The uploaded image contains a statistics problem regarding finding the mean of a frequency distribution. Let’s carefully analyze and solve it step-by-step.

Problem Details:

• The given table lists class intervals (C.I.), their frequencies (f), and the total frequency (∑f = 40).
• The mean (denoted as $\bar{x}$) is given as 14.7.

The frequency distribution:

We will use the assumed mean method for calculating the mean to find the missing value of $p$.

Step-by-Step Solution:

1. Identify the midpoints ($x_i$) of each class interval: The midpoint of a class interval is calculated as: $x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$ Midpoints:

   • 0–6: $x_1 = \frac{0 + 6}{2} = 3$
   • 6–12: $x_2 = \frac{6 + 12}{2} = 9$
   • 12–18: $x_3 = \frac{12 + 18}{2} = 15$
   • 18–24: $x_4 = \frac{18 + 24}{2} = 21$
   • 24–30: $x_5 = \frac{24 + 30}{2} = 27$
   • 30–36: $x_6 = \frac{30 + 36}{2} = 33$
   • 36–42: $x_7 = \frac{36 + 42}{2} = 39$

2. Assume a mean ($A$): From the table, use $A = 15$ (the midpoint of the class 12–18).

3. Calculate the deviations ($d_i = x_i - A$): Deviations from $A = 15$:

   • $d_1 = 3 - 15 = -12$
   • $d_2 = 9 - 15 = -6$
   • $d_3 = 15 - 15 = 0$
   • $d_4 = 21 - 15 = 6$
   • $d_5 = 27 - 15 = 12$
   • $d_6 = 33 - 15 = 18$
   • $d_7 = 39 - 15 = 24$

4. Calculate the frequency × deviation ($f_i \cdot d_i$): Using the given frequencies:

   • $f_1 \cdot d_1 = 10 \cdot (-12) = -120$
   • $f_2 \cdot d_2 = p \cdot (-6) = -6p$
   • $f_3 \cdot d_3 = 4 \cdot 0 = 0$
   • $f_4 \cdot d_4 = 7 \cdot 6 = 42$
   • $f_5 \cdot d_5 = 9 \cdot 12 = 108$
   • $f_6 \cdot d_6 = 4 \cdot 18 = 72$
   • $f_7 \cdot d_7 = 1 \cdot 24 = 24$

   Sum of $f_i \cdot d_i$: $\text{Sum} = -120 - 6p + 0 + 42 + 108 + 72 + 24 = 126 - 6p$

5. Apply the formula for the mean: The formula for the mean using deviations is: $\bar{x} = A + \frac{\sum(f_i \cdot d_i)}{\sum f_i}$ Substituting the known values: $14.7 = 15 + \frac{126 - 6p}{40}$

6. Solve for $p$: Simplify the equation: $14.7 - 15 = \frac{126 - 6p}{40}$ $-0.3 = \frac{126 - 6p}{40}$ Multiply through by 40: $-12 = 126 - 6p$ $-12 - 126 = -6p$ $-138 = -6p$ $p = \frac{138}{6} = 23$

Final Answer:

The missing frequency $p = 23$.

Would you like further clarification or a detailed explanation of any step?

Follow-Up Questions:

1. How do we calculate the midpoints for grouped data?
2. What is the assumed mean method, and why is it useful?
3. How does the choice of the assumed mean ($A$) affect the calculations?
4. How would the calculation change if the mean were a different value?
5. Can you explain how to verify the accuracy of the result?

Tip:

In frequency distributions, the assumed mean method simplifies calculations by reducing large deviations into manageable numbers.