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I have information of the order in which students were classified in regard to their scores in a SAT test. I know the distribution of scores for each student is uniform with support [a,b]. I also know that:

  • 17 students scored less than 3
  • 13 students scored between 3 and 5
  • 58 students scored more than 5

How can I form reasonable estimates for a and b?


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Please do not crosspost in the future. It is impolite and strongly discouraged. – cardinal Feb 3 '13 at 3:48
up vote 1 down vote accepted

Assume that $u$ students scored in $(x-r,x)$, $v$ students scored in $(x,x+s)$ and $w$ students scored in $(x+s,x+s+t)$, where $(u,v,w,x,s)$ are known and $(r,t)$ are unknown. The likelihood is $$ L(r,t)={u+v+w\choose u,v,w}\left(\frac{r}{r+s+t}\right)^u\left(\frac{s}{r+s+t}\right)^v\left(\frac{t}{r+s+t}\right)^w. $$ Thus, $L(r,t)$ is maximal when the partial derivatives of $L$ with respect to $r$ and $t$ are zero, that is, when $$ \frac{u}{r}=\frac{u+v+w}{r+s+t}=\frac{v}{t}. $$ Solving these yields $$ r=s\frac{u}v,\qquad t=s\frac{w}v, $$ a result which common sense could suggest directly. In the notations of the problem, $a=x-r$ and $b=x+s+t$ hence $a=3-\frac{17}{13}\cdot2$ and $b=5+\frac{58}{13}\cdot2$.

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Thanks, Did! But I don't understand the meaning of the first term you have in the lieklihood. I also don't get why you raise each term of the likelihood to the power of u, v, and w. Could you clarify that, please? – user1172558 Feb 2 '13 at 23:51
The multinomial factor accounts for the number of ways of assigning tags less than 3 or between 3 and 5 or more than 5 to the students. Once every student is tagged, the remaining product is the probability that the score of each agrees with their tag. Here is an analogy: you throw a coin twice with probabilities $h$ and $t$ to get heads and tails. Then the probability to get one head and one tail is $2ht$ (that is, $ht$ for heads then tails + $th$ for tails then heads). – Did Feb 3 '13 at 0:02
OK! But is there any easier way to solve this problem? I was trying to approach it just using the mean of a uniform distribution (a+b)/2. Any suggestion? – user1172558 Feb 3 '13 at 0:21

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