The wages of employees were wrongly noted as 120 instead of 100.The number of employees are 100 and the corrected mean 84.8. the previous variance and mean were respectively  16 and 85 what is the corrected variance?

To find the corrected variance, let's break down the problem step by step.

### Given:
- The wrong wage was recorded as 120 instead of 100 for one employee.
- Total number of employees, \( n = 100 \).
- Corrected mean: \( \bar{X}_{\text{corrected}} = 84.8 \).
- Previous mean: \( \bar{X}_{\text{previous}} = 85 \).
- Previous variance: \( \sigma_{\text{previous}}^2 = 16 \).

### Step 1: Calculate the sum of the values before correction
The formula for mean is:

\[
\text{Mean} = \frac{\sum X}{n}
\]

So, the sum of wages using the previous mean was:

\[
\sum X_{\text{previous}} = \bar{X}_{\text{previous}} \times n = 85 \times 100 = 8500
\]

### Step 2: Calculate the corrected sum of the values
Since one wage was incorrectly recorded as 120 instead of 100, the correct sum of wages is:

\[
\sum X_{\text{corrected}} = 8500 - 120 + 100 = 8480
\]

Now, check if this is consistent with the corrected mean:

\[
\bar{X}_{\text{corrected}} = \frac{\sum X_{\text{corrected}}}{n} = \frac{8480}{100} = 84.8
\]

This matches the given corrected mean.

### Step 3: Use the variance formula

The formula for variance is:

\[
\sigma^2 = \frac{\sum (X_i - \bar{X})^2}{n}
\]

We can also write it as:

\[
\sigma^2 = \frac{\sum X_i^2}{n} - \bar{X}^2
\]

Where \( \sum X_i^2 \) is the sum of the squares of the individual wages.

From the previous variance, we have:

\[
\sigma_{\text{previous}}^2 = \frac{\sum X_{\text{previous}}^2}{100} - 85^2 = 16
\]

So,

\[
\frac{\sum X_{\text{previous}}^2}{100} = 16 + 85^2 = 16 + 7225 = 7241
\]

Thus,

\[
\sum X_{\text{previous}}^2 = 7241 \times 100 = 724100
\]

### Step 4: Correct the sum of squares
Since one wage was wrongly noted, we need to adjust \( \sum X^2 \) as follows:

The wrong wage was recorded as 120, so \( 120^2 = 14400 \). The correct wage is 100, and \( 100^2 = 10000 \).

The corrected sum of squares is:

\[
\sum X_{\text{corrected}}^2 = 724100 - 14400 + 10000 = 719700
\]

### Step 5: Calculate the corrected variance
Now we use the corrected sum of squares to calculate the corrected variance:

\[
\sigma_{\text{corrected}}^2 = \frac{\sum X_{\text{corrected}}^2}{100} - (84.8)^2
\]

\[
\sigma_{\text{corrected}}^2 = \frac{719700}{100} - 84.8^2 = 7197 - 7193.04 = 3.96
\]

### Corrected Variance: \( 3.96 \)

Would you like any further details or have additional questions?

Here are 5 questions you could explore:
1. How does changing a single data point affect the mean?
2. Why is variance important in understanding data spread?
3. How does variance differ from standard deviation?
4. What is the impact of incorrect data recording in statistical analysis?
5. How do you calculate variance for grouped data?

**Tip:** Always double-check the accuracy of recorded data, especially outliers, as they can significantly affect statistical measures like mean and variance.

The wages of employees were wrongly noted as 120 instead of 100. The number of employees are 100 and the corrected mean 84.8. The previous variance and mean were respectively 16 and 85. What is the corrected variance?

Learn how to correct the variance of employee wages when an incorrect value is recorded. This statistical problem covers key concepts such as mean, variance, and correcting the sum of squares, with a step-by-step solution. Suitable for Grades 11-12.

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