Given Data:
- Mean = 52.4
- Class Intervals: 0-20, 20-40, 40-60, 60-80, 80-100, 100-120
- Frequencies: 14, P, 24, 32, 10, 2

To find: The value of missing frequency P.

Solution:
Step 1: Recall the formula for calculating the mean of a frequency distribution.

Mean = (Sum of (Class Mark × Frequency))/Total Frequency

Step 2: Calculate the class marks for each class interval.

Class Mark = (Upper Limit + Lower Limit)/2

For the given data, the class marks are:
- 0-20: (20+0)/2 = 10
- 20-40: (40+20)/2 = 30
- 40-60: (60+40)/2 = 50
- 60-80: (80+60)/2 = 70
- 80-100: (100+80)/2 = 90
- 100-120: (120+100)/2 = 110

Step 3: Substitute the given values in the formula for mean and simplify.

52.4 = [(10×14) + (30×P) + (50×24) + (70×32) + (90×10) + (110×2)]/(14+P+24+32+10+2)

Simplifying this equation, we get:

52.4 = (140 + 30P + 1200 + 2240 + 900 + 220)/(82+P)

52.4 = (4700 + 30P)/(82+P)

Multiplying both sides by (82+P), we get:

52.4(82+P) = 4700 + 30P

Expanding the brackets, we get:

4296.8 + 52.4P = 4700 + 30P

Subtracting 30P from both sides, we get:

22.4P = 403.2

Dividing both sides by 22.4, we get:

P = 18

Therefore, the missing frequency is 18.

Conclusion:
- The missing frequency in the given distribution is 18.
- The formula for calculating the mean of a frequency distribution is Mean = (Sum of (Class Mark × Frequency))/Total Frequency.
- Class marks are calculated as (Upper Limit + Lower Limit)/2.
- We can use the formula for mean to find the missing frequency in a frequency distribution.

Mean = sigma fıXı/sigma fı
52.4 = 4700+30p/82 +p
52.4p+4296.8 =4700 + 30p
22.4p = 403.2
p = 18