Question

If for some \[x\in R\], the frequency distribution of the marks by 20 students in a test is: -

Marks	2	3	5	7
Frequency	\[{{\left( x+1 \right)}^{2}}\]	\[\left( 2x-5 \right)\]	\[{{x}^{2}}-3x\]	\[x\]

then the mean of the marks is: -
(a) 2.8
(b) 3.2
(c) 3.0
(d) 2.5

Answer 1

Hint: Add all the frequencies and equate it with 20 to form a quadratic equation. Solve this quadratic equation to find the values of x. Neglect the negative value. Now, substitute the value of x to find the frequency of students for the given marks. Finally, apply the formula: - \[\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\] to find the mean. Here, \[{{x}_{i}}\] is the mark and \[{{f}_{i}}\] is the frequency of students for that particular mark.

Complete step by step answer:
We have been provided with the following frequency distribution table: -

Marks	2	3	5	7
Frequency	\[{{\left( x+1 \right)}^{2}}\]	\[\left( 2x-5 \right)\]	\[{{x}^{2}}-3x\]	\[x\]

It is given that there are 20 students. So, the sum of all the given frequencies will be 20.
\[\begin{align}
  & \Rightarrow \sum{{{f}_{i}}}=20 \\
 & \Rightarrow {{\left( x+1 \right)}^{2}}+\left( 2x-5 \right)+\left( {{x}^{2}}-3x \right)+x=20 \\
\end{align}\]
Using the identity, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\], we get,
\[\Rightarrow {{x}^{2}}+1+2x+2x-5+{{x}^{2}}-3x+x=20\]
\[\begin{align}
  & \Rightarrow 2{{x}^{2}}+2x-4=20 \\
 & \Rightarrow 2{{x}^{2}}+2x-24=0 \\
\end{align}\]
Dividing both sides by 2, we get,
\[\Rightarrow {{x}^{2}}+x-12=0\]
Using the middle term split method, we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}+4x-3x-12=0 \\
 & \Rightarrow x\left( x+4 \right)-3\left( x+4 \right)=0 \\
 & \Rightarrow \left( x+4 \right)\left( x-3 \right)=0 \\
\end{align}\]
Substituting each term equal to 0, we get,
\[\Rightarrow x=-4\] or 3.
Now, since the number of students cannot be negative, therefore x = -4 is neglected. So, we have x = 3.
Therefore, substituting the value x = 3 in the frequency distribution table, we get,

Marks	2	3	5	7
Frequency	16	1	0	3

This type of data is known as ungrouped data whose mean is given by the relation: -
Mean = \[\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\], where ‘\[{{x}_{i}}\]’ is the marks and ‘\[{{f}_{i}}\]’ is the frequency of students for that marks.
\[\Rightarrow \] Mean = \[\dfrac{\left( 16\times 2 \right)+\left( 1\times 3 \right)+\left( 0\times 5 \right)+\left( 3\times 7 \right)}{\left( 16+1+0+3 \right)}\]
\[\Rightarrow \] Mean = \[\dfrac{32+3+0+21}{20}\]
\[\Rightarrow \] Mean = \[\dfrac{56}{20}\]
\[\Rightarrow \] Mean = 2.8

Hence, option (a) is the correct answer.

Note:
One may note that, after solving the quadratic equation we found two values of ‘x’. Here, we neglected the negative value of ‘x’ because it denotes the number of students or frequency which cannot be negative. Also, note that the given distribution is known as ungrouped data because marks are not present as intervals. So, remember the formula for the mean of ungrouped data as: - Mean = \[\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\].