Difference between revisions of "Aufgaben:Exercise 2.5Z: Flower Meadow"
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| − | [[File:P_ID124__Sto_Z_2_5.gif|right|frame| | + | [[File:P_ID124__Sto_Z_2_5.gif|right|frame|Flower meadow – another example of the Poisson distribution]] |
A farmer is happy about the splendor of flowers on his land and wants to know how many dandelions are currently blooming on his meadow. | A farmer is happy about the splendor of flowers on his land and wants to know how many dandelions are currently blooming on his meadow. | ||
| − | *He knows that the meadow has an area of $5000$ square meters and he also knows from the agricultural school that the number of flowers in a small area is always | + | *He knows that the meadow has an area of $5000$ square meters and he also knows from the agricultural school that the number of flowers in a small area is always Poisson distributed. |
| − | *He stakes out ten squares, each with an edge length of $\text{25 cm}$ | + | *He stakes out ten squares, each with an edge length of $\text{25 cm}$, randomly distributed over the entire meadow and counts the flowers in each of these squares: |
::$$\rm 3, \ 4, \ 1, \ 5, \ 0, \ 3, \ 2, \ 4, \ 2, \ 6.$$ | ::$$\rm 3, \ 4, \ 1, \ 5, \ 0, \ 3, \ 2, \ 4, \ 2, \ 6.$$ | ||
Consider these numerical values as random results of the discrete random variable $z$. | Consider these numerical values as random results of the discrete random variable $z$. | ||
| − | It is obvious that the sample size | + | It is obvious that the sample size $(10)$ is very small but – this much is revealed – the farmer is lucky. First consider how you would proceed to solve this task, and then answer the following questions. |
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Hints: | Hints: | ||
| − | *This exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Poisson_Distribution|Poisson | + | *This exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Poisson_Distribution|Poisson Distribution]]. |
*Reference is also made to the chapter [[Theory_of_Stochastic_Signals/Moments_of_a_Discrete_Random_Variable|Moments of a Discrete Random Variable]]. | *Reference is also made to the chapter [[Theory_of_Stochastic_Signals/Moments_of_a_Discrete_Random_Variable|Moments of a Discrete Random Variable]]. | ||
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<quiz display=simple> | <quiz display=simple> | ||
| − | {Find the mean of $z$, that is, the mean number of flowers counted in the ten squares. | + | {Find the mean of $z$, that is, the mean number of flowers counted in each of the ten squares. |
|type="{}"} | |type="{}"} | ||
$m_z \ =$ { 3 3% } | $m_z \ =$ { 3 3% } | ||
| − | {Determine the | + | {Determine the standard deviation of the random variable $z$. |
|type="{}"} | |type="{}"} | ||
$\sigma_z\ = \ $ { 1.732 3% } | $\sigma_z\ = \ $ { 1.732 3% } | ||
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{Which of the following statements are true? | {Which of the following statements are true? | ||
|type="[]"} | |type="[]"} | ||
| − | + Actually, one would have to use considerably more than ten random numbers (squares) for the moment calculation. | + | + Actually, one would have to use considerably more than ten random numbers (squares) for the moment calculation. |
| − | + The random | + | + The random variable $z$ is in fact Poisson distributed. |
| − | - The rate $\lambda$ of the Poisson distribution is equal to the | + | - The rate $\lambda$ of the Poisson distribution is equal to the standard deviation $\sigma_z$. |
+ The rate $\lambda$ of the Poisson distribution is equal to the mean $m_z$. | + The rate $\lambda$ of the Poisson distribution is equal to the mean $m_z$. | ||
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</quiz> | </quiz> | ||
| − | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' The linear mean of these ten numbers gives $\underline{m_z = 3}$ | + | '''(1)''' The linear mean of these ten numbers gives |
| + | :$$\underline{m_z = 3}.$$ | ||
| − | '''(2)''' For the | + | '''(2)''' For the second moment of the random variable $z$ applies accordingly: |
:$$m_{\rm 2\it z}=\frac{1}{10}\cdot (0^2+1^2+ 2\cdot 2^2+ 2\cdot 3^2+2\cdot 4^2+ 5^2+6^2)=12.$$ | :$$m_{\rm 2\it z}=\frac{1}{10}\cdot (0^2+1^2+ 2\cdot 2^2+ 2\cdot 3^2+2\cdot 4^2+ 5^2+6^2)=12.$$ | ||
| − | *According to Steiner's theorem, the variance is | + | *According to Steiner's theorem, the variance is: |
| − | :$$\sigma_z^2 =12 -3^2 = 3$$ | + | :$$\sigma_z^2 =12 -3^2 = 3,$$ |
