Aufgaben:Exercise 3.6Z: Examination Correction - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^3.5 Gaussian Random Variable^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.5 Gaussian Random Variable^]]
	[[de:~~Zusatzaufgaben~~:3.~~6_Prüfungskorrektur~~]]		[[de:Aufgaben:Aufgabe 3.6Z: Prüfungskorrektur]]

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	'''(2)'''  One gets a  "1.0"  with  $82$  points or more.		'''(2)'''  One gets a  "1.0"  with  $82$  points or more.
	*Therefore,  with the mean  $m_z = 60$  and the standard deviation  $\sigma_z = 10$:		*Therefore,  with the mean  $m_z = 60$  and the standard deviation  $\sigma_z = 10$:
	:$$\rm Pr(\it z\ge \rm 82)=\rm Q\Bigg(\frac{\rm 82-60}{\rm 10}\Bigg)=\rm Q(\rm 2.2)		:$$\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}.$$
	\hspace{0.15cm}{=\rm 0.0139}.$$
	*For a thousand participants,  it follows  $N_\text{1.0}\hspace{0.15cm}\underline{= 14}$.		*For a thousand participants,  it follows  $N_\text{1.0}\hspace{0.15cm}\underline{= 14}$.

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	[[Category:Theory of Stochastic Signals: Exercises\|^3.5 Gaussian Random Variable^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.5 Gaussian Random Variable^]]
			[[de:Zusatzaufgaben:3.6_Prüfungskorrektur]]

Guenter at 15:05, 17 February 2022

2022-02-17T15:05:35Z

← Older revision		Revision as of 17:05, 17 February 2022
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	Further,  the exercise must take into account:		Further,  the exercise must take into account:
	*The maximum achievable score is  $100$.  The best student achieved  $88$  points.		*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  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Probability_density_function_.26_cumulative_density_function\|Gaussian distribution]]  with mean  $m_z = 60$  and standard deviation  ~~(standard deviation) ~~ $\sigma_z = 10$.		*Due to the relatively large number of participants,  the achieved score – this is the random variable  $z$  – has with good approximation a  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Probability_density_function_.26_cumulative_density_function\|Gaussian distribution]]  with mean  $m_z = 60$  and standard deviation  $\sigma_z = 10$.
	*In the correction,&~~nbsp~~; 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.		*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.

Bene: Text replacement - "rms value" to "standard deviation"

2022-02-17T11:11:39Z

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


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	'''(2)'''  One gets a  "1.0"  with  $82$  points or more.		'''(2)'''  One gets a  "1.0"  with  $82$  points or more.
	*Therefore,  with the mean  $m_z = 60$  and the ~~rms value~~  $\sigma_z = 10$:		*Therefore,  with the mean  $m_z = 60$  and the standard deviation  $\sigma_z = 10$:
	:$$\rm Pr(\it z\ge \rm 82)=\rm Q\Bigg(\frac{\rm 82-60}{\rm 10}\Bigg)=\rm Q(\rm 2.2)		:$$\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}.$$		\hspace{0.15cm}{=\rm 0.0139}.$$

Bene: Text replacement - "root mean square" to "standard deviation"

2022-02-17T11:00:51Z

Text replacement - "root mean square" to "standard deviation"

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	Further,  the exercise must take into account:		Further,  the exercise must take into account:
	*The maximum achievable score is  $100$.  The best student achieved  $88$  points.		*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  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Probability_density_function_.26_cumulative_density_function\|Gaussian distribution]]  with mean  $m_z = 60$  and ~~root mean square~~  (standard deviation)  $\sigma_z = 10$.		*Due to the relatively large number of participants,  the achieved score – this is the random variable  $z$  – has with good approximation a  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Probability_density_function_.26_cumulative_density_function\|Gaussian distribution]]  with mean  $m_z = 60$  and standard deviation  (standard deviation)  $\sigma_z = 10$.
	*In the correction,  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.		*In the correction,  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.

