Animated Solution for Mathematics - Statistics: 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 :

Welcome, fellow learners! Today, we are embarking on a journey into the heart of statistics. Often, we are presented with data that is neatly laid out, but in the world of JEE Advanced, data is rarely so cooperative.

Sometimes, it is masked, hidden behind algebraic expressions, waiting for a sharp mind to reveal its true form. Imagine you are a data detective. You have been handed a frequency distribution table where the marks are clear, but the frequencies are shrouded in an unknown variable $x$.

Your mission is to uncover the value of $x$ and then calculate the mean of the marks. This is not just a math problem; it is a test of your logical vigilance.

The first step in our investigation is to recognize the fundamental constraint of any frequency distribution: the sum of all individual frequencies must equal the total number of students. We are told there are 20 students in total.

Therefore, we can write the following equation:

Now, let us perform the algebraic expansion. Expanding the first term, $(x+1{)}^2$, gives us $x^2+2x+1$. The equation now looks like this:

It looks a bit messy, but do not be intimidated. Let us group the like terms. Combining the $x^2$ terms, we get $2x^2$. Combining the $x$ terms, we have $2x+2x-3x+x$, which simplifies to $2x$. Finally, combining the constants, $1-5$ gives us $-4$.

So, our equation becomes $2x^2+2x-4=20$. Bringing the 20 to the left side, we get $2x^2+2x-24=0$. To make our lives easier, we can divide the entire equation by 2, resulting in the elegant quadratic equation:

Now, we solve for $x$. We need two numbers that multiply to $-12$ and add to $1$. Those numbers are $4$ and $-3$. Thus, the factors are $(x+4)(x-3)=0$.

This gives us two potential values for $x$: $x=-4$ or $x=3$. But here is where the detective work truly begins. We must apply a physical constraint.

A frequency represents a count of students, and a count can never be negative. If we test $x=-4$, the frequency for 7 marks becomes $x=-4$, which is impossible. Therefore, we must reject $x=-4$ and accept $x=3$.

With $x=3$ firmly in our grasp, we can now find the actual frequencies. For 2 marks, $f_1=(3+1{)}^2=16$. For 3 marks, $f_2=2(3)-5=1$. For 5 marks, $f_3=3^2-3(3)=0$. And for 7 marks, $f_4=3$.

The sum is $16+1+0+3=20$, which confirms our work is correct. Now, we apply the mean formula:

We calculate the products: $2\times 16=32$, $3\times 1=3$, $5\times 0=0$, and $7\times 3=21$. Adding these together, we get $32+3+0+21=56$.

Finally, we divide by the total number of students:

And there you have it! Through careful algebra and logical validation, we have uncovered the mean. The final answer is $2.8$. Keep practicing, stay curious, and remember that every variable has a story waiting to be told.