| − | :and | + | :and thus the standard deviation: |
:$$\underline{\sigma_z \approx 1.732}.$$ | :$$\underline{\sigma_z \approx 1.732}.$$ | ||
| − | + | '''(3)''' Correct are <u>solutions 1, 2, and 4</u>: | |
| − | '''(3)''' Correct <u>solutions 1, 2, and 4</u>: | + | *Mean and standard deviation agree here. This is indicative of the Poisson distribution with rate $\lambda = 3$ <br>(equal to the mean and equal to the variance, not equal to the standard deviation). |
| − | *Mean and | + | *Naturally, it is questionable to make this statement on the basis of only ten values. |
| − | *Naturally, it is questionable to make this statement on the basis of only ten values. However, in the case of moments, a smaller sample number is less serious than, for example, in the case of probabilities. | + | *However, in the case of moments, a smaller sample number is less serious than, for example, in the case of probabilities. |
| − | '''(4)''' In total, there are $80000$ such squares, each with three flowers in the mean. | + | '''(4)''' In total, there are $80000$ such squares, each with three flowers in the mean. |
*This suggests a total of $\underline{B = 240}$ thousand flowers. | *This suggests a total of $\underline{B = 240}$ thousand flowers. | ||
| − | '''(5)''' According to the Poisson distribution, this probability results in | + | '''(5)''' According to the Poisson distribution, this probability results in |
| − | $${\rm Pr}(z = 0) = \frac{3^0}{0!} \cdot{\rm e}^{-3}\hspace{0.15cm}\underline{\approx 5\%}.$$ | + | :$${\rm Pr}(z = 0) = \frac{3^0}{0!} \cdot{\rm e}^{-3}\hspace{0.15cm}\underline{\approx 5\%}.$$ |
| − | *However, the small sample size $N = 10$ on which this task was based would have indicated probability ${\rm Pr}(z = 0) = { 10\%}$ since only in a single square no single flower was counted. | + | *However, the small sample size $N = 10$ on which this task was based would have indicated probability ${\rm Pr}(z = 0) = { 10\%}$ <br>since only in a single square no single flower was counted. |
{{ML-Fuß}} | {{ML-Fuß}} | ||
Latest revision as of 13:20, 18 January 2023
Flower meadow – another example of the Poisson distribution
A farmer is happy about the splendor of flowers on his land and wants to know how many dandelions are currently blooming on his meadow.
- He knows that the meadow has an area of $5000$ square meters and he also knows from the agricultural school that the number of flowers in a small area is always Poisson distributed.
- He stakes out ten squares, each with an edge length of $\text{25 cm}$, randomly distributed over the entire meadow and counts the flowers in each of these squares:
- $$\rm 3, \ 4, \ 1, \ 5, \ 0, \ 3, \ 2, \ 4, \ 2, \ 6.$$
Consider these numerical values as random results of the discrete random variable $z$.
It is obvious that the sample size $(10)$ is very small but – this much is revealed – the farmer is lucky. First consider how you would proceed to solve this task, and then answer the following questions.
Hints:
- This exercise belongs to the chapter Poisson Distribution.
- Reference is also made to the chapter Moments of a Discrete Random Variable.
Questions
Solution
(1) The linear mean of these ten numbers gives
- $$\underline{m_z = 3}.$$
(2) For the second moment of the random variable $z$ applies accordingly:
- $$m_{\rm 2\it z}=\frac{1}{10}\cdot (0^2+1^2+ 2\cdot 2^2+ 2\cdot 3^2+2\cdot 4^2+ 5^2+6^2)=12.$$
- According to Steiner's theorem, the variance is:
- $$\sigma_z^2 =12 -3^2 = 3,$$
- and thus the standard deviation:
- $$\underline{\sigma_z \approx 1.732}.$$
(3) Correct are solutions 1, 2, and 4:
- Mean and standard deviation agree here. This is indicative of the Poisson distribution with rate $\lambda = 3$
(equal to the mean and equal to the variance, not equal to the standard deviation). - Naturally, it is questionable to make this statement on the basis of only ten values.
- However, in the case of moments, a smaller sample number is less serious than, for example, in the case of probabilities.
(4) In total, there are $80000$ such squares, each with three flowers in the mean.
- This suggests a total of $\underline{B = 240}$ thousand flowers.
(5) According to the Poisson distribution, this probability results in
- $${\rm Pr}(z = 0) = \frac{3^0}{0!} \cdot{\rm e}^{-3}\hspace{0.15cm}\underline{\approx 5\%}.$$
- However, the small sample size $N = 10$ on which this task was based would have indicated probability ${\rm Pr}(z = 0) = { 10\%}$
since only in a single square no single flower was counted.
