Question

The mean and variance of $20$ observations are found to be $10$ and $4$, respectively. On rechecking, it was found that an observation $9$ was incorrect and the correct observation was $11$. Then the correct variance is:

• A. $4.01$
• B. $3.99$
• C. $3.98$
• D. $4.02$

Answer 1

The correct option is B

$3.99$

Explanation for the correct option:

Step 1: Write the equation for the given mean and variance.

Since, it is given that the mean of $20$ observations is $10$.

Therefore,

$$\begin{gathered} \frac{{\displaystyle \underset{i=1}{\overset{20}{\sum \left(x\right)}}}}{20}=10 \\ \Rightarrow \underset{i=1}{\overset{20}{\sum \left(x\right)}}=200 \end{gathered}$$

Also, it is given that the variance of $20$ observations is $4$.

$$\begin{gathered} \frac{{\displaystyle \underset{i=1}{\overset{20}{\sum \left(x^2\right)}}}}{20}-100=4 \\ \Rightarrow \frac{\underset{i=1}{\overset{20}{\sum \left(x^2\right)}}}{20}=104 \\ \Rightarrow \underset{i=1}{\overset{20}{\sum \left(x^2\right)}}=2080 \end{gathered}$$

Step 2: Compute the correct variance.

Since, an observation $9$ was incorrect and the correct observation was $11$.

So, correct mean is as follows:

$\frac{200-9+11}{20}=10.1$

Therefore, the correct mean is $10.1$.

Also, Correct variance is as follows:

$$\begin{gathered} \frac{2080-{\left(9\right)}^2+{\left(11\right)}^2}{20}-{(10.1)}^2=\frac{2080-81+121}{20}-{(10.1)}^2 \\ \Rightarrow \frac{2080-{\left(9\right)}^2+{\left(11\right)}^2}{20}-{(10.1)}^2=106-102.01 \\ \Rightarrow \frac{2080-{\left(9\right)}^2+{\left(11\right)}^2}{20}-{(10.1)}^2=3.99 \end{gathered}$$

Therefore, the correct variance is $3.99$.

Hence, option $B$ is the correct answer.