5 A manufacturer uses three types of capacitor in a particular electronic device. The capacitances, measured in suitable units, are modelled by independent Normal distributions with means and standard deviations as shown in the table.
| \cline { 2 - 3 }
\multicolumn{1}{c|}{} | Capacitance |
| Type | Mean | |
| A | 3.9 | 0.32 |
| B | 7.8 | 0.41 |
| C | 30.2 | 0.64 |
- Determine the probability that the total capacitance of a randomly chosen capacitor of Type B and two randomly chosen capacitors of Type A is at least 16 units.
- Determine the probability that the capacitance of a randomly chosen capacitor of Type C is within 1 unit of the total capacitance of four randomly chosen capacitors of Type B.
When the manufacturer gets a new batch of 1000 capacitors from the supplier, a random sample of 10 of them is tested to check the capacitances. For a new batch of Type C capacitors, summary statistics for the capacitances, \(x\) units, of the random sample are as follows.
\(n = 10\)
$$\sum x = 299.6 \quad \sum x ^ { 2 } = 8981.0$$
You should assume that the capacitances of the sample come from a Normally distributed population, but you should not assume that the standard deviation is 0.64 as for previous Type C capacitors. - In this question you must show detailed reasoning.
Carry out a hypothesis test at the \(5 \%\) significance level to check whether it is reasonable to assume that the capacitors in this batch have the specified mean capacitance for Type C of 30.2 units.