Guenter at 15:03, 1 February 2022

2022-02-01T15:03:33Z

← Older revision		Revision as of 17:03, 1 February 2022
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Correct are <u>the solutions 1 and 3</u>:		'''(1)'''  Correct are  <u>the solutions 1 and 3</u>:
	*According to the central limit theorem, a Gaussian distribution is obtained for the sum of many independent quantities.		*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.		*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).		*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.		*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.		'''(2)'''  One gets a  "1.0"  with  $82$  points or more.
	*Therefore, with the mean  $m_z = 60$  and the rms  $\sigma_z = 10$:		*Therefore,  with the mean  $m_z = 60$  and the rms value  $\sigma_z = 10$:
	:$$\rm Pr(\it z\ge \rm 82)=\rm Q\Bigg(\frac{\rm 82-60}{\rm 10}\Bigg)=\rm Q(\rm 2.2)		:$$\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}.$$		\hspace{0.15cm}{=\rm 0.0139}.$$
	*For a thousand participants, it follows  $N_\text{1.0}\hspace{0.15cm}\underline{= 14}$.		*For a thousand participants,  it follows  $N_\text{1.0}\hspace{0.15cm}\underline{= 14}$.


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	'''(3)'''  With less than  $46$  points, one has failed the exam:		'''(3)'''  With less than  $46$  points, one has failed the exam:
	:$$\rm Pr(\it z<\rm 46)=\rm Pr(\it z \le \rm 46)=\rm \phi\Bigg(\frac{\rm 46-60}{\rm 10}\Bigg)=\rm \phi(\rm -1.4)=\rm Q(\rm 1.4)=\rm 0.0807.$$		:$$\rm Pr(\it z<\rm 46)=\rm Pr(\it z \le \rm 46)=\rm \phi\Bigg(\frac{\rm 46-60}{\rm 10}\Bigg)=\rm \phi(\rm -1.4)=\rm Q(\rm 1.4)=\rm 0.0807.$$
	*So <u>81 students have to compete again</u>.		*So  <u>81 students have to compete again</u>.



	'''(4)'''  The difference in points  $82 - 46 = 36$  must be divided among nine grade intervals  $(1.3$, ... , $4.0)$~~ ~~ .		'''(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.		*Each interval thus comprises  $4$  points.
	*For example, one receives a grade of "3.0" if one  $58$  to  $62$  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.		*The probability that the grade is in this range is given by
	:$$\rm Pr(\rm 58 <\it z<\rm 62)=\rm \phi\Bigg(\frac{\rm 62-60}{\rm 10}\Bigg)-\rm \phi\Bigg(\frac{\rm 58-60}{\rm 10}\Bigg).$$		:$$\rm Pr(\rm 58 <\it z<\rm 62)=\rm \phi\Bigg(\frac{\rm 62-60}{\rm 10}\Bigg)-\rm \phi\Bigg(\frac{\rm 58-60}{\rm 10}\Bigg).$$

	Taking advantage of the symmetry, one obtains:		*Taking advantage of the symmetry,  one obtains:
	:$$\rm Pr(\rm 58 <\it z<\rm 62) = \rm \phi(\rm 0.2)-\rm \phi(\rm -0.2) = \rm 0.5792-\rm 0.4207=0.1587\hspace{0.2cm}\hspace{0.15cm}\underline{(159 \hspace{0.1cm}\rm participants)}.$$		:$$\rm Pr(\rm 58 <\it z<\rm 62) = \rm \phi(\rm 0.2)-\rm \phi(\rm -0.2) = \rm 0.5792-\rm 0.4207=0.1587\hspace{0.2cm}\hspace{0.15cm}\underline{(159 \hspace{0.1cm}\rm participants)}.$$

