Question 3 of 21intermediate🔧 ApplyNumerical3 marks

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency ff.

Daily pocket allowance (in ₹)11 - 1313 - 1515 - 1717 - 1919 - 2121 - 2323 - 25
Number of children76913ff54
Correct Answer

f = 20

Exercise: EXERCISE 13.1 | Q: 3 | (Page 11)
For More Understanding

Explanation

This is a direct application of the mean formula for grouped data. Since the mean is given and one frequency is missing, we set up an equation using Mean = Σ(fi × xi) / Σfi and solve for f. The class marks (midpoints) are calculated for each class interval, and the values are substituted into the formula.

Solution Steps

  1. Step 1: Find class marks (xi) for each class interval using midpoint formula: (Lower limit + Upper limit)/2

  2. Class marks: 12, 14, 16, 18, 20, 22, 24

  3. Step 2: Calculate fi × xi for each class:

  4. 7×127 \times 12=84, 6×146 \times 14=84, 9×169 \times 16=144, 13×1813 \times 18=234, f×20=20f, 5×225 \times 22=110, 4×244 \times 24=96

  5. Step 3: Find Σ(fi × xi) = 84+84+144+234+20f+110+96 = 752 + 20f

  6. Step 4: Find Σfi = 7+6+9+13+f+5+4 = 44 + f

  7. Step 5: Apply mean formula: Mean = Σ(fi × xi) / Σfi

  8. 18 = (752 + 20f) / (44 + f)

  9. Step 6: Solve for f

  10. 18(44 + f) = 752 + 20f

  11. 792 + 18f = 752 + 20f

  12. 792 - 752 = 20f - 18f

  13. 40 = 2f

  14. f = 20

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