If 38 percent of the departments in a certain research organization have 14 or fewer members each, what is the...

Understanding the Question

We need to find the median number of members per department in a research organization.

Given Information:

• 38% of departments have 14 or fewer members

What We Need:
The median is the middle value when all departments are arranged by size. Since the median represents the 50th percentile, we need to identify which department size corresponds to this position.

Key Insight: Since only 38% of departments have \(\leq 14\) members, and the median occurs at the 50th percentile, we can immediately deduce that the median must be \(> 14\). To find the exact value, we need information about the distribution of department sizes around the 50th percentile mark.

Analyzing Statement 1

Statement 1: 58 percent of the departments have 15 or fewer members each.

Let's see what this reveals:

• From the question: 38% have \(\leq 14\) members
• From Statement 1: 58% have \(\leq 15\) members

This means exactly \(58\% - 38\% = 20\%\) of departments have exactly 15 members.

Now we can map out the complete distribution:

• 0-38th percentile: departments with \(\leq 14\) members
• 38th-58th percentile: departments with exactly 15 members
• 58th-100th percentile: departments with \(\geq 16\) members

Since the median falls at the 50th percentile, and this percentile lies within the range where all departments have exactly 15 members (38th-58th percentile), the median = 15.

[STOP - Sufficient!]

Statement 1 is sufficient.

This eliminates choices B, C, and E.

Analyzing Statement 2

Now let's forget Statement 1 completely and analyze Statement 2 independently.

Statement 2: 42 percent of the departments have 16 or more members each.

Let's work through what this reveals:

• From the question: 38% have \(\leq 14\) members
• From Statement 2: 42% have \(\geq 16\) members

Since all percentages must total 100%, we have:
\(38\% + \text{(percentage with 15 members)} + 42\% = 100\%\)

Therefore, exactly 20% of departments have 15 members.

This gives us the same distribution as before:

• 0-38th percentile: departments with \(\leq 14\) members
• 38th-58th percentile: departments with exactly 15 members
• 58th-100th percentile: departments with \(\geq 16\) members

Again, the 50th percentile falls in the range where all departments have exactly 15 members, so the median = 15.

[STOP - Sufficient!]

Statement 2 is sufficient.

The Answer: D

Both statements independently lead us to the same conclusion that the median is 15. Each statement reveals that exactly 20% of departments have 15 members, and the 50th percentile (median) falls squarely within this group.

Answer Choice D: Each statement alone is sufficient.