The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking, it was found that an observation 8 was incorrect. Find the new standard deviation if wrong item is omitted.

Options

A. 1.99
B. 2.76
C. 2.99
D. 3.46

Answer

To solve the problem step by step, we will follow the procedure outlined in the video transcript.

### Step 1: Understand the given data
We have:
- Mean (x̄) = 10
- Standard Deviation (σ) = 2
- Number of observations (n) = 20
- Incorrect observation = 8

### Step 2: Calculate the total sum of observations
The total sum of observations (Σxi) can be calculated using the formula:
\[ \text{Sum of observations} = \text{Mean} \times \text{Number of observations} \]
\[ \Sigma xi = 10 \times 20 = 200 \]

### Step 3: Correct the total sum of observations
Since the observation 8 is incorrect, we need to subtract it from the total sum:
\[ \text{Corrected Sum} = \Sigma xi - 8 = 200 - 8 = 192 \]

### Step 4: Calculate the new mean
Now that we have 19 observations left (after omitting the incorrect observation), we can find the new mean:
\[ \text{New Mean} = \frac{\text{Corrected Sum}}{\text{Number of observations left}} = \frac{192}{19} \approx 10.1053 \]

### Step 5: Calculate the original sum of squares
Using the standard deviation formula:
\[ \sigma^2 = \frac{1}{n} \left( \Sigma xi^2 - \bar{x}^2 \cdot n \right) \]
We can rearrange it to find Σxi²:
\[ \sigma^2 \cdot n + \bar{x}^2 \cdot n = \Sigma xi^2 \]
Substituting the known values:
\[ 2^2 \cdot 20 + 10^2 \cdot 20 = \Sigma xi^2 \]
\[ 4 \cdot 20 + 100 \cdot 20 = \Sigma xi^2 \]
\[ 80 + 2000 = \Sigma xi^2 \]
\[ \Sigma xi^2 = 2080 \]

### Step 6: Correct the sum of squares
Now we need to correct this sum of squares by removing the square of the incorrect observation:
\[ \text{Corrected } \Sigma xi^2 = 2080 - 8^2 = 2080 - 64 = 2016 \]

### Step 7: Calculate the new standard deviation
Now we can calculate the new standard deviation using the corrected values:
\[ \sigma' = \sqrt{\frac{1}{n-1} \left( \Sigma xi^2 - \bar{x}'^2 \cdot (n-1) \right)} \]
Where \( n - 1 = 19 \) and \( \bar{x}' \approx 10.1053 \):
\[ \sigma' = \sqrt{\frac{1}{19} \left( 2016 - (10.1053)^2 \cdot 19 \right)} \]
Calculating \( (10.1053)^2 \):
\[ (10.1053)^2 \approx 102.11 \]
Now substituting:
\[ \sigma' = \sqrt{\frac{1}{19} \left( 2016 - 102.11 \cdot 19 \right)} \]
Calculating \( 102.11 \cdot 19 \):
\[ 102.11 \cdot 19 \approx 1940.09 \]
Now substituting back:
\[ \sigma' = \sqrt{\frac{1}{19} \left( 2016 - 1940.09 \right)} = \sqrt{\frac{75.91}{19}} \approx \sqrt{3.996} \approx 1.99 \]

### Final Answer
The new standard deviation is approximately **2**.