Math Problem Statement

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?

Solution

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=100n = 100.
  • Corrected mean: Xˉcorrected=84.8\bar{X}_{\text{corrected}} = 84.8.
  • Previous mean: Xˉprevious=85\bar{X}_{\text{previous}} = 85.
  • Previous variance: σprevious2=16\sigma_{\text{previous}}^2 = 16.

Step 1: Calculate the sum of the values before correction

The formula for mean is:

Mean=Xn\text{Mean} = \frac{\sum X}{n}

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

Xprevious=Xˉprevious×n=85×100=8500\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:

Xcorrected=8500120+100=8480\sum X_{\text{corrected}} = 8500 - 120 + 100 = 8480

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

Xˉcorrected=Xcorrectedn=8480100=84.8\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:

σ2=(XiXˉ)2n\sigma^2 = \frac{\sum (X_i - \bar{X})^2}{n}

We can also write it as:

σ2=Xi2nXˉ2\sigma^2 = \frac{\sum X_i^2}{n} - \bar{X}^2

Where Xi2\sum X_i^2 is the sum of the squares of the individual wages.

From the previous variance, we have:

σprevious2=Xprevious2100852=16\sigma_{\text{previous}}^2 = \frac{\sum X_{\text{previous}}^2}{100} - 85^2 = 16

So,

Xprevious2100=16+852=16+7225=7241\frac{\sum X_{\text{previous}}^2}{100} = 16 + 85^2 = 16 + 7225 = 7241

Thus,

Xprevious2=7241×100=724100\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 X2\sum X^2 as follows:

The wrong wage was recorded as 120, so 1202=14400120^2 = 14400. The correct wage is 100, and 1002=10000100^2 = 10000.

The corrected sum of squares is:

Xcorrected2=72410014400+10000=719700\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:

σcorrected2=Xcorrected2100(84.8)2\sigma_{\text{corrected}}^2 = \frac{\sum X_{\text{corrected}}^2}{100} - (84.8)^2

σcorrected2=71970010084.82=71977193.04=3.96\sigma_{\text{corrected}}^2 = \frac{719700}{100} - 84.8^2 = 7197 - 7193.04 = 3.96

Corrected Variance: 3.963.96

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.

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Math Problem Analysis

Mathematical Concepts

Statistics
Variance
Mean

Formulas

Mean = (Sum of values) / (Number of values)
Variance = (Sum of squares of deviations) / (Number of values)
Corrected sum = Previous sum - (incorrect value)^2 + (correct value)^2

Theorems

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Suitable Grade Level

Grades 11-12