	''Notes:''		Notes:
	*$z$  is to be taken as a continuous random variable. Therefore, the score  $62$  is simultaneously the upper bound for the "3.0"&~~ndash~~;~~range~~ as well as the lower bound for the grade "2.7."		*$z$  is to be taken as a continuous valued random variable.
	*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.		*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 ~~sample~~ solution of the subtask  '''(4)'''  applies to the grade "2.7":		'''(5)'''  Analogous to the solution of the subtask  '''(4)'''  applies to the grade  "2.7":
	:$$\rm Pr(\rm 62 <\it z<\rm 66)=\rm \phi(\rm 0.6)-\rm \phi(\rm 0.2)=\rm 0.7257-\rm 0.5792=0.1465.$$		:$$\rm Pr(\rm 62 <\it z<\rm 66)=\rm \phi(\rm 0.6)-\rm \phi(\rm 0.2)=\rm 0.7257-\rm 0.5792=0.1465.$$

	*For reasons of symmetry, the same value is obtained for the grade "3.3":		*For reasons of symmetry,  the same value is obtained for the grade  "3.3":
	:$$\rm Pr(\rm 54 <\it z<\rm 58)=\rm \phi(-\rm 0.2)-\rm \phi(-\rm 0.6)= \rm Q(\rm 0.2)-\rm Q(\rm 0.6)=\rm 0.1465.$$		:$$\rm Pr(\rm 54 <\it z<\rm 58)=\rm \phi(-\rm 0.2)-\rm \phi(-\rm 0.6)= \rm Q(\rm 0.2)-\rm Q(\rm 0.6)=\rm 0.1465.$$

	*So <u>each 146 participants receive a grade of "2.7" or "3.3"</u>.		*So  <u>each 146 participants receive a grade of  "2.7"  or  "3.3"</u>.



	'''(6)'''  With the points~~–scores–~~assignment made here, not only ~~are~~ the points distributed around  $m_z = 60$  symmetrically, but also the scores around  "3.0".  There are		'''(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  "2.7"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  "2.3 "s  as  "3.7"s  $(3.0 ±0.7)$, and
	*exactly as many "1.0 "s as "5.0 "s.		*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.		Therefore,  the  $\rm mean\hspace{0.15cm} grade\hspace{0.15cm}\underline{ 3.0}$  results.

	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 14:50, 1 February 2022

2022-02-01T14:50:26Z

← Older revision		Revision as of 16:50, 1 February 2022
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	}}		}}

	[[File:P_ID148__Sto_Z_3_6.png\|right\|frame\|~~table~~ for  ${\rm \phi}(x)$  and  ${\rm Q}(x)$]]		[[File:P_ID148__Sto_Z_3_6.png\|right\|frame\|Table for  ${\rm \phi}(x)$  and  ${\rm Q}(x)$]]
	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:		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:
	:$$1.0, \ 1.3, \ 1.7, \ 2.0, \ 2.3, \ 2.7, \ 3.0, \ 3.3, \ 3.7, \ 4.0, \ 4.3, \ 4.7, \ 5.0.$$		:$$1.0, \ 1.3, \ 1.7, \ 2.0, \ 2.3, \ 2.7, \ 3.0, \ 3.3, \ 3.7, \ 4.0, \ 4.3, \ 4.7, \ 5.0.$$
	Further, the exercise must take into account:		Further,  the exercise must take into account:
	*The maximum achievable score is  $100$.  The best student achieved  $88$  points.		*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  $z$  – with good approximation a  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Probability_density_function_.26_cumulative_density_function\|Gaussian distribution]]  with mean  $m_z = 60$  and ~~rms~~ (standard deviation)  $\sigma_z = 10$.		*Due to the relatively large number of participants,  the achieved score – this is the random variable  $z$  – has with good approximation a  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Probability_density_function_.26_cumulative_density_function\|Gaussian distribution]]  with mean  $m_z = 60$  and root mean square  (standard deviation)  $\sigma_z = 10$.
	*In the correction, not only whole scores were assigned, but also (arbitrary) intermediate values, so that the random variable  $z$  can be taken as "continuous" with good approximation.		*In the correction,  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:		For scoring,  the guidelines given are:
	*Even with six points less than the best  $($so from  $82$  points$)$  one shall get "1.0".		*Even with six points less than the best  $($so from  $82$  points$)$  one shall get  "1.0".
	*If one  $46\%$  of the total score, one has passed the exam.		*If one reaches  $46\%$  of the total score,  one has passed the exam.
	*The points/grades assignment shall be linear.		*The points/grades assignment shall be linear.

