So I recently got back an exam score and I found myself wondering if I got the lowest score in the class...this led me to formulate an interesting problem:

Suppose an exam is scored from $0$ to $100$. What is the lowest possible score in the class given the number of students $N$, the class average $\mu$, standard deviation $\sigma$ and your grade $x$? What is the highest possible grade?

Obviously for large enough $N$, the lowest possible grade is $0$ and the highest is $100$, but what are the limits for general $N$?

  • 1
    $\begingroup$ This doesn't really make any sense. Even if there was one student, the minimum score would be 0, and the maximum would be 100. Knowing the distribution would only give you an indication of how likely it was to score less than, e.g. 10 out of 100. $\endgroup$
    – mardat
    Commented Dec 5, 2014 at 7:01
  • $\begingroup$ No I am not assuming that the grades have a certain distribution. I am interested in the min/max grades over all possible distributions that satisfy the mean and standard deviation criterion. $\endgroup$ Commented Dec 5, 2014 at 7:03
  • $\begingroup$ @mardat, we are given mean and standard deviation, so if there was only 1 student, then that student must have scored $\mu$. If there are $>2$ students, then the score distribution must satisfy both $\mu$ and $\sigma$ to be valid. In that case, what is the lowest possible and highest possible score? $\endgroup$ Commented Dec 5, 2014 at 7:07
  • $\begingroup$ Okay in that case just misunderstood your question. I'll give an answer below. $\endgroup$
    – mardat
    Commented Dec 5, 2014 at 7:48
  • $\begingroup$ Nope, mardat is right, you misunderstand the meaning of $\mu$. $\mu$ is the population mean. If there's only one student, the lowest possible score is 0. But the lowest most probable score is $\mu$. If you have $N$ students, the answer to the first question is still 0. But the answer to the second is different and can be found in this article. $\endgroup$ Commented Dec 5, 2014 at 8:06

1 Answer 1


If we call the test scores $y_i$, then we'll begin by noting that in order to find the lowest (or highest) score, that $y_1$ can be our lowest (or highest) score, and $y_2=y_3=...=y_n$ must be all the other scores. Since we know that

$$\frac{1}{n}(y_1+\sum_2^n y_i) = \mu$$



we can replace our sums to get $\frac{1}{n}(y_1+(n-1)y_2)=\mu$ and $\frac{1}{n}(y_1^2+(n-1)y_2^2)-\mu^2=\sigma^2$. Since we've now got two equations in two variables, we can just solve to find the values for $y_1$ and $y_2$, to get (skipping the intermediary math):




to get either the minimum score possible, or the maximum score possible.

Edit: To clarify, this is the lowest score possible given $\mu$ and $\sigma$ for the population. It does not necessarily mean that it was the actual lowest score (and in fact it's highly unlikely that it was the lowest score). It just means that given those two parameters, it's not possible for anyone to achieve a lower score.

  • $\begingroup$ Thanks! But two questions: 1)Note that we know my score, $x$, as well. How does that affect the bounds? 2) How do you ensure that the scores fall into the $0$ and $100$ range? For example, when $y_1 = \mu - \sigma\sqrt{n-1} > 0$ is the lowest score, how do you ensure that all other scores $y_2,...,y_n = \mu + \frac{\sigma}{\sqrt{n-1}}$ are $<100$? If this bound is greater than $100$ then I think the answer for $y_1$ would look different. $\endgroup$ Commented Dec 5, 2014 at 8:38
  • $\begingroup$ The easiest way to do it is to just find a $\mu_1$ and $sigma_1$ that have removed the effects of $x$. For $\mu_1$, this is quite easy - $\mu=\frac{(n-1)\mu_1+x}{n}$, which gives $\mu_1=\frac{n\mu-x}{n-1}$. Using a similar strategy for $\sigma$ gives $\sigma_1=\sqrt{\frac{n}{n-1}(\sigma^2+\mu^2-\frac{x^2}{n})-\mu_1^2}$. We then find y_1 and y_2 based off those values (and using $n-2$ instead of $n-1$). $\endgroup$
    – mardat
    Commented Dec 5, 2014 at 9:53
  • $\begingroup$ As far as ensuring the scores fall between 0 and 100; I haven't bothered to deal with that case because it's quite trivial. If we found that $y_2$ was greater than 100, than we can simply find $y_1=n\mu-(n-1)y_2$ but we substitute $y_2=100$, to get $y_1=n\mu-100(n-1)$. $\endgroup$
    – mardat
    Commented Dec 5, 2014 at 10:03

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