Question

Mean of 100 items is 49. It was discovered that three items which should have been $60,70,80$ were wrongly read as $40,20,50$ respectively. The correct mean is
A. 48
B. 49
C. 50
D. 80

Answer 1

Hint: We first find the general form of mean for those n observations. We subtract the wrong inputs from the summation and add the right ones. We find the actual sum of the numbers. Then we find the correct mean of the numbers.

Complete step-by-step solution:
Let a set of n observations be ${{x}_{i}},i=1\left( 1 \right)n$. The mean of the set of the observations will be \[\overline{x}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}\].
It is given that three items which should have been $60,70,80$ were wrongly read as $40,20,50$ respectively.
Therefore, we first find the sum of those numbers including those which have been considered wrong.
Putting the values \[\overline{x}=49,n=100\] in the equation \[\overline{x}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}\], we get
\[\begin{align}
  & 49=\dfrac{1}{100}\sum\limits_{i=1}^{n}{{{x}_{i}}} \\
 & \Rightarrow \sum\limits_{i=1}^{n}{{{x}_{i}}}=4900 \\
\end{align}\].
Now the numbers $40,20,50$ that have been added in the summation of 4900. We first delete them to find the sum of 97 numbers as $4900-40-20-50=4790$.
Now we add the actual numbers which should have been added to 4790 $60,70,80$.
So, the actual sum of 100 numbers should have been $4790+60+70+80=5000$.
The correct mean of the numbers will be \[\overline{x}=\dfrac{1}{100}\sum\limits_{i=1}^{n}{{{x}_{i}}}=\dfrac{5000}{100}=50\].
The correct option is C.

Note: We also could have adjusted the numbers for wrong and right inputs. We are omitting the numbers $40,20,50$ and adding $60,70,80$ which means the adjusted values are
$60+70+80-40-20-50=100$,.
We are adding 100 to the previous sum to find the correct mean. This will increase the mean by $\dfrac{100}{100}=1$ which gives the corrected mean as $49+1=50$.