Question

The mean of the distribution, in which the values of \[X\] are \[1,2,...,n\] the frequency of each being unity is:
A. \[\dfrac{{n\left( {n + 1} \right)}}{2}\]
B. \[\dfrac{n}{2}\]
C. \[\dfrac{{\left( {n + 1} \right)}}{2}\]
D. None of these

Answer 1

Hint: We will solve this problem by considering the values of distribution of given data as \[{x_i}\] and the frequencies as \[{f_i}\] . Then we will use the formula of mean of the data which is given as \[Mean,{\text{ }}\overline x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\]. After that we will simplify it and find the required result.

Complete step by step answer:
We are given that the values of \[X\] are \[1,2,...,n\] with the frequency of each being unity which means the frequencies are \[1,1,....,1\] (n times)
Now let us assume the values of distribution of given data as \[{x_i}\] and the frequencies as \[{f_i}\]
Therefore, the values of \[{x_i}\] from \[1 < i < n\] we have
\[{x_1} = 1,{\text{ }}{x_2} = 2,{\text{ }}{x_3} = 3,{\text{ }}......,{\text{ }}{x_n} = n\]
Similarly, the values of \[{f_i}\] from \[1 < i < n\] we have
\[{f_1} = 1,{\text{ }}{f_2} = 1,{\text{ }}{f_3} = 1,{\text{ }}......,{\text{ }}{f_n} = 1\]
Now we know that the formula of mean is given as
\[Mean,{\text{ }}\overline x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\]
Now let’s find the numerator of above formula as
\[\sum\limits_{i = 1}^n {{f_i}{x_i}} = {f_1}{x_1} + {f_2}{x_2} + {f_3}{x_3} + ..... + {f_n}{x_n}\]

On substituting values, we get
\[ \Rightarrow \sum\limits_{i = 1}^n {{f_i}{x_i}} = 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 + ..... + 1 \cdot n\]
On simplifying, we get
\[ \Rightarrow \sum\limits_{i = 1}^n {{f_i}{x_i}} = 1 + 2 + 3 + ..... + n\]
Here we can see that it forms as A.P.
As we know, sum of the n terms of an A.P. is given by \[\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Therefore, on substituting the values we have
\[ \Rightarrow \sum\limits_{i = 1}^n {{f_i}{x_i}} = \dfrac{n}{2}\left[ {2\left( 1 \right) + \left( {n - 1} \right)1} \right]\]
On simplifying it, we get
\[ \Rightarrow \sum\limits_{i = 1}^n {{f_i}{x_i}} = \dfrac{n}{2}\left[ {n + 1} \right]{\text{ }} - - - \left( i \right)\]

Now let’s find the denominator of above formula as
\[\sum\limits_{i = 1}^n {{f_i}} = {f_1} + {f_2} + {f_3} + ..... + {f_n}\]
On substituting values, we get
\[ \Rightarrow \sum\limits_{i = 1}^n {{f_i}} = 1 + 1 + 1 + ..... + 1{\text{ }}\left( {n{\text{ }}times} \right)\]
On simplifying, we get
\[ \Rightarrow \sum\limits_{i = 1}^n {{f_i}} = n{\text{ }} - - - \left( {ii} \right)\]
So, on substituting the values from equation \[\left( i \right)\] and \[\left( {ii} \right)\] in the formula, we get
\[Mean,{\text{ }}\overline x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }} = \dfrac{{\dfrac{{n\left( {n + 1} \right)}}{2}}}{n}\]
On cancelling we get
\[\therefore Mean,{\text{ }}\overline x = \dfrac{{n + 1}}{2}\]

Hence, option C is the correct answer.

Note: Students may do mistake in solving the problem, that is they might calculate the mean by using the formula,
\[\overline x = \dfrac{{\sum {{x_i}} }}{{\sum {{f_i}} }}\]
This formula is wrong for this type of problem because, in the mean formula, the numerator is the total sum of distributions. But in the above formula, the numerator does not lead to the total sum of distributions. So, take care of this thing while solving. Also, instead of using the formula of sum of an A.P. one can also use the formula of sum of first $n$ natural number i.e., Sum of $n$ natural numbers is \[\dfrac{{n\left( {n + 1} \right)}}{2}\]. And proceed further to get the required result.