In an inter-state Mathematics Olympiad, the distribution of the scores

Question

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

Show Answer

Official Answer

D

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

Suhasi · Answer

chetan2u  can you please help with this one's explanation? I got the right answer but I'm not sure of my approach.

I assumed the total set to be approx. 500. Since 68% lies within 1 SD, so that's approximately 350 students. According to the question, the number of students who scored less than Ricky are > 350, so he definitely scored more than the mean score and more than 1 SD.

Since 5% of the students (approx 25) had a score of more than 2 SD, and because 70 students scored more than Ricky, so he definitely didn't score more than 2 SD.

Therefore, his score must lie between m+d and m+2d.

My approach might be faulty so please feel free to correct me!

ermohit86 · Answer

Total students = 70+1+428 = 499 (lets assume 500)
Ricky's score is greater than 428 which is = 68% (1 SD) +14% (left) +2% (left) = 84% of Total = 420
m+2d = 14% of 500 = 70
So Ricky's score must lie between (m+d) and (m+2d).

MT1302 · Answer

Bunuel  ,  chetan2u  ,  KarishmaB  ,  ScottTargetTestPrep   CrackVerbal 
Kindly throw some light on this question.

PandCduo · Answer

Assume total at 500
428 students less than Ricky
So 428/500 i.e. approx 85.6% which is greater than 84% (m+d)
And 70/500 i.e. 14% So ricky's score must be between 84% to 86% which is between m+d and m+2d.

chetan2u · Answer

Standard Deviation is used to measure the dispersion around the mean.

A normal curve has 68% around the mean. It is well dispersed and therefore 34% on either side. => m to m+d has 34% and m to m-d has remaining 34%.
Similarly 95% within m+2d means  \frac{95-68}{2}  or 13% on either side. => m+d to m+2d has 13% and m-d to m-2d has remaining 13%.

Now, let us see where does Ricky stand. 
428….Ricky……70, so total 499, and he is the 71 st. => 100 * 70/499 ~ 7000/500 = 14%

Thus, Ricky is ahead of 86% students or behind nearly 14% of total.
 The distr of SD is
….(2.5%)….. m-2d …..(13%)…… m-d …….34%……. m ………34%…… m+d ……..13%….. m+2d ……2.5%…..

Thus, 14% from right lies in m+d to m+2d zone.

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In an inter-state Mathematics Olympiad, the distribution of the scores

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