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Difficulty: Very hardMeasures of Central Tendency (Mean, Median, Mode)

Group A consists of 1515 test scores with an arithmetic mean of 8080 and a median of 8282. Group B consists of 2525 test scores with an arithmetic mean of 9090 and a median of 8686. The two groups are combined into a single dataset of 4040 test scores. Which of the following statements MUST be true about the combined dataset? Indicate all such statements.

  1. The mean of the combined dataset is equal to 86.2586.25.Answer
  2. The median of the combined dataset is at least 8282.Answer
  3. The median of the combined dataset is at most 8686.Answer
  4. D
    The median of the combined dataset is equal to 8484.
  5. E
    The mode of the combined dataset must lie between 8080 and 9090.

Answer

The statements asserting that the combined mean is 86.2586.25, that the combined median is at least 8282, and that the combined median is at most 8686 must all be true.
The combined mean is calculated directly from the total sum divided by total elements, giving exactly 86.2586.25. For the median of 4040 combined values (the average of the 20th20\text{th} and 21st21\text{st} values in sorted order), counting how many values are guaranteed to be above or below specific thresholds shows that at least 2121 values are ≥82\ge 82 and at least 2121 values are ≤86\le 86. Therefore, the median must lie within [82,86][82, 86].

Step-by-Step Solution

1
Calculate the weighted arithmetic mean for the combined dataset.
Combined Mean = 15×80+25×9015+25=1200+225040=345040=86.25\frac{15 \times 80 + 25 \times 90}{15 + 25} = \frac{1200 + 2250}{40} = \frac{3450}{40} = 86.25.
The mean of combined sets is the total sum of all values divided by the total number of items.
2
Analyze the lower bound for the median of the combined set.
At least 2121 values in the combined set of 4040 items are greater than or equal to 8282.
Since the median of Group A (1515 items) is 8282, at least 88 items are ≥82\ge 82. Since the median of Group B (2525 items) is 8686, at least 1313 items are ≥86≥82\ge 86 \ge 82. In total, 8+13=218 + 13 = 21 items are ≥82\ge 82. In a sorted list of 4040 items, having at least 2121 items ≥82\ge 82 implies that the 20th20\text{th} and 21st21\text{st} items must be ≥82\ge 82, so the median ≥82\ge 82.
3
Analyze the upper bound for the median of the combined set.
At least 2121 values in the combined set of 4040 items are less than or equal to 8686.
Group A has at least 88 items ≤82≤86\le 82 \le 86. Group B has at least 1313 items ≤86\le 86. Thus, at least 8+13=218 + 13 = 21 items are ≤86\le 86. In a sorted list of 4040 items, having at least 2121 items ≤86\le 86 ensures the 20th20\text{th} and 21st21\text{st} items are ≤86\le 86, so the median ≤86\le 86.
4
Evaluate the remaining statements regarding the exact median and mode.
Neither the exact median value of 8484 nor the existence/location of a mode is guaranteed.
Medians do not combine linearly like means, and mode depends entirely on specific frequency counts which are not given.

Key Concept

Weighted averages and order statistic bounds for combined datasets
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Topics

Measures of Central Tendency (Mean, Median, Mode)

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