Question

The variate of a distribution takes the values \[1,2,3,......,n\] with the frequencies
\[n,n-1,n-2,......,2,1\], then the mean value of the distribution is
(a) \[\dfrac{n\left( n+2 \right)}{3}\]
(b) \[\dfrac{n\left( n+1 \right)\left( n+2 \right)}{6}\]
(c) \[\dfrac{\left( n+2 \right)}{3}\]
(d) \[\dfrac{\left( n+1 \right)\left( n+2 \right)}{6}\]

Answer 1

Hint: We solve this problem by considering the values of distribution of given data as \[{{x}_{i}}\] and the frequencies as \[{{f}_{i}}\], then the formula of mean of data is given as
\[\bar{x}=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{1}}}}\]
By using the formula of mean we calculate the mean value of given data.

Complete step-by-step solution:
We are given that the variate of a distribution takes the values \[1,2,3,......,n\] with the frequencies
\[n,n-1,n-2,......,2,1\].
Let us assume the values of distribution of given data as \[{{x}_{i}}\] and the frequencies as \[{{f}_{i}}\]
We know that the formula of mean is given as
\[\bar{x}=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{1}}}}\]
Now, let us find the numerator of above formula as
\[\Rightarrow \sum{{{f}_{i}}{{x}_{i}}}=\left( n\times 1 \right)+\left( \left( n-1 \right)\times 2 \right)+\left( \left( n-2 \right)\times 3 \right)+......+\left( 1\times n \right)\]
Now, let us convert the above equation into summation form as follows
\[\Rightarrow \sum{{{f}_{i}}{{x}_{i}}}=\sum\limits_{k=1}^{n}{\left( n-k+1 \right)k}\]
Now by separating the terms in above equation we get
\[\Rightarrow \sum{{{f}_{i}}{{x}_{i}}}=n\sum{k}-\sum{{{k}^{2}}}+\sum{k}..........equation(i)\]
We know that standard results of summation as
\[\begin{align}
  & \sum\limits_{k=1}^{n}{k}=\dfrac{n\left( n+1 \right)}{2} \\
 & \sum\limits_{k=1}^{n}{{{k}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \\
\end{align}\]
By using the above standard results in equation (i) we get
\[\Rightarrow \sum{{{f}_{i}}{{x}_{i}}}=n\left( \dfrac{n\left( n+1 \right)}{2} \right)-\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{n\left( n+1 \right)}{2}\]
Now, by taking the common term out in each term we get
\[\begin{align}
  & \Rightarrow \sum{{{f}_{i}}{{x}_{i}}}=\left[ \dfrac{n\left( n+1 \right)}{2} \right]\left[ n-\dfrac{2n+1}{3}+1 \right] \\
 & \Rightarrow \sum{{{f}_{i}}{{x}_{i}}}=\left[ \dfrac{n\left( n+1 \right)}{2} \right]\left[ \dfrac{3n-2n-1+3}{3} \right] \\
 & \Rightarrow \sum{{{f}_{i}}{{x}_{i}}}=\left[ \dfrac{n\left( n+1 \right)}{2} \right]\left[ \dfrac{n+2}{3} \right] \\
\end{align}\]
Now let us find the denominator in the mean formula mentioned above as
\[\Rightarrow \sum{{{f}_{i}}}=n+\left( n-1 \right)+\left( n-2 \right)+.......+2+1\]
Now, let us convert the above equation into summation form as follows
\[\Rightarrow \sum{{{f}_{i}}}=\sum\limits_{k=1}^{n}{k}\]
By using the above standard results we get
\[\Rightarrow \sum{{{f}_{i}}}=\dfrac{n\left( n+1 \right)}{2}\]
Now, by substituting the required values in the mean formula we get
\[\begin{align}
  & \Rightarrow \bar{x}=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{1}}}} \\
 & \Rightarrow \bar{x}=\dfrac{\left[ \dfrac{n\left( n+1 \right)}{2} \right]\left[ \dfrac{n+2}{3} \right]}{\left[ \dfrac{n\left( n+1 \right)}{2} \right]} \\
 & \Rightarrow \bar{x}=\left[ \dfrac{n+2}{3} \right] \\
\end{align}\]
Therefore the mean value of given data is\[\left[ \dfrac{n+2}{3} \right]\].
So, option (c) is correct answer.

Note: Students will do mistake in solving the problem, that is they calculate the mean by using the formula
\[\Rightarrow \bar{x}=\dfrac{\sum{{{x}_{i}}}}{\sum{{{f}_{i}}}}\]
This formula is wrong for this type of problem because, in the average formula, the numerator is the total sum of distributions. But in the above formula, the numerator doesn’t lead to the total sum of distributions. But if the formula is taken as
\[\Rightarrow \bar{x}=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\]
Here, the numerator gives the total sum of distributions. So, this is the correct formula to use.