If for some x in R,the frequency distribution | vedclass

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(A) The sum of frequencies is given by $\sum f_i = (x+1)^2 + (2x-5) + (x^2-3x) + x = 20$.Expanding this,we get $x^2 + 2x + 1 + 2x - 5 + x^2 - 3x + x =

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$2.8$

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(A) The sum of frequencies is given by $\sum f_i = (x+1)^2 + (2x-5) + (x^2-3x) + x = 20$.Expanding this,we get $x^2 + 2x + 1 + 2x - 5 + x^2 - 3x + x = 20$.$2x^2 + 2x - 4 = 20$ $\Rightarrow 2x^2 + 2x - 24 = 0$ $\Rightarrow x^2 + x - 12 = 0$.Factoring the quadratic equation: $(x+4)(x-3) = 0$.Since frequency cannot be negative,$2x-5 > 0$ implies $x > 2.5$,so we take $x = 3$.Substituting $x = 3$ into the frequencies:Marks $2$: $f_1 = (3+1)^2 = 16$Marks $3$: $f_2 = 2(3)-5 = 1$Marks $5$: $f_3 = 3^2 - 3(3) = 0$Marks $7$: $f_4 = 3$Total students: $16 + 1 + 0 + 3 = 20$.The mean $\bar{x} = \frac{\sum x_i f_i}{\sum f_i} = \frac{(2 	imes 16) + (3 	imes 1) + (5 	imes 0) + (7 	imes 3)}{20} = \frac{32 + 3 + 0 + 21}{20} = \frac{56}{20} = 2.8$.

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