# Exercise 3.6Z: Examination Correction

In an exam at the TU Munich $1000$ students participated. From the grade "4.0" upwards up to "1.0" the exam is considered to be passed. The exam regulations provide for the following grades:

- $$1.0, \ 1.3, \ 1.7, \ 2.0, \ 2.3, \ 2.7, \ 3.0, \ 3.3, \ 3.7, \ 4.0, \ 4.3, \ 4.7, \ 5.0.$$

Further, the exercise must take into account:

- The maximum achievable score is $100$. The best student achieved $88$ points.
- Due to the relatively large number of participants, the achieved score – this is the random variable $z$ – has with good approximation a Gaussian distribution with mean $m_z = 60$ and standard deviation $\sigma_z = 10$.
- In the correction,&nbs

p; not only whole scores were assigned, but also (arbitrary) intermediate values, so that the random variable $z$ can be taken as "continuous valued" with good approximation.

For scoring, the guidelines given are:

- Even with six points less than the best $($so from $82$ points$)$ one shall get "1.0".
- If one reaches $46\%$ of the total score, one has passed the exam.
- The points/grades assignment shall be linear.

Hints:

- The exercise belongs to the chapter Gaussian distributed random variables.
- The "Gaussian distribution" is often called "normal distribution". This is not quite correct:
- A normally distributed random variable $z$ does have a Gaussian PDF and CDF, but always with mean $m_z = 0$ and standard deviation $\sigma_z = 1$.

### Questions

### Solution

**(1)**Correct are

__the solutions 1 and 3__:

- According to the central limit theorem, a Gaussian distribution is obtained for the sum of many independent quantities.
- Conversely, if there are only a few dependent tasks, there is no Gaussian distribution.
- A single yes/no question leads to a two-point distribution ($0$ points or maximum number of points).
- Even if these imperatives are followed, a Gaussian distribution will not be expected for very few participants.

**(2)** One gets a "1.0" with $82$ points or more.

- Therefore, with the mean $m_z = 60$ and the standard deviation $\sigma_z = 10$:

- $$\rm Pr(\it z\ge \rm 82)=\rm Q\Bigg(\frac{\rm 82-60}{\rm 10}\Bigg)=\rm Q(\rm 2.2) \hspace{0.15cm}{=\rm 0.0139}.$$

- For a thousand participants, it follows $N_\text{1.0}\hspace{0.15cm}\underline{= 14}$.

**(3)** With less than $46$ points, one has failed the exam:

- $$\rm Pr(\it z<\rm 46)=\rm Pr(\it z \le \rm 46)=\rm \phi\Bigg(\frac{\rm 46-60}{\rm 10}\Bigg)=\rm \phi(\rm -1.4)=\rm Q(\rm 1.4)=\rm 0.0807.$$

- So
__81 students have to compete again__.

**(4)** The difference in points $82 - 46 = 36$ must be divided among nine grade intervals $(1.3$, ... , $4.0)$.

- Each interval thus comprises $4$ points.
- For example, one receives a grade of "3.0" if one has $58$ to $62$ points.
- The probability that the grade is in this range is given by

- $$\rm Pr(\rm 58 <\it z<\rm 62)=\rm \phi\Bigg(\frac{\rm 62-60}{\rm 10}\Bigg)-\rm \phi\Bigg(\frac{\rm 58-60}{\rm 10}\Bigg).$$

- Taking advantage of the symmetry, one obtains:

- $$\rm Pr(\rm 58 <\it z<\rm 62) = \rm \phi(\rm 0.2)-\rm \phi(\rm -0.2) = \rm 0.5792-\rm 0.4207=0.1587\hspace{0.2cm}\hspace{0.15cm}\underline{(159 \hspace{0.1cm}\rm participants)}.$$

Notes:

- $z$ is to be taken as a continuous valued random variable.
- Therefore, the score $62$ is simultaneously the upper bound for the grade "3.0" as well as the lower bound for the grade "2.7."
- If $z$ were only an integer, then $62$ would have to be assigned to either the "2.7" grade or the "3.0" grade, depending on the mood of the corrector.
- Of course, this would have to be done in the same way for all examinees.

**(5)** Analogous to the solution of the subtask **(4)** applies to the grade "2.7":

- $$\rm Pr(\rm 62 <\it z<\rm 66)=\rm \phi(\rm 0.6)-\rm \phi(\rm 0.2)=\rm 0.7257-\rm 0.5792=0.1465.$$

- For reasons of symmetry, the same value is obtained for the grade "3.3":

- $$\rm Pr(\rm 54 <\it z<\rm 58)=\rm \phi(-\rm 0.2)-\rm \phi(-\rm 0.6)= \rm Q(\rm 0.2)-\rm Q(\rm 0.6)=\rm 0.1465.$$

- So
__each 146 participants receive a grade of "2.7" or "3.3"__.

**(6)** With the points/grade assignment made here, not only the points are distributed around $m_z = 60$ symmetrically, but also the scores around "3.0". There are

- exactly as many "2.7"s as "3.3"s $($around $±0.3$ away from $3.0$ $)$,
- exactly as many "2.3 "s as "3.7"s $(3.0 ±0.7)$, and
- exactly as many "1.0 "s as "5.0 "s.

Therefore, the $\rm mean\hspace{0.15cm} grade\hspace{0.15cm}\underline{ 3.0}$ results.