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

A company has 8 departments. The dataset of the number of employees in these 8 departments has a median of 42, a range of 25, and a unique mode of 38, which appears exactly 3 times. If no department has more than 55 employees, what is the maximum possible arithmetic mean of the number of employees across all 8 departments?

Answer: 44.25

Answer

The maximum possible arithmetic mean of the number of employees across all 8 departments is 44.25.
To maximize the mean, the sum of the 8 department sizes must be maximized under all given constraints. Ordering the dataset as x1≤x2≤x3≤x4≤x5≤x6≤x7≤x8x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8, the median requirement gives x4+x5=84x_4 + x_5 = 84. Since no element exceeds 55 and the range is 25, x1x_1 cannot be 38 because 38+25=63>5538 + 25 = 63 > 55. Hence, the three 38s must be x2=x3=x4=38x_2 = x_3 = x_4 = 38, which forces x5=46x_5 = 46. To maximize the sum, x8x_8 is set to its maximum limit of 55, forcing x1=55−25=30x_1 = 55 - 25 = 30. Next, x7x_7 is set to 55, and x6x_6 is set to 54 so that 55 appears only twice and 38 remains the unique mode. The maximum sum is 30+38+38+38+46+54+55+55=35430 + 38 + 38 + 38 + 46 + 54 + 55 + 55 = 354, giving a maximum mean of 354/8=44.25354 / 8 = 44.25.

Step-by-Step Solution

1
Order the dataset variables and define constraints.
Let the department sizes be ordered as x1≤x2≤x3≤x4≤x5≤x6≤x7≤x8x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8.
Arranging values in ascending order allows direct analysis of median, range, and mode bounds.
2
Use the median to form an equation for the middle two elements.
x4+x52=42  ⟹  x4+x5=84\frac{x_4 + x_5}{2} = 42 \implies x_4 + x_5 = 84.
For n=8n=8 elements, the median is the arithmetic mean of the 4th and 5th terms.
3
Determine the exact position of the three occurrences of 38.
x2=x3=x4=38x_2 = x_3 = x_4 = 38, forcing x5=84−38=46x_5 = 84 - 38 = 46.
If x1=38x_1 = 38, then x8=38+25=63x_8 = 38 + 25 = 63, exceeding the upper bound of 55. Thus 38 cannot start at x1x_1, so it must occupy x2,x3,x4x_2, x_3, x_4.
4
Maximize the remaining elements x1,x6,x7,x8x_1, x_6, x_7, x_8.
x8=55x_8 = 55, x1=30x_1 = 30, x7=55x_7 = 55, and x6=54x_6 = 54.
To maximize the sum, set x8=55x_8 = 55, which fixes x1=55−25=30x_1 = 55 - 25 = 30. Set x7=55x_7 = 55. x6x_6 can be at most 54 because setting x6=55x_6 = 55 would give 55 a frequency of 3, violating the unique mode requirement.
5
Calculate the maximum sum and arithmetic mean.
Sum =30+38+38+38+46+54+55+55=354= 30 + 38 + 38 + 38 + 46 + 54 + 55 + 55 = 354; Mean =354/8=44.25= 354 / 8 = 44.25.
Dividing the maximum total sum of 354 by 8 gives the maximum possible arithmetic mean.

Key Concept

Optimization of Means Subject to Central Tendency and Range Constraints
Estimated Time:2m 30s
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Topics

Measures of Central Tendency (Mean, Median, Mode)

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