The data represent the capacity of fruit juice (in litres) in 8 different Containers arranged in ascending order: Pt 4, 11, 12, 15, 3p-1, 1p-2, 27, P2-6. a) If the range is 20, find the i) value of P, ii) Mean of Distribution.

The range is the difference between the maximum and minimum values. Since the data is in ascending order, the minimum value is $P+4$ and the maximum value is $P^2-6$. The given range is 20.

Rearrange the equation to form a standard quadratic equation:

Factor the quadratic equation:

This gives two possible values for P: $P=6$ or $P=-5$.

If $P=6$, the data points are:
$6+4 = 10$
11
12
15
$3(6)-1 = 18-1 = 17$
$4(6)-2 = 24-2 = 22$
27
$(6)^2-6 = 36-6 = 30$
The sequence is $10, 11, 12, 15, 17, 22, 27, 30$. This is in ascending order.

If $P=-5$, the data points are:
$-5+4 = -1$
11
12
15
$3(-5)-1 = -15-1 = -16$
$4(-5)-2 = -20-2 = -22$
27
$(-5)^2-6 = 25-6 = 19$
The sequence is $-1, 11, 12, 15, -16, -22, 27, 19$. This is not in ascending order (e.g., $15 > -16$).
Therefore, $P=-5$ is not a valid solution.

The value of P is $\boxed{6}$.

a) ii) Mean of distribution

The data points are: $10, 11, 12, 15, 17, 22, 27, 30$.
The number of data points is $n=8$.

The mean of the distribution is $\boxed{18}$.