A class of 45 students sat a test. The marks and their frequencies are shown in the table. Find the value of x and calculate the mean mark.

The total number of students is given as 45. This means the sum of all frequencies must be equal to 45.

$$6 + (x^2-1) + 3x + 10 + (3x-2) + 5 = 45$$

Combine the constant terms and the terms with $x$:

$$x^2 + (3x+3x) + (6-1+10-2+5) = 45$$ $$x^2 + 6x + 18 = 45$$

Subtract 45 from both sides to form a quadratic equation:

$$x^2 + 6x - 27 = 0$$

Factor the quadratic equation:

$$(x+9)(x-3) = 0$$

This gives two possible values for $x$:

$$x = -9 \quad \text{or} \quad x = 3$$

Frequencies cannot be negative. Let's check the frequencies for each value of $x$:
If $x = -9$:
Frequency for mark 17 is $3x = 3(-9) = -27$, which is not possible.
If $x = 3$:
Frequency for mark 16 is $x^2-1 = (3)^2-1 = 9-1 = 8$.
Frequency for mark 17 is $3x = 3(3) = 9$.
Frequency for mark 19 is $3x-2 = 3(3)-2 = 9-2 = 7$.
All these frequencies are positive, so $x=3$ is the correct value.

First, update the frequencies using $x=3$:

• Mark 15: Frequency = 6
• Mark 16: Frequency = $3^2-1 = 8$
• Mark 17: Frequency = $3(3) = 9$
• Mark 18: Frequency = 10
• Mark 19: Frequency = $3(3)-2 = 7$
• Mark 20: Frequency = 5

The mean ($\bar{m}$) is calculated using the formula $\bar{m} = \frac{\sum (m \cdot f)}{\sum f}$, where $m$ is the mark and $f$ is the frequency.
Calculate the sum of (mark $\times$ frequency):

$$\sum (m \cdot f) = (15 \times 6) + (16 \times 8) + (17 \times 9) + (18 \times 10) + (19 \times 7) + (20 \times 5)$$ $$\sum (m \cdot f) = 90 + 128 + 153 + 180 + 133 + 100$$ $$\sum (m \cdot f) = 784$$

The total frequency is $\sum f = 6+8+9+10+7+5 = 45$.
Now, calculate the mean:

$$\bar{m} = \frac{784}{45}$$ $$\bar{m} \approx 17.4222...$$

Rounding the mean to 1 decimal place:

$$\bar{m} \approx 17.4$$

The value of $x$ is $\boxed{3}$
The mean, corrected to 1 decimal place, is $\boxed{17.4}$