An examination is marked out of $100$. It is taken by a large number of candidates. The mean mark, for all candidates, is $72.1$ and the standard deviation is $15.2$. Give a reason why a normal distribution, with this mean and standard deviation, would not give a good approximation to the distribution of marks.

My answer: Since the standard deviation is quite large ($=15.2)$, the normal curve will disperse wildly. Hence, it is not a good approximation.

Is my answer acceptable?

  • $\begingroup$ The maximum of $100$ is about $1.836$ standard deviation units from the mean. A not insignificant part of a normal is more than $1.836$ standard deviation units above the mean. $\endgroup$ – André Nicolas Jul 7 '16 at 4:00
  • $\begingroup$ @AndréNicolas: Can elaborate further? I don't quite get your answer. $\endgroup$ – Idonknow Jul 7 '16 at 4:06
  • $\begingroup$ One possible reason is that the grader flipped a coin and assigned a score of $72.1 - 15.2$ or $72.1 + 15.2$ with equal probability. (You didn't specify that it had to be a good reason :-) $\endgroup$ – Bungo Jul 7 '16 at 4:40

The maximum possible mark of $100$ is about $1.836$ standard deviation units above the mean.

There is a probability of about $3.4\%$ that a normally distributed random variable is more than $1.836$ standard deviation units above the mean. So at the upper end of the range at least, the normal with the given mean and standard deviation gives a poor fit.

Remark: Without additional detail, your proposed answer is not sufficient. If the mean were something like $55$, with a standard deviation of $15.2$, that would be compatible with the normal providing a reasonably good fit.

  • $\begingroup$ May I know how you obtain the figure $1.836$? $\endgroup$ – Idonknow Jul 8 '16 at 1:42
  • $\begingroup$ @Idonknow: $\frac{100-72.1}{15.2}\approx 1.8355$. $\endgroup$ – André Nicolas Jul 8 '16 at 1:55
  • $\begingroup$ How to obtain the figure $3.4\%$? $\endgroup$ – Idonknow Jul 8 '16 at 2:39
  • $\begingroup$ @Idonknow: Go to the normal table. There you will find that $\Pr(Z\le 1.8356)\approx 0.966$. So the probability $Z$ is $\gt 1.8356$ is about $1-0.966$, that is, $0.034$. $\endgroup$ – André Nicolas Jul 8 '16 at 2:45

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