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Zorluk: ZorMean, Median, and Mode

A class of 20 students took a 10-point mathematics quiz. The frequency table below records the quiz scores achieved by 18 of the students:

ScoreFrequency
63
75
84
94
102

The scores of the remaining 2 students were recorded later. If the score of every student is an integer from 0 to 10, inclusive, and adding the 2 missing scores causes the median score of the entire class of 20 students to be 8 and the arithmetic mean score to be an integer, what is the score of the higher-scoring student among the 2 remaining students?

  1. A
    6
  2. B
    7
  3. C
    8
  4. D
    9
  5. 10Cevap

Cevap

10
The sum of the 18 known scores is 141. For the overall mean of 20 scores to be an integer, the total sum of all 20 scores must be a multiple of 20. Since each score is at most 10, the maximum possible total sum is 141+10+10=161141 + 10 + 10 = 161. The only multiple of 20 between 141 and 161 is 160, requiring the sum of the two missing scores to be 160−141=19160 - 141 = 19. The only valid integer scores bounded by 10 that sum to 19 are 9 and 10. Adding scores of 9 and 10 places the 10th and 11th ordered values at 8, giving a median of 8. Thus, the higher missing score is 10.

Adım Adım Çözüm

1
Calculate the total sum and count of the 18 known student scores.
Known count = 3+5+4+4+2=183 + 5 + 4 + 4 + 2 = 18 students. Known sum = (6×3)+(7×5)+(8×4)+(9×4)+(10×2)=18+35+32+36+20=141(6 \times 3) + (7 \times 5) + (8 \times 4) + (9 \times 4) + (10 \times 2) = 18 + 35 + 32 + 36 + 20 = 141.
Establishing baseline sum and count is essential before analyzing missing values.
2
Set up the equation for the total sum of all 20 student scores and apply the integer mean condition.
Let the missing scores be aa and bb with 0≤a≤b≤100 \le a \le b \le 10. The total sum for 20 students is S20=141+a+bS_{20} = 141 + a + b. The arithmetic mean is 141+a+b20\frac{141 + a + b}{20}.
Since the mean must be an integer, 141+a+b141 + a + b must be a multiple of 20.
3
Determine the required sum of the two missing scores a+ba + b.
Since 0≤a≤100 \le a \le 10 and 0≤b≤100 \le b \le 10, we have 0≤a+b≤200 \le a + b \le 20. The range for S20S_{20} is [141,161][141, 161]. The only multiple of 20 in this range is 160. Thus, 141+a+b=160  ⟹  a+b=19141 + a + b = 160 \implies a + b = 19.
160 is the unique multiple of 20 reachable given score bounds.
4
Find the unique integer pair (a,b)(a, b) satisfying a+b=19a + b = 19 with a,b≤10a, b \le 10.
Since a≤10a \le 10 and b≤10b \le 10, the only integer solution with a≤ba \le b is a=9a = 9 and b=10b = 10.
No other pair of integers between 0 and 10 sums to 19.
5
Verify that adding scores 9 and 10 maintains a median score of 8.
With 9 and 10 added, the frequencies are: Score 6 (3), Score 7 (5), Score 8 (4), Score 9 (5), Score 10 (3). The 10th and 11th values in order are both 8, so the median is 8+82=8\frac{8 + 8}{2} = 8.
Confirms the median constraint is fully satisfied.

Anahtar Kavram

Properties of Weighted Means and Median Constraints in Frequency Tables
Tahmini Süre:2m 0s
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Konular

Mean, Median, and Mode

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