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Difficulty: HardMeasures of Central Tendency

A logistics specialist is inventorying six shipping crates. The median weight of the crates is 4545 pounds, and the mode is 4040 pounds. The weights of five of the crates are 5050, 4040, 5555, 4242, and 4040 pounds. What is the weight of the sixth crate?

  1. A
    43
  2. 48Answer
  3. C
    6
  4. D
    45

Answer

The weight of the sixth crate is 4848 pounds.
The correct answer is 4848. To find the missing weight, first arrange the known weights in ascending order: 4040, 4040, 4242, 5050, 5555. Since there are 66 crates in total, the median is the average of the 3rd and 4th values. Based on the known numbers, the 3rd value must be 4242 and the 4th value must be the unknown weight, xx (if xx were smaller or larger, the middle values would average to 4141 or 4646, not 4545). Setting up the equation: 42+x2=45\frac{42 + x}{2} = 45. Multiplying both sides by 22 gives 42+x=9042 + x = 90. Subtracting 4242 from both sides yields x=48x = 48. Adding 4848 to the set also ensures 4040 remains the mode.

Step-by-Step Solution

1
Sort the known weights in ascending order.
4040, 4040, 4242, 5050, 5555
The median depends on the ordered numerical position of values in a data set.
2
Identify how the median is calculated for 66 items.
The median is the average of the 3rd and 4th items.
Because there is an even number of crates (66), there is no single middle value. The median is the mean of the two central numbers.
3
Determine the position of the missing value, xx.
The 3rd item must be 4242, and the 4th item must be xx.
If xx were smaller than 4242, the middle two numbers would be 4040 and 4242 (median 4141). If xx were 5050 or larger, the middle two numbers would be 4242 and 5050 (median 4646). To get a median of 4545, xx must be the 4th value.
4
Set up an equation for the median and solve for xx.
42+x2=45⇒42+x=90⇒x=48\frac{42 + x}{2} = 45 \Rightarrow 42 + x = 90 \Rightarrow x = 48
The average of the 3rd (4242) and 4th (xx) numbers must equal the given median (4545).
5
Verify that the mode remains 4040.
The final set is 4040, 4040, 4242, 4848, 5050, 5555. The number 4040 appears twice and is the most frequent.
Ensures all conditions stated in the problem are met.

Key Concept

Measures of Central Tendency (Median of an even-numbered data set)

Alternative Method

You can logically deduce the answer without full algebra: The median is exactly halfway between the two middle numbers. Since the median is 4545 and the lower middle number is 4242 (a distance of 33), the upper middle number must be exactly 33 units above 4545, which is 4848.
Estimated Time:2m 0s
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Topics

Measures of Central Tendency

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