To solve the problem, let's first summarize the given information:

The numbers are: 3, 6, 12, and 14.

Their respective frequencies are:

• For 3: $x + 6$
• For 6: $x + 2$
• For 12: $x - 3$
• For 14: $x$

The mean is given as 5.97.

Formula for the mean:

The mean $\mu$ of a data set with frequencies is given by:

$\mu = \frac{\sum ( \text{value} \times \text{frequency} )}{\sum \text{frequency}}$

Step 1: Write the equation for the mean

Using the information provided, the equation for the mean becomes:

$5.97 = \frac{(3(x + 6) + 6(x + 2) + 12(x - 3) + 14(x))}{(x + 6) + (x + 2) + (x - 3) + x}$

Step 2: Simplify the numerator

The numerator is the sum of the product of the numbers and their respective frequencies:

$3(x + 6) + 6(x + 2) + 12(x - 3) + 14x$ Expanding each term:

$= 3x + 18 + 6x + 12 + 12x - 36 + 14x$ $= (3x + 6x + 12x + 14x) + (18 + 12 - 36)$ $= 35x - 6$

Step 3: Simplify the denominator

The denominator is the sum of all the frequencies:

$(x + 6) + (x + 2) + (x - 3) + x$ $= 4x + 5$

Step 4: Substitute into the mean equation

Now substitute both the numerator and denominator into the mean equation:

$5.97 = \frac{35x - 6}{4x + 5}$

Step 5: Solve for $x$

Multiply both sides by $4x + 5$:

$5.97(4x + 5) = 35x - 6$ Expanding both sides:

$23.88x + 29.85 = 35x - 6$ Now, bring all terms involving $x$ to one side and constants to the other side:

$23.88x - 35x = -6 - 29.85$ $-11.12x = -35.85$ Solve for $x$:

$x = \frac{-35.85}{-11.12} \approx 3.224$

Step 6: Find the next highest integer

The next highest integer is $\boxed{4}$.

Would you like any further details or clarifications on the steps?

Here are some related questions to explore further:

1. How do you compute the mean for grouped frequency data?
2. What are the different types of means in statistics (e.g., arithmetic, geometric)?
3. How does changing one frequency affect the overall mean?
4. Can you explain the difference between mean, median, and mode?
5. How do you solve for unknowns in weighted averages?

Tip: Always double-check your expansions and simplifications in frequency problems to avoid small errors that could lead to incorrect results.