If for some positive x in R,the frequency dis | vedclass

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(C) Given the total number of students is $20$,the sum of frequencies must be $20$:$(x+1)^2 + (2x-5) + (x^2-3x) + x = 20$$(x^2+2x+1) + 2x - 5 + x^2 -

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$2.8$

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(C) Given the total number of students is $20$,the sum of frequencies must be $20$:$(x+1)^2 + (2x-5) + (x^2-3x) + x = 20$$(x^2+2x+1) + 2x - 5 + x^2 - 3x + x = 20$$2x^2 + 2x - 4 = 20$$2x^2 + 2x - 24 = 0$$x^2 + x - 12 = 0$$(x+4)(x-3) = 0$Since $x > 0$,we have $x = 3$.Now,substitute $x=3$ into the frequency distribution:Marks $(x_i)$: $2, 3, 5, 7$Frequencies $(f_i)$: $(3+1)^2=16, (2(3)-5)=1, (3^2-3(3))=0, 3$Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2(16) + 3(1) + 5(0) + 7(3)}{20}$$= \frac{32 + 3 + 0 + 21}{20} = \frac{56}{20} = 2.8$

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

If for some positive $x \in R$,the frequency distribution of the marks obtained by $20$ students in a certain test is as follows:
Marks $2$ $3$ $5$ $7$
Frequency $(x+1)^2$ $2x-5$ $x^2-3x$ $x$

Then the mean of the marks is:

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If for some positive $x \in R$,the frequency distribution of the marks obtained by $20$ students in a certain test is as follows:Marks$2$$3$$5$$7$Frequency$(x+1)^2$$2x-5$$x^2-3x$$x$Then the mean of the marks is: