Question

Marks of 12 students in a unit test are given as 4, 21, 13, 17, 5, 9, 10, 20, 19, 12, 20 and 14. Assume a mean and calculate arithmetic mean of the data. Assume the other number as the mean and calculate the arithmetic mean again. Do you get the same result? Comment

Answer 1

Hint: We solve this problem by using the formula of arithmetic mean of a data.
If the mean of the data is \['x'\] then the arithmetic mean is given as
\[A.M=x+\sum{\dfrac{{{x}_{i}}-x}{n}}\]
Where, \[{{x}_{i}}\] is the value of frequency of the given data and \['n'\] is the total number of numbers in the data.
By using the above formula we assume that the mean of given data as one of the values of the data and calculate the arithmetic mean as we are told to.
Again we calculate the arithmetic mean by assuming the different values as the mean and check whether the two values we get are equal or not.

Complete step by step answer:
We are given that the data of marks of 12 students as
4, 21, 13, 17, 5, 9, 10, 20, 19, 12, 20, 14
First let us assume that the mean of the data as 13 that is
\[\Rightarrow x=13\]
Now, let us assume that the arithmetic mean in this case as \[{{A}_{1}}\]
We know that the formula that is if the mean of the data is \['x'\] then the arithmetic mean is given as
\[A.M=x+\sum{\dfrac{{{x}_{i}}-x}{n}}\]
Where, \[{{x}_{i}}\] is the value of frequency of the given data and \['n'\] is the total number of numbers in the data.
By using the above formula we get the value of arithmetic mean as
\[\Rightarrow {{A}_{1}}=13+\sum{\dfrac{{{x}_{i}}-13}{12}}\]
Now, let us expand the summation for each and every value of the given data then we get
\[\begin{align}
  & \Rightarrow {{A}_{1}}=13+\left( \dfrac{\begin{align}
  & \left( 4-13 \right)+\left( 21-13 \right)+\left( 13-13 \right)+\left( 17-13 \right)+\left( 5-13 \right)+\left( 9-13 \right)+ \\
 & \left( 10-13 \right)+\left( 20-13 \right)+\left( 19-13 \right)+\left( 12-13 \right)+\left( 20-13 \right)+\left( 14-13 \right) \\
\end{align}}{12} \right) \\
 & \Rightarrow {{A}_{1}}=13+\left( \dfrac{-9+8+0+4-8-4-3+7+6-1+7+1}{12} \right) \\
 & \Rightarrow {{A}_{1}}=13+\dfrac{8}{12}=13.67 \\
\end{align}\]
Now, let us assume that the mean of the data as 14 that is
\[\Rightarrow x=14\]
Now, let us assume that the arithmetic mean in this case as \[{{A}_{2}}\]
By using the arithmetic mean formula we get the value of arithmetic mean as
\[\Rightarrow {{A}_{2}}=14+\sum{\dfrac{{{x}_{i}}-14}{12}}\]
Now, let us expand the summation for each and every value of the given data then we get
\[\begin{align}
  & \Rightarrow {{A}_{2}}=14+\left( \dfrac{\begin{align}
  & \left( 4-14 \right)+\left( 21-14 \right)+\left( 13-14 \right)+\left( 17-14 \right)+\left( 5-14 \right)+\left( 9-14 \right)+ \\
 & \left( 10-14 \right)+\left( 20-14 \right)+\left( 19-14 \right)+\left( 12-14 \right)+\left( 20-14 \right)+\left( 14-14 \right) \\
\end{align}}{12} \right) \\
 & \Rightarrow {{A}_{2}}=14+\left( \dfrac{-10+7-1+3-9-5-4+6+5-2+6+0}{12} \right) \\
 & \Rightarrow {{A}_{2}}=14+\dfrac{-4}{12}=13.67 \\
\end{align}\]

Therefore, we can conclude that both the arithmetic values in two cases are equal.

Note: Students may make mistakes in assuming the mean of the data.
We are told to assume the mean of the data from the given values. So, we can assume the mean of the data as any of the value from
4, 21, 13, 17, 5, 9, 10, 20, 19, 12, 20, 14
But students may make mistakes by taking the mean by calculating the mean of the data.
The formula of the mean of the data is given as
\[\Rightarrow Mean=\dfrac{\text{sum of observations}}{\text{number of observations}}\]
They calculate the mean from this formula and assume it as mean.
This process is correct but it is not suitable for this question because we are told to assume the mean from the given data.