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	Hints:		Hints:
	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables\|Gaussian distributed random variables]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables\|Gaussian distributed random variables]].
			*The  "Gaussian distribution"  is often called  "normal distribution".  This is not quite correct:
	*~~Especially in the school curriculum, 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 rms value  $\sigma_z = 1$.
	*A normally distributed random variable  $z$  does have a Gaussian PDF and CDF,  but always with mean  $m_z = 0$  and rms  $\sigma_z = 1$.


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	<quiz display=simple>		<quiz display=simple>
	{What criteria should be considered in problem creation so that the grades will result in "approximately a normal distribution"?		{What criteria should be considered in problem creation so that the grades will result in  "approximately a normal distribution"?
	\|type="[]"}		\|type="[]"}
	+ There are many exam participants.		+ There are many exam participants.
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	{How many examinees are expected to score "1.0"?		{How many examinees are expected to score  "1.0"?
	\|type="{}"}		\|type="{}"}
	$N_\text{1.0} \ = \ $ { 14 3% }		$N_\text{1.0} \ = \ $ { 14 3% }


	{How many examinees are likely to fail the exam?  Take into account that  $z$  can be taken as a continuous random variable.		{How many examinees are likely to fail the exam?  Take into account that  $z$  can be taken as a continuous valued random variable.
	\|type="{}"}		\|type="{}"}
	$N_\text{4.3 ... 5.0} \ = \ $ { 81 3% }		$N_\text{4.3 ... 5.0} \ = \ $ { 81 3% }


	{Specify the points/grades~~–~~assignment.  At what point do you get a "3.0"?  How many examinees will get this grade?		{Specify the points/grades assignment.  At what point do you get a  "3.0"?  How many examinees will get this grade?
	\|type="{}"}		\|type="{}"}
	$N_\text{3.0} \ = \ $ { 159 3% }		$N_\text{3.0} \ = \ $ { 159 3% }


	{How many examinees are expected to receive the grade "2.7"?  Justify why exactly that many examinees will receive the grade "3.3".		{How many examinees are expected to receive the grade "2.7"?  Justify why exactly that many examinees will receive the grade  "3.3".
	\|type="{}"}		\|type="{}"}
	$N_\text{2.7} \ = \ $ { 146 3% }		$N_\text{2.7} \ = \ $ { 146 3% }

Noah at 23:40, 1 January 2022

2022-01-01T23:40:59Z

Show changes

Noah at 19:41, 1 January 2022

2022-01-01T19:41:02Z

← Older revision		Revision as of 21:41, 1 January 2022
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	{{quiz-Header\|Buchseite=~~Stochastische Signaltheorie~~/~~Gaußverteilte Zufallsgröße~~		{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables
	}}		}}

