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Difficulty: HardMeasures of Central Tendency (Mean, Median, Mode)

A beverage manufacturing company tracks the monthly production counts (in thousands of units) for its facilities located in two separate regions. Region A consists of 66 facilities with monthly production counts of 1212, 1515, 1515, 1818, 2020, and 2828. Region B consists of 44 facilities with monthly production counts of 1010, 1414, 2222, and 3434.

Which of the following statements must be true regarding the individual regional datasets and the combined dataset of all 1010 facilities? Select all such statements.

  1. The arithmetic mean of the combined dataset of all 1010 facilities is greater than the arithmetic mean of Region A.Answer
  2. The median of the combined dataset of all 1010 facilities is equal to the median of Region A.Answer
  3. C
    The median of the combined dataset of all 1010 facilities is equal to the arithmetic mean of the median of Region A and the median of Region B.
  4. D
    The mode of the combined dataset of all 1010 facilities is greater than the mode of Region A.
  5. E
    The arithmetic mean of the combined dataset of all 1010 facilities is equal to the simple average of the mean of Region A and the mean of Region B.

Answer

The correct statements are the one asserting that the combined arithmetic mean is greater than Region A's mean, and the one asserting that the combined median is equal to Region A's median.
The mean of the combined dataset (18.818.8) is greater than the mean of Region A (18.018.0), making the statement comparing the combined mean to Region A's mean correct. Additionally, both Region A's median and the combined dataset's median evaluate to 16.516.5, making the statement asserting equality between these two medians correct.

Step-by-Step Solution

1
Calculate the sum, mean, and median for Region A.
Region A sum =12+15+15+18+20+28=108= 12 + 15 + 15 + 18 + 20 + 28 = 108. Mean =1086=18= \frac{108}{6} = 18. Sorted values are 12,15,15,18,20,2812, 15, 15, 18, 20, 28, so Median =15+182=16.5= \frac{15 + 18}{2} = 16.5. Mode =15= 15.
Establishing base central metrics for the first subgroup.
2
Calculate the sum, mean, and median for Region B.
Region B sum =10+14+22+34=80= 10 + 14 + 22 + 34 = 80. Mean =804=20= \frac{80}{4} = 20. Sorted values are 10,14,22,3410, 14, 22, 34, so Median =14+222=18= \frac{14 + 22}{2} = 18.
Establishing base central metrics for the second subgroup.
3
Combine and order all 10 values to calculate combined metrics.
Combined ordered set: 10,12,14,15,15,18,20,22,28,3410, 12, 14, 15, 15, 18, 20, 22, 28, 34. Total sum =108+80=188= 108 + 80 = 188. Combined Mean =18810=18.8= \frac{188}{10} = 18.8. Combined Median =15+182=16.5= \frac{15 + 18}{2} = 16.5. Combined Mode =15= 15.
Necessary to evaluate combined properties accurately against subgroup properties.
4
Evaluate each candidate statement against calculated values.
Combined mean (18.818.8) > Region A mean (1818) is TRUE. Combined median (16.516.5) = Region A median (16.516.5) is TRUE. Average of medians (17.2517.25) = Combined median (16.516.5) is FALSE. Combined mode (1515) > Region A mode (1515) is FALSE. Simple average of means (1919) = Combined mean (18.818.8) is FALSE.
Determining which statements satisfy the required conditions.

Key Concept

Combining Datasets and Weighted Measures of Central Tendency
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Topics

Measures of Central Tendency (Mean, Median, Mode)

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