Analyzing Sample Data Correction

Let's analyze how correcting an observation in a sample affects its mean and median. We are given a sample of 5 observations with an initial mean of 32 and a median of 33. An observation was incorrectly recorded as 40 instead of the correct value 35.

Calculating the Impact on the Mean

The mean of a sample is calculated as the sum of all observations divided by the number of observations.

Given:

• Number of observations (n) = 5
• Original Mean = 32

The original sum of observations can be calculated using the formula:

\(\text{Sum} = \text{Mean} \times n\)

\(\text{Original Sum} = 32 \times 5 = 160\)

Now, we need to correct the sum of observations. The incorrect value 40 needs to be removed, and the correct value 35 needs to be added.

\(\text{Corrected Sum} = \text{Original Sum} - \text{Incorrect Value} + \text{Correct Value}\)

\(\text{Corrected Sum} = 160 - 40 + 35\)

\(\text{Corrected Sum} = 160 - 5 = 155\)

The number of observations remains 5. The new mean is calculated using the corrected sum:

\(\text{New Mean} = \frac{\text{Corrected Sum}}{n}\)

\(\text{New Mean} = \frac{155}{5} = 31\)

The original mean was 32, and the new mean is 31. Therefore, the mean will decrease.

Analyzing the Impact on the Median

The median is the middle value in a dataset when it is ordered from least to greatest. For a sample of 5 observations (an odd number), the median is the \(\frac{n+1}{2}\)th observation.

\(\text{Median Position} = \frac{5+1}{2} = \frac{6}{2} = 3\text{rd observation}\)

The original median is given as 33. This means that when the original 5 observations were sorted, the 3rd observation was 33.

Let the original sorted observations be \(y_1, y_2, y_3, y_4, y_5\), where \(y_1 \le y_2 \le y_3 \le y_4 \le y_5\). We know \(y_3 = 33\).

The incorrect value was 40, and the correct value is 35.

Consider the position of the incorrect value 40 relative to the median 33. Since \(40 > 33\), the value 40 must have been one of the observations greater than or equal to the median (i.e., it could be \(y_3\), \(y_4\), or \(y_5\)). However, if 40 was \(y_3\), the median would be 40, which contradicts the given median of 33. So, 40 must have been either \(y_4\) or \(y_5\).

• If 40 was \(y_4\), the original sorted list was \(y_1, y_2, 33, 40, y_5\) with \(y_1 \le y_2 \le 33 \le 40 \le y_5\). Replacing 40 with 35 gives the values \(y_1, y_2, 33, 35, y_5\). Since \(33 \le 35\), and assuming \(35 \le y_5\), the new sorted list is \(y_1, y_2, 33, 35, y_5\). The 3rd observation is still 33.
• If 40 was \(y_5\), the original sorted list was \(y_1, y_2, 33, y_4, 40\) with \(y_1 \le y_2 \le 33 \le y_4 \le 40\). Replacing 40 with 35 gives the values \(y_1, y_2, 33, y_4, 35\). Since \(33 \le y_4 \le 40\), the new value 35 will be positioned relative to \(y_4\). If \(y_4 > 35\), the new sorted list is \(y_1, y_2, 33, 35, y_4\). If \(y_4 \le 35\), the new sorted list is \(y_1, y_2, 33, y_4, 35\). In both cases, the 3rd observation is still 33.

Since the incorrect value 40 (which was greater than the median 33) is replaced by 35 (which is also greater than or equal to the median 33 but smaller than 40), and the median itself (33) is not the value being corrected, the relative order of the 3rd observation (which is 33) is preserved. The 3rd observation in the sorted list remains 33. Therefore, the median remains the same.

Conclusion on Mean and Median Change

Based on our calculations and analysis:

• The mean decreases from 32 to 31.
• The median remains the same at 33.

This matches the statement that the median remains the same but the mean will decrease.

Revision Table: Key Statistical Measures

Additional Information: Measures of Central Tendency

Mean, median, and mode are common measures of central tendency used to describe the center point of a dataset.

• Mean: Good for symmetrical data without outliers. It uses all data points.
• Median: Good for skewed data or data with outliers, as it is not affected by extreme values.
• Mode: The most frequent value. Useful for categorical or discrete data.

In this problem, correcting an outlier-like value (40) to a less extreme value (35) shows how the mean is directly impacted by the change in the sum, while the median's position-based definition makes it more robust to such changes, especially if the median value itself is not the one being corrected and the change doesn't cross the median's position.