The mean and variance of 20 observations are found to be 10 and 4 respectively. On rechecking, it was found that an observation 8 is incorrect. If the wrong observation is omitted, then the correct variance is

Options

A. 7
B. `(100)/(16)`
C. `(1400)/(361)`
D. `(1440)/(361)`

Answer

To solve the problem, we need to find the correct variance after omitting the incorrect observation of 8 from the original set of 20 observations. The original mean and variance are given as 10 and 4, respectively.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Number of observations (n) = 20
- Mean (μ) = 10
- Variance (σ²) = 4
- Incorrect observation = 8

2. **Calculate the Total Sum of Observations:**
The mean is calculated as:
\[ \text{Mean} = \frac{\text{Total Sum}}{n} \]
Therefore, the total sum of observations (S) can be calculated as:
\[ S = \text{Mean} \times n = 10 \times 20 = 200 \]

3. **Calculate the Sum of Squares of Observations:**
The variance is calculated as:
\[ \text{Variance} = \frac{\text{Sum of Squares}}{n} - \left(\frac{\text{Total Sum}}{n}\right)^2 \]
Rearranging gives:
\[ \text{Sum of Squares} = n \times \text{Variance} + \left(\text{Total Sum}\right)^2/n \]
Substituting the known values:
\[ \text{Sum of Squares} = 20 \times 4 + \frac{200^2}{20} \]
\[ = 80 + \frac{40000}{20} = 80 + 2000 = 2080 \]

4. **Adjust for the Incorrect Observation:**
Since the observation 8 is incorrect, we need to remove it from the total sum and the sum of squares:
- New total sum after removing 8:
\[ S' = S - 8 = 200 - 8 = 192 \]
- New sum of squares after removing \(8^2 = 64\):
\[ \text{Sum of Squares}' = \text{Sum of Squares} - 64 = 2080 - 64 = 2016 \]

5. **Calculate the New Number of Observations:**
After omitting one observation, the new number of observations (n') is:
\[ n' = 20 - 1 = 19 \]

6. **Calculate the New Mean:**
The new mean (μ') is:
\[ \mu' = \frac{S'}{n'} = \frac{192}{19} \]

7. **Calculate the New Variance:**
The new variance (σ²') is given by:
\[ \sigma'^2 = \frac{\text{Sum of Squares}'}{n'} - \left(\mu'\right)^2 \]
Substituting the values:
\[ \sigma'^2 = \frac{2016}{19} - \left(\frac{192}{19}\right)^2 \]
- Calculate \(\left(\frac{192}{19}\right)^2\):
\[ \left(\frac{192}{19}\right)^2 = \frac{36864}{361} \]
- Now calculate \(\frac{2016}{19}\):
\[ \frac{2016}{19} = \frac{38304}{361} \]
- Therefore, the new variance becomes:
\[ \sigma'^2 = \frac{38304}{361} - \frac{36864}{361} = \frac{1440}{361} \]

8. **Final Answer:**
The correct variance after omitting the wrong observation is:
\[ \sigma'^2 = \frac{1440}{361} \]