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After a 16-member team at Company A saw a certain software demonstration, each of the team members gave the software exactly one of the ratings 1, 2, 3, 4, or 5, where greater numbers indicate higher ratings. The team's manager intended to record the 16 ratings but accidentally recorded one team member's rating twice. The ratings she recorded show two ratings of 1, two ratings of 2, two ratings of 3, two ratings of 4, and nine ratings of 5, for a total of 17 ratings.
The manager determined that the extra rating was \(\mathrm{P}\), and after this extra rating was removed, the average (arithmetic mean) of the remaining 16 ratings was exactly \(\mathrm{Q}\). Select for P and for Q the options that create a statement that is consistent with the information provided.
1
2
3
4
5
Let's create a table showing the recorded ratings:
| Rating | Count Recorded |
| 1 | 2 |
| 2 | 2 |
| 3 | 2 |
| 4 | 2 |
| 5 | 9 |
| Total | 17 |
Key insight: One rating was recorded twice, so we have 17 entries instead of 16.
We need to find:
Let's calculate the sum of all 17 recorded ratings:
After removing rating P, we'll have 16 ratings with sum = \(65 - \mathrm{P}\)
The average Q = \(\frac{65 - \mathrm{P}}{16}\)
Since Q must be one of our answer choices (1, 2, 3, 4, or 5), let's check each possibility:
If P = 1:
If P = 2:
If P = 3:
If P = 4:
If P = 5:
Stop here - we found our answer.
If P = 1 (the extra rating), after removal we have:
Sum = \(1 + 2(2) + 2(3) + 2(4) + 9(5) = 1 + 4 + 6 + 8 + 45 = 64\)
Average = \(\frac{64}{16} = 4\) ✓
The only combination that produces a valid average from our answer choices is when the extra rating of 1 is removed, resulting in an average of exactly 4.