2.10 Deviation Method (Assumed Mean Method) for Fast Calculation

Figure: Work with small deviations from a convenient assumed mean instead of the raw numbers.

Instead of adding all raw numbers, pick a convenient assumed average (A) close to the expected mean, find the deviation of each number from A, average the deviations, and add back to A.

Actual Average = A + (Sum of deviations from A)/(n)

This is extremely fast for large or awkward numbers, and forms the backbone of several shortcuts in Section 3.

Solved Example 27: Find the average of 41, 43, 47, 49, and 53 using the deviation method.

Assume A = 47 (middle value estimate). Deviations: 41−47=−6, 43−47=−4, 47−47=0, 49−47=+2, 53−47=+6. Sum of deviations = −6−4+0+2+6 = −2. Average = 47 + (−2/5) = 47 − 0.4 = 46.6.

Solved Example 28: Find the average of 198, 202, 205, 210, and 215 using the deviation method.

Assume A = 200. Deviations: −2, +2, +5, +10, +15. Sum = 30. Average = 200 + 30/5 = 200+6 = 206.

Solved Example 28(a): Find the average of 1005, 1012, 998, 1020, 995, and 1010 using the deviation method.

Assume A = 1000. Deviations: +5, +12, −2, +20, −5, +10. Sum = 5+12−2+20−5+10 = 40. Average = 1000 + 40/6 = 1000 + 6.67 = 1006.67 (approx).

Why this method saves time: Adding six four-digit numbers directly (1005+1012+998+1020+995+1010) is error-prone under exam pressure. Working with small deviations (single or double-digit numbers, some negative) from a convenient round assumed mean is dramatically faster and far less likely to produce an arithmetic slip. This technique is especially valuable when the actual numbers are large (salaries, populations, distances in a DI table) but cluster tightly around a round figure.

Solved Example 28(b): Find the average of 72, 75, 68, 80, 65, and 74 using the deviation method, taking the assumed mean as 70.

Deviations from A=70: 72−70=+2, 75−70=+5, 68−70=−2, 80−70=+10, 65−70=−5, 74−70=+4. Sum of deviations = 2+5−2+10−5+4 = 14. Average = 70 + 14/6 = 70 + 2.33 = 72.33 (approx). (Direct check: 72+75+68+80+65+74 = 434; 434/6 = 72.33 ✓.)

Solved Example 28(c): Find the average of 2998, 3005, 2990, 3012, 3001, 2996, and 3008 using the deviation method.

This is the same idea applied to larger, harder-to-add numbers — exactly where the technique pays off most. Assume A = 3000. Deviations: −2, +5, −10, +12, +1, −4, +8. Sum of deviations = −2+5−10+12+1−4+8 = 10. Average = 3000 + 10/7 = 3000 + 1.43 = 3001.43 (approx). (Direct check: 2998+3005+2990+3012+3001+2996+3008 = 21010; 21010/7 = 3001.43 ✓.)