We are tasked with finding the missing frequency $x$ in the given frequency distribution, given that the arithmetic mean of the data is 17.8.

Steps to solve:

1. Midpoints of class intervals:

   • For each class interval, calculate the midpoint $m$ as: $m = \frac{\text{lower limit} + \text{upper limit}}{2}$ Midpoints:
   • $m_1 = \frac{0 + 8}{2} = 4$
   • $m_2 = \frac{8 + 16}{2} = 12$
   • $m_3 = \frac{16 + 24}{2} = 20$
   • $m_4 = \frac{24 + 32}{2} = 28$
   • $m_5 = \frac{32 + 40}{2} = 36$

2. Total Frequency: The total frequency $N$ is the sum of all frequencies, including $x$: $N = 6 + x + 8 + 6 + 5 = 25 + x$

3. Mean formula: The mean is given by the formula: $\bar{x} = \frac{\sum{f_i m_i}}{N}$ where $f_i$ is the frequency and $m_i$ is the midpoint.

   Using the known mean $\bar{x} = 17.8$, we can set up the equation for $\sum{f_i m_i}$:

   $17.8 = \frac{(6 \times 4) + (x \times 12) + (8 \times 20) + (6 \times 28) + (5 \times 36)}{25 + x}$

   Let's now compute each term and solve for $x$. I'll calculate this.The missing frequency $x$ is 15.

Would you like further details on the steps? Here are some related questions to explore:

1. How do you calculate the arithmetic mean from a frequency distribution?
2. What would the total frequency be once the missing frequency is found?
3. How can midpoints in a class interval affect the mean calculation?
4. What is the difference between mean, median, and mode in frequency distributions?
5. How would the result change if the arithmetic mean were different?

Tip: Always ensure to recheck your class intervals and frequencies when solving problems involving grouped data to avoid mistakes in midpoints or frequency sums.