Question

The average height of 30 students is 150 cm. It was detected later that one value 165 cm was wrongly copied as 135 cm for the computation of mean. Find the correct mean.

Answer 1

Hint: First let us assign the variables to the 29 students and the $30^{th}$ student whose height was noted incorrectly, now use the mean formula $\left( \overline{x} \right)=\dfrac{\text{sum of the total terms}}{\text{number of terms}}$, to find the sum of the heights of the 29 students. Finally, use the sum of the heights of 29 students and the original height of the 30th person to find the corrected mean.

Complete step-by-step solution
We have been given an average height of 30 students which is 150 cm, one of the person’s data was incorrectly noted as 135 cm but originally it should be 165 cm.
Let us consider the sum of the heights of the 29 students is $x$, the height of the 30th student be $y$.
We know,
Mean $\left( \overline{x} \right)=\dfrac{\text{sum of the total terms}}{\text{number of terms}}$
$\overline{x}=\dfrac{x+y}{30}$
First, let us take the $y$ as 135 cm and $\overline{x}$ as 150 cm, hence the value of $x$ will be
$150=\dfrac{x+135}{30}$
Now, let us multiply by 30 on both the sides of the equation, we get
$\begin{align}
  & 150\times 30=\dfrac{x+135}{30}\times 30 \\
 &\Rightarrow 4500=x+135 \\
\end{align}$
Subtract by 135 on both the sides of the equation, we get
$4500 – 135 = x + 135 – 1354$
$\Rightarrow 4365 = x$
Therefore, the sum of the heights of 29 students is 4365 cm.
Let us use the mean formula one more time but this time let us take the height of the $30^{th}$ person as 165 cm.
$\begin{align}
  & \overline{x}=\dfrac{4365+165}{30} \\
 & =\dfrac{4530}{30} \\
 & =151
\end{align}$
Hence, the corrected mean of the heights of 30 students is 151 cm.

Note: Mean is the arithmetic average of the data given. Mean is also used to calculate the variance and standard deviation of the data in statistics. The mean is affected by extremely high or low values, called outliers.