Prepare a grouped frequency distribution table and compute the mean for the given religious education test scores.

First, arrange the given scores in ascending order:
0, 4, 5, 6, 7, 8, 9, 9, 10, 10, 10, 11, 12, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 17, 18, 19, 19, 20, 23, 24, 24

The total number of data points is $n = 31$.
The minimum value is 0 and the maximum value is 24.
The range is $24 - 0 = 24$.

To determine the number of classes, we can use Sturges' formula: $k = 1 + 3.322 \log_{10}(n)$.

$$k = 1 + 3.322 \log_{10}(31) \approx 1 + 3.322 \times 1.491 \approx 1 + 4.956 \approx 5.956$$

We can choose 5 or 6 classes. Let's choose 5 classes for a clear distribution.
The class width ($w$) is calculated as $\frac{\text{Range}}{\text{Number of classes}}$.

$$w = \frac{24}{5} = 4.8$$

We round up to 5 to ensure all data points are covered and to have convenient integer class limits.

The class intervals will be:
0-4
5-9
10-14
15-19
20-24

For each class interval, we count the frequency ($f$) of scores falling within it and calculate the midpoint ($x$). The midpoint is the average of the lower and upper class limits. Then, we calculate the product $f \cdot x$.

Class Interval	Tally	Frequency ($f$)	Midpoint ($x$)	$f \cdot x$
0-4	0, 4	2	2	4
5-9	5, 6, 7, 8, 9, 9	6	7	42
10-14	10, 10, 10, 11, 12, 12, 12, 13, 13, 14, 14	11	12	132
15-19	15, 15, 16, 16, 17, 18, 19, 19	8	17	136
20-24	20, 23, 24, 24	4	22	88
Total		31		402

The formula for the mean ($\bar{x}$) of grouped data is:

$$\bar{x} = \frac{\sum (f \cdot x)}{\sum f}$$

From the table, we have:
$\sum (f \cdot x) = 402$
$\sum f = 31$

Substitute these values into the formula:

$$\bar{x} = \frac{402}{31}$$ $$\bar{x} \approx 12.9677419$$

Rounding to two decimal places:

$$\bar{x} \approx 12.97$$

The grouped frequency distribution table is shown above.
The computed mean is $\text{12.97}$.