Understanding the Impact of a Data Recording Error on Average

This problem asks us to determine how correcting a recording error in one observation affects the average of a set of observations. We are given that the average of 15 observations was calculated, but one observation had an error in the tens place digit.

Analyzing the Recording Error

The error was that the digit in the tens place was recorded as 8 instead of the correct digit 3. Let's consider the effect this error has on the value of the observation:

• Incorrect tens digit: 8
• Correct tens digit: 3

The difference in the tens place value is \(8 \times 10 - 3 \times 10 = 80 - 30 = 50\). Since the recorded digit (8) is greater than the correct digit (3), the recorded value of that observation was 50 higher than its actual correct value.

Let the incorrect observation be \(x_{incorrect}\) and the correct observation be \(x_{correct}\). Then, according to the error described:

\(x_{incorrect} = x_{correct} + 50\)

Effect of the Error on the Sum of Observations

The average of a set of observations is calculated by dividing the sum of the observations by the number of observations. Let the sum of the 15 observations with the incorrect recording be \( \Sigma X_{incorrect} \). Let the sum of the other 14 correct observations be \( \Sigma X_{other} \). The incorrect sum is:

\( \Sigma X_{incorrect} = \Sigma X_{other} + x_{incorrect} \)

After correcting the observation, the sum of the observations becomes \( \Sigma X_{correct} \). This correct sum is:

\( \Sigma X_{correct} = \Sigma X_{other} + x_{correct} \)

Now, let's find the difference between the incorrect sum and the correct sum:

\( \Sigma X_{incorrect} - \Sigma X_{correct} = (\Sigma X_{other} + x_{incorrect}) - (\Sigma X_{other} + x_{correct}) \)

\( \Sigma X_{incorrect} - \Sigma X_{correct} = x_{incorrect} - x_{correct} \)

We know that \(x_{incorrect} = x_{correct} + 50\), so \(x_{incorrect} - x_{correct} = 50\). Therefore:

\( \Sigma X_{incorrect} - \Sigma X_{correct} = 50 \)

This means that the incorrect sum was 50 greater than the correct sum. Correcting the error reduces the total sum of the 15 observations by 50.

\( \Sigma X_{correct} = \Sigma X_{incorrect} - 50 \)

Calculating the Change in Average

The average is calculated as Sum / Number of observations. The number of observations is 15.

The initial (incorrect) average is \( \bar{X}_{incorrect} = \dfrac{\Sigma X_{incorrect}}{15} \).

The corrected average is \( \bar{X}_{correct} = \dfrac{\Sigma X_{correct}}{15} \).

Substitute the relationship between the sums:

\( \bar{X}_{correct} = \dfrac{\Sigma X_{incorrect} - 50}{15} \)

\( \bar{X}_{correct} = \dfrac{\Sigma X_{incorrect}}{15} - \dfrac{50}{15} \)

\( \bar{X}_{correct} = \bar{X}_{incorrect} - \dfrac{50}{15} \)

Simplifying the fraction \( \dfrac{50}{15} \):

\( \dfrac{50}{15} = \dfrac{10 \times 5}{3 \times 5} = \dfrac{10}{3} \)

So, the corrected average is:

\( \bar{X}_{correct} = \bar{X}_{incorrect} - \dfrac{10}{3} \)

This equation shows that the corrected average is less than the incorrect average by \( \dfrac{10}{3} \).

Conclusion on Average Correction

Since the recorded value was 50 higher than the actual value, the total sum of the observations was 50 higher than it should have been. When this total sum is used to calculate the average, the average will also be higher than the correct average. By correcting the observation, the total sum decreases by 50, and consequently, the average decreases by 50 divided by the number of observations (15).

The average is reduced by \( \dfrac{50}{15} = \dfrac{10}{3} \).

Revision Table: Average Calculation and Correction

Additional Information: Properties of Average

The average, or mean, is a measure of central tendency. It is sensitive to every value in the dataset, including extreme values or errors. A change in any single observation directly affects the sum, and thus the average. The impact of a change in a single observation on the average is distributed equally among all observations. If a value increases by \(k\), the sum increases by \(k\), and the average increases by \(k/n\), where \(n\) is the number of observations. If a value decreases by \(k\), the sum decreases by \(k\), and the average decreases by \(k/n\).

In this problem, the recorded value was effectively decreased by 50 to get the correct value. This decrease in the observation leads to a decrease in the total sum, and subsequently, a decrease in the average. The magnitude of the decrease in average is the total decrease in sum divided by the number of observations.