Aufgaben:Exercise 2.5Z: Flower Meadow

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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

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 standard deviation 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 standard deviation  $\sigma_z$.
	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

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.