2$ $2x-5$ $x^2-3x$ $x$

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If for some $x \in R^{+} \cup \{0\}$,the frequency distribution of the marks obtained by $20$ students in a test is given by the table below,then find the mean of the marks.Marks:$2$$3$$5$$7$Frequency:$(x+1)^2$$2x-5$$x^2-3x$$x$