Question

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On respectively, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

• A. \( \sqrt{3.86} \)
• B. 1.8
• C. \( \sqrt{3.96} \)
• D. 1.94

Answer 1

The Correct Option is C

The given mean is:

\[ \bar{x} = 10 \implies \frac{\Sigma x_i}{20} = 10. \]

Thus:

\[ \Sigma x_i = 10 \times 20 = 200. \]

When the incorrect observation (8) is replaced with the correct value (12):

\[ \Sigma x_i = 200 - 8 + 12 = 204. \]

The corrected mean is:

\[ \bar{x} = \frac{\Sigma x_i}{20} = \frac{204}{20} = 10.2. \]

The standard deviation (S.D.) is given as:

\[ \text{S.D.}^2 = \text{Variance} = 2^2 = 4. \]

From the variance formula:

\[ \frac{\Sigma x_i^2}{20} - \left(\frac{\Sigma x_i}{20}\right)^2 = 4. \]

Substitute:

\[ \frac{\Sigma x_i^2}{20} - 10^2 = 4. \] \[ \frac{\Sigma x_i^2}{20} = 104 \implies \Sigma x_i^2 = 2080. \]

After replacing 8 with 12:

\[ \Sigma x_i^2 = 2080 - 8^2 + 12^2 = 2080 - 64 + 144 = 2160. \]

The corrected variance is:

\[ \frac{\Sigma x_i^2}{20} - \left(\frac{\Sigma x_i}{20}\right)^2. \] \[ \frac{2160}{20} - (10.2)^2. \] \[ \frac{\Sigma x_i^2}{20} = 108, \quad (10.2)^2 = 104.04. \] \[ \text{Variance} = 108 - 104.04 = 3.96. \]

The corrected standard deviation is:

\[ \text{S.D.} = \sqrt{3.96}. \]