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

A research study records five numerical observations: 1212, 1616, 2020, 2020, and xx, where xx is a positive integer. If the arithmetic mean of these five observations is equal to their median, which of the following values could be xx? Select all such values.

  1. 1212Answer
  2. B
    1515
  3. 1717Answer
  4. D
    2424
  5. 3232Answer

Answer

The possible values of xx are 1212, 1717, and 3232.
The mean of the dataset is given by 68+x5\frac{68+x}{5}. Depending on where xx falls relative to the sorted known values (12,16,20,2012, 16, 20, 20), the median can be 1616 (if x≤16x \le 16), xx (if 16≤x≤2016 \le x \le 20), or 2020 (if x≥20x \ge 20). Equating the mean to the median in each case yields three valid solutions: x=12x = 12, x=17x = 17, and x=32x = 32.

Step-by-Step Solution

1
Express the arithmetic mean in terms of xx.
The sum of the five numbers is 12+16+20+20+x=68+x12 + 16 + 20 + 20 + x = 68 + x. Therefore, the mean is 68+x5\frac{68 + x}{5}.
The mean of nn numbers is the total sum divided by nn.
2
Analyze Case 1 where x≤16x \le 16.
The sorted dataset is x,12,16,20,20x, 12, 16, 20, 20 (or 12,x,16,20,2012, x, 16, 20, 20). The median (3rd value) is 1616. Setting mean = median gives 68+x5=16  ⟹  68+x=80  ⟹  x=12\frac{68 + x}{5} = 16 \implies 68 + x = 80 \implies x = 12. Since 12≤1612 \le 16, x=12x = 12 is valid.
The median of a 5-element set is the 3rd element when ordered.
3
Analyze Case 2 where 16≤x≤2016 \le x \le 20.
The sorted dataset is 12,16,x,20,2012, 16, x, 20, 20. The median is xx. Setting mean = median gives 68+x5=x  ⟹  68+x=5x  ⟹  4x=68  ⟹  x=17\frac{68 + x}{5} = x \implies 68 + x = 5x \implies 4x = 68 \implies x = 17. Since 16≤17≤2016 \le 17 \le 20, x=17x = 17 is valid.
When xx lies between 16 and 20, xx itself becomes the 3rd element of the ordered set.
4
Analyze Case 3 where x≥20x \ge 20.
The sorted dataset is 12,16,20,20,x12, 16, 20, 20, x. The median is 2020. Setting mean = median gives 68+x5=20  ⟹  68+x=100  ⟹  x=32\frac{68 + x}{5} = 20 \implies 68 + x = 100 \implies x = 32. Since 32≥2032 \ge 20, x=32x = 32 is valid.
When x≥20x \ge 20, 20 is the 3rd element of the ordered set.

Key Concept

Solving for missing observations where mean equals median requires analyzing how the position of the variable affects the sorted order and the resulting median value.
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Topics

Measures of Central Tendency (Mean, Median, Mode)

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