S2 June 2006 Q2
2. The continuous random variable \(L\) represents the error, in mm, made when a machine cuts rods to a target length. The distribution of \(L\) is continuous uniform over the interval \([-4.0, 4.0]\).
Find
(a) \(\mathrm{P}(L \lt -2.6)\), (1)
(b) \(\mathrm{P}(L \lt -3.0 \text{ or } L \gt 3.0)\). (2)
A random sample of 20 rods cut by the machine was checked.
(c) Find the probability that more than half of them were within 3.0 mm of the target length. (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(L \lt -2.6) = 1.4 \times \dfrac{1}{8} = \underline{\dfrac{7}{40}}\) or 0.175 or equivalent | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(L \lt -3.0 \text{ or } L \gt 3.0) = 2 \times \left(1 \times \dfrac{1}{8}\right) = \dfrac{1}{4}\) | M1; A1 |
| (2) |
Notes
M1 for \(1/8\) seen
| Scheme | Marks |
|---|---|
| P(within 3mm) \(= 1 - \dfrac{1}{4} = 0.75 \qquad \mathrm{B}(20, 0.75)\) | B1 |
| Let \(X\) represent number of rods within 3mm | M1 |
| \(\mathrm{P}(X \leqslant 9 / p = 0.25)\) or \(1 - \mathrm{P}(X \leqslant 10 / p = 0.75)\) | M1 |
| \(= 0.9861\) | A1 |
| (4) | |
| (7 marks) |
Notes
B1 recognises binomial
1st M1 using \(\mathrm{B}(20, p)\)
A1 awrt 0.9861