Difference between revisions of "Aufgaben:Exercise 2.5Z: Flower Meadow"

From LNTwww
Line 17: Line 17:
  
 
Hints:
 
Hints:
*This exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Poisson_Distribution|Poisson distribution]].
+
*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]].
  
Line 60: Line 60:
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; The linear mean of these ten numbers gives&nbsp; $\underline{m_z = 3}$.
+
'''(1)'''&nbsp; The linear mean of these ten numbers gives&nbsp;  
 +
:$$\underline{m_z = 3}.$$
  
  
Line 68: Line 69:
 
:$$m_{\rm 2\it z}=\frac{1}{10}\cdot (0^2+1^2+ 2\cdot 2^2+ 2\cdot 3^2+2\cdot 4^2+ 5^2+6^2)=12.$$
 
:$$m_{\rm 2\it z}=\frac{1}{10}\cdot (0^2+1^2+ 2\cdot 2^2+ 2\cdot 3^2+2\cdot 4^2+ 5^2+6^2)=12.$$
  
*According to Steiner's theorem, the variance is thus equal to.
+
*According to Steiner's theorem, the variance is
:$$\sigma_z^2 =12 -3^2 = 3$$
+
:$$\sigma_z^2 =12 -3^2 = 3,$$
:and accordingly the rms
+
:and thus the rms value:
 
:$$\underline{\sigma_z \approx 1.732}.$$
 
:$$\underline{\sigma_z \approx 1.732}.$$
  
  
 +
'''(3)'''&nbsp; Correct are&nbsp; <u>solutions 1, 2, and 4</u>:
 +
*Mean and rms agree here.&nbsp; This is indicative of the Poisson distribution with rate&nbsp; $\lambda = 3$&nbsp; <br>(equal to the mean and equal to the variance,&nbsp; not equal to the rms value).
 +
*Naturally,&nbsp; it is questionable to make this statement on the basis of only ten values.&nbsp;
 +
*However,&nbsp; in the case of moments,&nbsp; a smaller sample number is less serious than,&nbsp; for example,&nbsp; in the case of probabilities.
  
'''(3)'''&nbsp; Correct <u>solutions 1, 2, and 4</u>:
 
*Mean and rms agree here.&nbsp; This is indicative of the Poisson distribution with rate&nbsp; $\lambda = 3$&nbsp; (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)'''&nbsp; In total,&nbsp; there&nbsp; are $80000$&nbsp; such squares,&nbsp; each with three flowers in the mean.&nbsp;  
'''(4)'''&nbsp; In total, there&nbsp; are $80000$&nbsp; such squares, each with three flowers in the mean.&nbsp;  
 
 
*This suggests a total of &nbsp; $\underline{B = 240}$&nbsp; thousand flowers.
 
*This suggests a total of &nbsp; $\underline{B = 240}$&nbsp; thousand flowers.
  
  
  
'''(5)'''&nbsp; According to the Poisson distribution, this probability results in  
+
'''(5)'''&nbsp; According to the Poisson distribution,&nbsp; 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&nbsp; $N = 10$&nbsp; on which this task was based would have indicated probability&nbsp; ${\rm Pr}(z = 0) = { 10\%}$&nbsp; since only in a single square no single flower was counted.
+
*However,&nbsp; the small sample size&nbsp; $N = 10$&nbsp; on which this task was based would have indicated probability&nbsp; ${\rm Pr}(z = 0) = { 10\%}$&nbsp; <br>since only in a single square no single flower was counted.
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}

Revision as of 16:26, 18 December 2021

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:




Questions

1

Find the mean of  $z$,  that is,  the mean number of flowers counted in each of the ten squares.

$m_z \ =$

2

Determine the rms value of the random variable  $z$.

$\sigma_z\ = \ $

3

Which of the following statements are true?

Actually,  one would have to use considerably more than ten random numbers  (squares)  for the moment calculation.
The random variable  $z$  is in fact Poisson distributed.
The rate  $\lambda$  of the Poisson distribution is equal to the rms value  $\sigma_z$.
The rate  $\lambda$  of the Poisson distribution is equal to the mean  $m_z$.

4

Predict the total number  $B$  of all flowers in the meadow.

$B\ = \ $

$\ \text{thousand}$

5

What is the probability of a square without any flowers?

${\rm Pr}(z = 0) \ = \ $

$\ \%$


Solution

(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:
$$\sigma_z^2 =12 -3^2 = 3,$$
and thus the rms value:
$$\underline{\sigma_z \approx 1.732}.$$


(3)  Correct are  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 value).
  • 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.