Question

For a group of 200 candidates, the mean and S.D were found to be 40 and 15 respectively. Later on it was found that the score 43 was misread as 34. Find the correct mean and correct S.D.

Answer 1

Hint: For solving this question, we will use following terminology:
$\overline{X}=\text{ mean}$, $\sum{x}$ is sum of all values, N is number of terms, $\sigma $ is standard deviation.
Formula for calculating the mean that will be used is $\overline{X}=\dfrac{\sum{x}}{N}$.
Formula for calculating standard deviation that will be used is $\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{N}-{{\left( \overline{X} \right)}^{2}}}$.
We will first use the mean formula to find incorrect $\sum{x}$ and correct it by adding the right term and subtracting the wrong term and then find corrected mean. After that, we will use the SD formula to find incorrect ${{\sum{x}}^{2}}$ and correct it by adding squares of right term corrected standard deviation.

Complete step-by-step answer:
Let us first use the formula of mean and find incorrect $\sum{x}$.
As we know, $\overline{X}=\dfrac{\sum{x}}{N}$.
We are given an incorrect mean as 40 and N is 200, therefore, incorrect $\sum{x}$ becomes $40=\dfrac{\sum{x}}{200}$.
Incorrect $\sum{x}=8000$.
As 8000 is the wrong value, so we will correct it by adding the correct item and subtracting the incorrect item. Hence,
Correct $\sum{x}=8000-$ incorrect item + correct item.
$\Rightarrow 8000-34+43=8009$.
Hence, we can find correct mean as:
\[\text{Correct}\overline{X}=\dfrac{\text{Correct }\sum{x}}{N}=\dfrac{8009}{200}=40.045\].
Thus, correct mean = 40.045
Now, let us use the formula of standard deviation and find incorrect ${{\sum{x}}^{2}}$ using incorrect $\overline{X}$ and incorrect $\sigma $.
\[\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{N}-{{\left( \overline{X} \right)}^{2}}}\]
Putting value of incorrect $\overline{X}=40$ we get:
\[\begin{align}
  & 15=\sqrt{\dfrac{\sum{{{x}^{2}}}}{200}-{{\left( 40 \right)}^{2}}} \\
 & \Rightarrow 15=\sqrt{\dfrac{\sum{{{x}^{2}}}}{200}-1600} \\
\end{align}\]
Squaring both sides, we get:
\[\begin{align}
  & 225=\dfrac{\sum{{{x}^{2}}}}{200}-1600 \\
 & \Rightarrow 225+1600=\dfrac{\sum{{{x}^{2}}}}{200} \\
 & \Rightarrow \sum{{{x}^{2}}}=1825\times 200 \\
 & \Rightarrow \sum{{{x}^{2}}}=365000 \\
\end{align}\]
Which is incorrect.
To find correct ${{\sum{x}}^{2}}$ we will add square of correct term and subtract square of incorrect term, we get:
\[\begin{align}
  & \text{Correct }{{\sum{x}}^{2}}=\text{Incorrect }{{\sum{x}}^{2}}-{{\left( \text{Incorrect item} \right)}^{2}}+{{\left( \text{Correct item} \right)}^{2}} \\
 & \Rightarrow 365000-{{\left( 34 \right)}^{2}}+{{\left( 43 \right)}^{2}} \\
 & \Rightarrow 365000-1156+1849 \\
 & \Rightarrow 365693 \\
\end{align}\]
Hence, $\text{Correct }{{\sum{x}}^{2}}=365693$.
Let us find the correct $\sigma $ using correct ${{\sum{x}}^{2}}$ correct $\overline{X}$ and N.
Therefore,
\[\begin{align}
  & \text{Correct }\sigma =\sqrt{\dfrac{\text{Correct}\sum{{{x}^{2}}}}{N}-{{\left( \text{Correct}\overline{X} \right)}^{2}}} \\
 & \Rightarrow \sqrt{\dfrac{365693}{200}-{{\left( 40.045 \right)}^{2}}} \\
 & \Rightarrow \sqrt{1828.465-1603.602} \\
 & \Rightarrow \sqrt{224.863} \\
 & \Rightarrow 14.99 \\
\end{align}\]
Hence, correct standard deviation = 14.99

Note: Students should not get confused between correct and incorrect items. While taking incorrect standard deviation, use incorrect mean only. Students should avoid mistakes while calculating squares and square roots. It should be noted that, N remains the same as the number of correct terms is equal to the number of incorrect terms.