Difference between revisions of "Aufgaben:Exercise 2.5Z: Flower Meadow"
From LNTwww
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| − | {Determine the | + | {Determine the rms of the random variable $z$. |
|type="{}"} | |type="{}"} | ||
$\sigma_z\ = \ $ { 1.732 3% } | $\sigma_z\ = \ $ { 1.732 3% } | ||
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+ 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 size $z$ is in fact Poisson distributed. | + The random size $z$ 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 rms $\sigma_z$. |
+ 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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===Musterlösung=== | ===Musterlösung=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' | + | '''(1)''' The linear mean of these ten numbers gives $\underline{m_z = 3}$. |
| − | '''(2)''' | + | '''(2)''' For the quadratic mean 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 thus equal to. |
:$$\sigma_z^2 =12 -3^2 = 3$$ | :$$\sigma_z^2 =12 -3^2 = 3$$ | ||
| − | : | + | :and accordingly the rms |
| − | :$$\underline{\sigma_z | + | :$$\underline{\sigma_z \approx 1.732}.$$ |
| − | '''(3)''' | + | '''(3)''' Correct <u>solutions 1, 2, and 4</u>: |
| − | * | + | *Mean and rms agree here. This is indicative of the Poisson distribution with rate $\lambda = 3$ (equal to the mean and equal to the variance, not equal to the rms). |
| − | * | + | *Naturally, it is questionable to make this statement on the basis of only ten values. However, in the case of moments, a smaller sample number is less serious than, for example, in the case of probabilities. |
| − | '''(4)''' | + | '''(4)''' In total, there are $80000$ such squares, each with three flowers in the mean. |
| − | * | + | *This suggests a total of $\underline{B = 240}$ thousand flowers. |
| − | '''(5)''' | + | '''(5)''' According to the Poisson distribution, this probability results in |
| − | + | $${\rm Pr}(z = 0) = \frac{3^0}{0!} \cdot{\rm e}^{-3}\hspace{0.15cm}\underline{\approx 5\%}.$$ | |
| + | |||
| + | *However, the small sample size $N = 10$ on which this task was based would have indicated probability ${\rm Pr}(z = 0) = { 10\%}$ since only in a single square no single flower was counted. | ||
| − | |||
{{ML-Fuß}} | {{ML-Fuß}} | ||
Revision as of 23:46, 15 December 2021
flower meadow – another
example of the Poisson distribution
example of the Poisson distribution
A farmer is happy about the splendor of flowers on his land and wants to know how many dandelions are currently blooming on his meadow.
- He knows that the meadow has an area of $5000$ square meters and he also knows from the agricultural school that the number of flowers in a small area is always poisson distributed.
- He stakes out ten squares, each with an edge length of $\text{25 cm}$ , randomly distributed over the entire meadow and counts the flowers in each of these squares:
- $$\rm 3, \ 4, \ 1, \ 5, \ 0, \ 3, \ 2, \ 4, \ 2, \ 6.$$
Consider these numerical values as random results of the discrete random variable $z$.
It is obvious that the sample size is very small at $10$ but – this much is revealed – the farmer is lucky. First consider how you would proceed to solve this task, and then answer the following questions.
Hints:
- This exercise belongs to the chapter Poisson distribution.
- Reference is also made to the chapter Moments of a Discrete Random Variable.
Questions
Musterlösung
(1) The linear mean of these ten numbers gives $\underline{m_z = 3}$.
(2) For the quadratic mean of the random variable $z$ applies accordingly:
- $$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.$$
- According to Steiner's theorem, the variance is thus equal to.
- $$\sigma_z^2 =12 -3^2 = 3$$
- and accordingly the rms
- $$\underline{\sigma_z \approx 1.732}.$$
(3) Correct solutions 1, 2, and 4:
- Mean and rms agree here. This is indicative of the Poisson distribution with rate $\lambda = 3$ (equal to the mean and equal to the variance, not equal to the rms).
- Naturally, it is questionable to make this statement on the basis of only ten values. However, in the case of moments, a smaller sample number is less serious than, for example, in the case of probabilities.
(4) In total, there are $80000$ such squares, each with three flowers in the mean.
- This suggests a total of $\underline{B = 240}$ thousand flowers.
(5) According to the Poisson distribution, this probability results in $${\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.
