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In an inter-state Mathematics Olympiad, the distribution of the scores obtained by the participating students is symmetric about
the mean m. 68 percent of the distribution lies within one standard deviation d of the mean and 95 percent of the distribution lies
within 2 standard deviations of the mean. If there were 70 students who scored more than Ricky, 428 students who scored less than Ricky and none that scored equal to him, his score must lie between
  • a)
    m - 2d and m – d
  • b)
    m - d and m
  • c)
    m and m + d
  • d)
    m + d and m + 2d
  • e)
    None of the above
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
In an inter-state Mathematics Olympiad, the distribution of the scores...
Given:
  • 70 students scored more than Ricky
  • 428 students scored less than Ricky
  • None scored equal to Ricky
    • Number of students = 70 + 428 + 1 = 499
    • When arranged in ascending order, Ricky’s score would be at 429 place.
  • Mean score = m
  • Standard deviation of the scores = d
To Find:: The range in which Ricky’s score should lie
Approach:
  1. As the options are given in terms of m and d, we first need to understand the distribution curve. 
  2. The above distribution is symmetric about the mean m
    • 68% of the distribution lies within 1 standard deviation, the distribution would have   68% / 2 = 34  of data sets on either sides of the mean.
    • Similarly, 95% of the distribution lies within 2 standard deviation, the distribution would have 95 % / 2 = 47.5 % of data sets on either sides of the mean.
  3. Since we know the position of Ricky’s score and we know the number of students who appeared in the Olympiad, we can calculate the number of students between consecutive standard deviations of the mean score.
Working out:
  1. Since, 499 students appeared in the Olympiad, the mean score would lie at position 250
    • So, number of students whose score lies within 1 standard deviation = 68% of 500 = 340
    • So, number of students whose score is between m and m+d = 340 / 2 = 170  i.e. position of scores of students who lie between m and m + d will be between 250 and (250+170) = 420
    • We need not bother with scores of students whose position is less than 250, as we are concerned about the position of student who is at 429th place.
  2. As Ricky’s score is at 429 place, his score does not lie between m and m+ d. Let’s see if he lies between m + d and m + 2d.
  3. Number of students whose score lie within 2 standard deviations = 95% of 500 = 475
    • Number of students whose score lie between m and m + 2d = 475/2 = 237
  4. So, number of students whose score lie between m + d and m + 2d = 237– 170 = 67
  5. So, the position of students whose score lie between m + d and m + 2d will be between 420 and 420 + 67 = 487
    • As Ricky’s score lies at a position of 429, his score would lie between m + d and m + 2d
Answer : D 
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Most Upvoted Answer
In an inter-state Mathematics Olympiad, the distribution of the scores...
Given Information:
- The distribution of scores is symmetric about the mean m.
- 68 percent of the distribution lies within one standard deviation d of the mean.
- 95 percent of the distribution lies within 2 standard deviations of the mean.
- There are 70 students who scored more than Ricky.
- There are 428 students who scored less than Ricky.
- No student scored equal to Ricky.

To Find:
The range within which Ricky's score must lie.

Solution:
Let's analyze the information given step by step:

1. Symmetry of the Distribution:
Since the distribution is symmetric about the mean m, we can infer that the number of students scoring above the mean m is equal to the number of students scoring below the mean m.

2. 68 Percent within One Standard Deviation:
Given that 68 percent of the distribution lies within one standard deviation d of the mean, we can conclude that the remaining 32 percent of the distribution is outside this range. However, since the distribution is symmetric, 16 percent of the distribution lies above one standard deviation d and 16 percent below one standard deviation d.

3. 95 Percent within Two Standard Deviations:
Given that 95 percent of the distribution lies within two standard deviations of the mean, we can infer that the remaining 5 percent of the distribution is outside this range. However, since the distribution is symmetric, 2.5 percent of the distribution lies above two standard deviations and 2.5 percent lies below two standard deviations.

4. Students Scoring Above and Below Ricky:
70 students scored more than Ricky, and 428 students scored less than Ricky. Since the distribution is symmetric, we can conclude that 70 students also scored less than Ricky, and 428 students also scored more than Ricky.

Analysis:
- Since 70 students scored more than Ricky and the distribution is symmetric, 70 students also scored less than Ricky.
- Similarly, since 428 students scored less than Ricky, 428 students also scored more than Ricky.
- We can conclude that there are a total of 70 + 428 + 1 (Ricky himself) = 499 students in the distribution.

Calculating Percentiles:
- From the given information, we can infer that 16 percent of the distribution lies above one standard deviation d and 2.5 percent lies above two standard deviations.
- Using these percentages, we can calculate the number of students corresponding to each percentile.

1. Students above one standard deviation:
16 percent of 499 students = (16/100) * 499 = 79.84 ≈ 80 students
Since the distribution is symmetric, this is also the number of students below one standard deviation.

2. Students above two standard deviations:
2.5 percent of 499 students = (2.5/100) * 499 = 12.475 ≈ 12.5 students
Similarly, this is also the number of students below two standard deviations.

Conclusion:
- Ricky's score lies within one standard deviation d and two standard deviations d of the mean.
- Since there are 80 students above and below one standard deviation, Ricky's score lies between m - d and m + d.
- Since there are 12.5 students above and below two standard deviations, Ricky's score lies between m - 2d and m +
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In an inter-state Mathematics Olympiad, the distribution of the scores obtained by the participating students is symmetric aboutthe mean m. 68 percent of the distribution lies within one standard deviation d of the mean and 95 percent of the distribution lieswithin 2 standard deviations of the mean. If there were 70 students who scored more than Ricky, 428 students who scored less than Ricky and none that scored equal to him, his score must lie betweena)m - 2d and m – db)m - d and mc)m and m + dd)m + d and m + 2de)None of the aboveCorrect answer is option 'D'. Can you explain this answer?
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