3 The volume, \(X\) litres, of orange juice in a 1-litre carton may be modelled by a normal distribution with unknown mean \(\mu\).
The volumes, \(x\) litres, recorded to the nearest 0.01 litre, in a random sample of 100 cartons are shown in the table.
| Volume ( \(\boldsymbol { x }\) litres) | Number of cartons (f) |
| 0.95-0.97 | 2 |
| 0.98-1.00 | 7 |
| 1.01-1.03 | 15 |
| 1.04-1.06 | 32 |
| 1.07-1.09 | 22 |
| 1.10-1.12 | 14 |
| 1.13-1.15 | 7 |
| 1.16-1.18 | 1 |
| Total | 100 |
- For the group ' \(0.98 - 1.00\) ':
- show that it has a mid-point of 0.99 litres;
- state the minimum and the maximum values of \(x\) that could be included in this group.
- Calculate, to three decimal places, estimates of the mean and the standard deviation of these 100 volumes.
- Construct an approximate \(99 \%\) confidence interval for \(\mu\).
- State why use of the Central Limit Theorem was not required when calculating this confidence interval.
- Give a reason why the confidence interval is approximate rather than exact.
- Give a reason in support of the claim that:
- \(\mu > 1\);
- \(\mathrm { P } ( 0.94 < X < 1.16 )\) is approximately 1 .
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