The given table shows the number of people in different age groups. Find the mean age of the given number of people. Class interval of age (in years) | Number of people (f) 25 - 29 | 10 30 - 34 | 12 35 - 39 | 8 40 - 44 | 5 45 - 49 | 15 50 - 54 | 10 55 - 59 | 15

Correct — D. Give everyone in a class its class mark, the midpoint: (25 + 29) ⁄ 2 = 27, and so on up the table.

Class marks: 27, 32, 37, 42, 47, 52, 57
Take 42 as the working mean, d = mark − 42: −15, −10, −5, 0, 5, 10, 15
Multiply by f: −150, −120, −40, 0, 75, 100, 225

Add the frequencies: Σf = 10 + 12 + 8 + 5 + 15 + 10 + 15 = 75
Add the products: Σfd = −310 + 400 = 90
Mean = 42 + 90 ⁄ 75 = 42 + 1.2 = 43.2 years → option (d)

Why the others are wrong

• (a)
  38.6 years — 38.6 is below the working mean 42, which would need Σfd = (38.6 − 42) × 75 = −255. The table's deviations sum to +90, so the mean sits above 42.
• (b)
  32.4 years — 32.4 is far too low. 40 of the 75 people are 45 or older, and Σfd = +90 puts the mean above 42, nowhere near the youngest classes.
• (c)
  54.2 years — 54.2 would need Σfd = (54.2 − 42) × 75 = 915, more than ten times the table's +90. It lies above every class mark except the top one, 57.

Grouped data hide the exact ages, so each class is represented by its class mark, the midpoint of the interval. The mean is then Σfx ⁄ Σf.

The assumed-mean shortcut subtracts a convenient mark A from each x: mean = A + Σfd ⁄ Σf. Picking the middle mark, 42, keeps the deviations at ±5, ±10 and ±15, so the products stay small.

The direct route agrees: Σfx = 3240, and 3240 ⁄ 75 = 43.2.

Key facts

• Class mark = (lower limit + upper limit) ⁄ 2.
• Mean of grouped data = Σfx ⁄ Σf, with x the class marks.
• Assumed-mean method: mean = A + Σfd ⁄ Σf, where d = x − A.

Common traps

• Using the lower limits 25, 30, 35 … instead of the class marks, which gives 41.2.
• Dividing Σfx by 7, the number of classes, instead of by Σf = 75.

No directly related past PYQ was found.