Q.The daily wages (in ₹) of 20 workers in a small Ahmedabad workshop are: 210, 245, 198, 260, 230, 215, 250, 205, 240, 225, 235, 255, 200, 248, 220, 232, 262, 218, 244, 228. Form a continuous frequency distribution with class width 20, starting from 190, using the exclusive method.

Group the 20 wage values into classes of width 20 starting at 190, and count how many wages fall in each class using tally marks.

Step 1 — decide the classes. Class width $h = 20$, starting from 190: 190–210, 210–230, 230–250, 250–270. (Under the exclusive method, a wage exactly equal to a class's upper limit is counted in the next class — e.g., a wage of 230 goes into 230–250, not 210–230.)

Step 2 — tally each value. Sorting the 20 wages in ascending order: 198, 200, 205, 210, 215, 218, 220, 225, 228, 230, 232, 235, 240, 244, 245, 248, 250, 255, 260, 262.

Verification (dual-check — re-classify from the original, unsorted list, independently of the sort above): going through the wages in their original order — 210→210–230, 245→230–250, 198→190–210, 260→250–270, 230→230–250, 215→210–230, 250→250–270, 205→190–210, 240→230–250, 225→210–230, 235→230–250, 255→250–270, 200→190–210, 248→230–250, 220→210–230, 232→230–250, 262→250–270, 218→210–230, 244→230–250, 228→210–230 — tallying this independent pass gives the identical counts: 190–210: 3, 210–230: 6, 230–250: 7, 250–270: 4, total $3+6+7+4=20$, matching all 20 workers exactly. Both passes agree.

The most frequent error when forming a grouped frequency distribution by the exclusive method is mis-assigning a value that exactly equals a class boundary (like 230 here) — it always belongs to the class where it is the lower limit, never the class where it is the upper limit. A second common error is losing or double-counting a value while tallying a long list — always re-count the total frequency against the original number of observations as a check, exactly as done above.