4 A machine produces plastic strip in a continuous process. Occasionally there is a flaw at some point along the strip. The length of strip (in hundreds of metres) between successive flaws is modelled by a continuous random variable \(X\) with probability density function \(\mathrm { f } ( x ) = \frac { 18 } { ( 3 + x ) ^ { 3 } }\) for \(x > 0\). The table below gives the frequencies for 100 randomly chosen observations of \(X\). It also gives the probabilities for the class intervals using the model.
| Length \(x\) (hundreds of metres) | Observed frequency | Probability |
| \(0 < x \leqslant 0.5\) | 21 | 0.2653 |
| \(0.5 < x \leqslant 1\) | 24 | 0.1722 |
| \(1 < x \leqslant 2\) | 12 | 0.2025 |
| \(2 < x \leqslant 3\) | 15 | 0.1100 |
| \(3 < x \leqslant 5\) | 13 | 0.1094 |
| \(5 < x \leqslant 10\) | 9 | 0.0874 |
| \(x > 10\) | 6 | 0.0532 |
- Examine the fit of this model to the data at the \(5 \%\) level of significance.
You are given that the median length between successive flaws is 124 metres. At a later date the following random sample of ten lengths (in metres) between flaws is obtained.
$$\begin{array} { l l l l l l l l l l }
239 & 77 & 179 & 221 & 100 & 312 & 52 & 129 & 236 & 42
\end{array}$$
- Test at the \(10 \%\) level of significance whether the median length may still be assumed to be 124 metres.