	[[File:P_ID148__Sto_Z_3_6.png\|right\|frame\|~~Tabelle für~~  ${\rm \phi}(x)$  ~~und~~  ${\rm Q}(x)$]]		[[File:P_ID148__Sto_Z_3_6.png\|right\|frame\|table for  ${\rm \phi}(x)$  and  ${\rm Q}(x)$]]
	~~An einer Prüfung~~ an ~~der~~ TU ~~München haben~~  $1000$  ~~Studentinnen und Studenten teilgenommen~~.  ~~Ab der Note~~ "4.0" ~~gilt die Prüfung als bestanden~~.  ~~Die Prüfungsordnung sieht folgende Noten vor~~:		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:
	:$$1.0, \ 1.3, \ 1.7, \ 2.0, \ 2.3, \ 2.7, \ 3.0, \ 3.3, \ 3.7, \ 4.0, \ 4.3, \ 4.7, \ 5.0.$$		:$$1.0, \ 1.3, \ 1.7, \ 2.0, \ 2.3, \ 2.7, \ 3.0, \ 3.3, \ 3.7, \ 4.0, \ 4.3, \ 4.7, \ 5.0.$$
	~~Weiter ist bei der Aufgabe zu berücksichtigen~~:		Further, the exercise must take into account:
	*~~Die maximal erreichbare Punktzahl beträgt~~  $100$.  ~~Der beste Student erreichte~~  $88$  ~~Punkte~~.		*The maximum achievable score is  $100$.  The best student achieved  $88$  points.
	*~~Aufgrund der relativ großen Teilnehmerzahl ergibt sich für die erreichte Punktzahl~~ – ~~dies sei die Zufallsgröße~~  $z$  – ~~mit guter Näherung eine~~  [[Theory_of_Stochastic_Signals/~~Gaußverteilte_Zufallsgröße~~#~~Wahrscheinlichkeitsdichte-_und_Verteilungsfunktion~~\|~~Gaußverteilung~~]]  ~~mit Mittelwert~~  $m_z = 60$  ~~und Streuung~~ (~~Standardabweichung~~)  $\sigma_z = 10$.		*Due to the relatively large number of participants, the achieved score – this is the random variable  $z$  – with good approximation a  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Probability_density_function_.26_cumulative_density_function\|Gaussian_Distribution]]  with mean  $m_z = 60$  and rms (standard deviation)  $\sigma_z = 10$.
	*~~Bei der Korrektur wurden nicht nur ganze Punktzahlen vergeben~~, ~~sondern auch~~ (~~beliebige~~) ~~Zwischenwerte~~, so ~~dass man die Zufallsgröße~~  $z$  ~~mit guter Näherung als~~ "~~kontinuierlich~~" ~~auffassen kann~~.		*In the correction, not only whole scores were assigned, but also (arbitrary) intermediate values, so that the random variable  $z$  can be taken as "continuous" with good approximation.


	~~Für die Bewertung werden als Richtlinien vorgegeben~~:		For scoring, the guidelines given are:
	*~~Auch mit sechs Punkten weniger als der Beste~~  $($~~also ab~~  $82$  ~~Punkten~~$)$  ~~soll man~~ "1.0" ~~bekommen~~.		*Even with six points less than the best  $($so from  $82$  points$)$  one shall get "1.0".
	*~~Hat man~~  $46\%$  ~~der Gesamtpunktzahl erreicht~~, ~~so hat man die Prüfung bestanden~~.		*If one  $46\%$  of the total score, one has passed the exam.
	*~~Die Punkte~~/~~Noten-Zuordnung soll~~ linear ~~erfolgen~~.		*The points/grades assignment shall be linear.


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	~~''Hinweise~~:''		Hints:
	*~~Die Aufgabe gehört zum Kapitel~~  [[Theory_of_Stochastic_Signals/~~Gaußverteilte_Zufallsgröße~~\|~~Gaußverteilte Zufallsgrößen~~]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables\|Gaussian distributed random variables]].

	*~~Gerade im Schulbereich wird die~~  "~~Gaußverteilung~~"  ~~oft als~~  "~~Normalverteilung~~"  ~~bezeichnet.  Dies ist nicht ganz korrekt~~:		*Especially in the school curriculum, the  "Gaussian distribution"  is often called  "normal distribution"  This is not quite correct:
	*~~Eine normalverteilte Zufallsgröße~~  $z$  ~~hat zwar eine Gaußsche WDF und VTF~~,  ~~jedoch stets mit Mittelwert~~  $m_z = 0$  ~~und Streuung~~  $\sigma_z = 1$.		*A normally distributed random variable  $z$  does have a Gaussian PDF and CDF,  but always with mean  $m_z = 0$  and rms  $\sigma_z = 1$.


