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 \(\mathrm { F } ( t ) = 1 - \mathrm { e } ^ { - \lambda t }\).
The table below shows some values of \(\mathrm { F } ( t )\) for the case when the mean is 20 . Find the missing value.
| \(t\) | 0 | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 |
| \(\mathrm {~F} ( t )\) | 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.
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.