S2 June 2008 Q4
4. Each cell of a certain animal contains 11000 genes. It is known that each gene has a probability 0.0005 of being damaged.
A cell is chosen at random.
(a) Suggest a suitable model for the distribution of the number of damaged genes in the cell. (2)
(b) Find the mean and variance of the number of damaged genes in the cell. (2)
(c) Using a suitable approximation, find the probability that there are at most 2 damaged genes in the cell. (4)
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{B}(11000, 0.0005)\) | M1 A1 |
| (2) |
Notes
M1 for Binomial,
A1 fully correct
These cannot be awarded unless seen in part a
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = 11000 \times 0.0005 = 5.5\) | B1 |
| \(\mathrm{Var}(X) = 11000 \times 0.0005 \times (1 - 0.0005)\) \(= 5.49725\) | B1 |
| (2) |
Notes
B1 cao
B1 also allow 5.50, 5.497, 5.4973, do not allow 5.5
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{Po}(5.5)\) | M1 A1 |
| \(\mathrm{P}(X \leqslant 2) = 0.0884\) | dM1 A1 |
| (4) | |
| (8 marks) |
Notes
M1 for Poisson
A1 for using Po (5.5)
M1 this is dependent on the previous M mark. It is for attempting to find \(\mathrm{P}(X \leqslant 2)\)
A1 awrt 0.0884
Special case If they use normal approximation they could get M0 A0 M1 A0 if they use 2.5 in their standardisation.
NB exact binomial is 0.0883