	===~~Fragebogen~~===		===Questions===

	<quiz display=simple>		<quiz display=simple>
	{~~Welche Kriterien sind bei der Aufgabenerstellung zu beachten, damit die Punktezahl~~ "~~etwa eine Normalverteilung~~" ~~ergeben wird~~?		{What criteria should be considered in problem creation so that the score will result in "approximately a normal distribution"?
	\|type="[]"}		\|type="[]"}
	+ ~~Es gibt viele Prüfungsteilnehmer~~.		+ There are many exam participants.
	- ~~Die Teilaufgaben hängen in starkem Maße voneinander ab~~.		- The subproblems depend on each other to a large extent.
	+ ~~Es gibt viele unabhängige Aufgaben~~.		+ There are many independent problems.
	- ~~Die Prüfung besteht aus einer einzigen Frage mit Ja~~/~~Nein-Antwort~~.		- The exam consists of a single yes/no question.


	{~~Wieviele Teilnehmer werden voraussichtlich mit „1~~.~~0“ abschließen~~?		{How many examinees are expected to score "1.0"?
	\|type="{}"}		\|type="{}"}
	$N_\text{1.0} \ = \ $ { 14 3% }		$N_\text{1.0} \ = \ $ { 14 3% }


	{~~Wieviele Teilnehmer werden die Prüfung wohl nicht bestehen~~?  ~~Berücksichtigen Sie, dass man~~  $z$  ~~als kontinuierliche Zufallsgröße auffassen kann~~.		{How many examinees are likely to fail the exam?  Take into account that  $z$  can be taken as a continuous random variable.
	\|type="{}"}		\|type="{}"}
	$N_\text{4.3 ... 5.0} \ = \ $ { 81 3% }		$N_\text{4.3 ... 5.0} \ = \ $ { 81 3% }


	{~~Legen Sie die Punkte~~/~~Noten~~–~~Zuordnung fest~~.  ~~Ab wann bekommt man eine „3~~.0“?  ~~Wieviele Prüfungsteilnehmer werden diese Note erhalten~~?		{Specify the points/grades–assignment.  At what point do you get a "3.0"?  How many examinees will get this grade?
	\|type="{}"}		\|type="{}"}
	$N_\text{3.0} \ = \ $ { 159 3% }		$N_\text{3.0} \ = \ $ { 159 3% }


	{~~Wieviele Teilnehmer erhalten voraussichtlich die Note „2~~.7“?  ~~Begründen Sie, warum genau so viele Prüflinge die Note „3~~.~~3“ bekommen werden~~.		{How many examinees are expected to receive the grade "2.7"?  Justify why exactly that many examinees will receive the grade "3.3".
	\|type="{}"}		\|type="{}"}
	$N_\text{2.7} \ = \ $ { 146 3% }		$N_\text{2.7} \ = \ $ { 146 3% }


	{~~Welche Mittelnote wird sich bei dieser Prüfung ergeben~~?  ~~Berücksichtigen Sie zur Lösung dieser Teilaufgabe das Ergebnis von Teilaufgabe~~  '''(5)'''.		{What will be the mean grade on this exam?  Consider the result of subtask  '''(5)''' to solve this subtask.
	\|type="{}"}		\|type="{}"}
	$\rm ~~Mittelnote~~ \ = \ $ { 3 3% }		$\rm mean grade \ = \ $ { 3 3% }


	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Richtig sind <u>die Lösungsvorschläge 1 und 3</u>:		'''(1)'''  Richtig sind <u>die Lösungsvorschläge 1 und 3</u>: