S2 January 2009 Q1
1. A botanist is studying the distribution of daisies in a field. The field is divided into a number of equal sized squares. The mean number of daisies per square is assumed to be 3. The daisies are distributed randomly throughout the field.
Find the probability that, in a randomly chosen square there will be
(a) more than 2 daisies, (3)
(b) either 5 or 6 daisies. (2)
The botanist decides to count the number of daisies, \(x\), in each of 80 randomly selected squares within the field. The results are summarised below
\[\textstyle\sum x = 295 \qquad \sum x^2 = 1386\](c) Calculate the mean and the variance of the number of daisies per square for the 80 squares. Give your answers to 2 decimal places. (3)
(d) Explain how the answers from part (c) support the choice of a Poisson distribution as a model. (1)
(e) Using your mean from part (c), estimate the probability that exactly 4 daisies will be found in a randomly selected square. (2)
| Scheme | Marks |
|---|---|
| The random variable \(X\) is the number of daisies in a square. Poisson(3) | B1 |
| \(1 - \mathrm{P}(X \leqslant 2) = 1 - 0.4232\) \(1 - \mathrm{e}^{-3}\left(1 + 3 + \dfrac{3^2}{2!}\right)\) | M1 |
| \(= 0.5768\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \leqslant 6) - \mathrm{P}(X \leqslant 4) = 0.9665 - 0.8153\) \(\mathrm{e}^{-3}\left(\dfrac{3^5}{5!} + \dfrac{3^6}{6!}\right)\) | M1 |
| \(= 0.1512\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mu = 3.69\) | B1 |
| \(\mathrm{Var}(X) = \dfrac{1386}{80} - \left(\dfrac{295}{80}\right)^2\) | M1 |
| \(= 3.73/3.72/3.71\) accept \(s^2 = 3.77\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| For a Poisson model, Mean = Variance ; For these data \(3.69 \approx 3.73\) \(\Rightarrow\) Poisson model | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{e}^{-3.6875}\,3.6875^4}{4!} = 0.193\) allow their mean or var | M1 |
| Awrt 0.193 or 0.194 | A1 ft |
| (2) | |
| (11 marks) |