Answer only one of the following two alternatives.
**EITHER**
A particle of mass 0.1 kg lies on a smooth horizontal table on the line between two points \(A\) and \(B\) on the table, which are 6 m apart. The particle is joined to \(A\) by a light elastic string of natural length 2 m and modulus of elasticity 60 N, and to \(B\) by a light elastic string of natural length 1 m and modulus of elasticity 20 N. The mid-point of \(AB\) is \(M\), and \(O\) is the point between \(M\) and \(B\) at which the particle can rest in equilibrium. Show that \(MO = 0.2\) m. [4]
The particle is held at \(M\) and then released. Show that the equation of motion is
$$\frac{\mathrm{d}^2y}{\mathrm{d}t^2} = -500y,$$
where \(y\) metres is the displacement from \(O\) in the direction \(OB\) at time \(t\) seconds, and state the period of the motion. [5]
For the instant when the particle is 0.3 m from \(M\) for the first time, find
- the speed of the particle, [2]
- the time taken, after release, to reach this position. [3]
**OR**
The continuous random variable \(T\) has a negative exponential distribution with probability density function given by
$$\mathrm{f}(t) = \begin{cases} \lambda\mathrm{e}^{-\lambda t} & t \geqslant 0, \\ 0 & \text{otherwise.} \end{cases}$$
Show that for \(t \geqslant 0\) the distribution function is given by F\((t) = 1 - \mathrm{e}^{-\lambda t}\). [2]
The table below shows some values of F\((t)\) for the case when the mean is 20. Find the missing value. [2]
| \(t\) | 0 | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 |
| F\((t)\) | 0 | 0.2212 | 0.3935 | | 0.6321 | 0.7135 | 0.7769 | 0.8262 | 0.8647 |
It is thought that the lifetime of a species of insect under laboratory conditions has a negative exponential distribution with mean 20 hours. When observation starts there are 100 insects, which have been randomly selected. The lifetimes of the insects, in hours, are summarised in the table below.
| Lifetime (hours) | \(0-5\) | \(5-10\) | \(10-15\) | \(15-20\) | \(20-25\) | \(25-30\) | \(30-35\) | \(35-40\) | \(\geqslant 40\) |
| Frequency | 20 | 20 | 11 | 9 | 9 | 8 | 5 | 1 | 17 |
Calculate the expected values for each interval, assuming a negative exponential model with a mean of 20 hours, giving your values correct to 2 decimal places. [3]
Perform a \(\chi^2\)-test of goodness of fit, at the 5% level of significance, in order to test whether a negative exponential distribution, with a mean of 20 hours, is a suitable model for the lifetime of this species of insect under laboratory conditions. [7]