Mean — Assumed Mean and Step-Deviation Methods

Why We Need Shortcuts

The direct method is fine for small numbers, but if the class marks are large (like $510, 530, 550, \dots$) the products $f_i x_i$ become huge and error-prone. Two shortcuts make the arithmetic much lighter — the assumed-mean method and the step-deviation method. Both give exactly the same answer as the direct method.

The Assumed-Mean Method

Pick any convenient class mark as the assumed mean $a$ (often a middle one). Measure each class mark's deviation from it: $$d_i = x_i - a.$$ Then the true mean is $a$ corrected by the average deviation: $$\boxed{\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}}$$

The deviations $d_i$ are small numbers (often with mixed signs), so $\sum f_i d_i$ is easy to compute.

Key Point: The final mean does not depend on which $a$ you choose — pick one that makes the $d_i$ small.

The Step-Deviation Method

When all the classes have the same width $h$ and the deviations $d_i$ are all multiples of $h$, go one step further. Let $$u_i = \dfrac{x_i - a}{h}.$$ The $u_i$ are tiny whole numbers like $-2, -1, 0, 1, 2$. Then $$\boxed{\bar{x} = a + h\left(\dfrac{\sum f_i u_i}{\sum f_i}\right)}$$

Key Point: Step-deviation is the assumed-mean method with the extra division by $h$ — use it when class marks are large and equally spaced. It gives the smallest, easiest numbers.

Which Method When?

• Direct method: small class marks and frequencies.
• Assumed-mean method: large numbers, but you would rather not divide by $h$.
• Step-deviation method: large, equally spaced class marks (equal class width $h$) — the lightest arithmetic.

All three are algebraically identical; the answer is the same. The shortcut methods just reduce the size of the numbers you handle.

[Board Important] In the step-deviation method, do not forget to multiply the average of $u_i$ by $h$ and then add $a$.

Solved Examples

Example 1: Assumed-mean method

Find the mean of the marks table (class marks $17.5,32.5,47.5,62.5,77.5,92.5$; frequencies $2,3,7,6,6,6$) using $a=62.5$.

Solution:

1. $d_i = x_i - 62.5 = -45,-30,-15,0,15,30$.
2. $f_i d_i = -90,-90,-105,0,90,180$; $\sum f_i d_i = -15$; $\sum f_i = 30$.
3. $\bar{x} = 62.5 + \dfrac{-15}{30} = 62.5 - 0.5 = 62$.

Final Answer: Mean $= 62$ (same as the direct method).

Takeaway: Choosing $a$ as a middle class mark keeps the $d_i$ small.

Example 2: Step-deviation method

Same table, with $a=62.5$ and $h=15$.

Solution:

1. $u_i = \dfrac{x_i-62.5}{15} = -3,-2,-1,0,1,2$.
2. $f_i u_i = -6,-6,-7,0,6,12$; $\sum f_i u_i = -1$; $\sum f_i = 30$.
3. $\bar{x} = 62.5 + 15\times\dfrac{-1}{30} = 62.5 - 0.5 = 62$.

Final Answer: Mean $= 62$.

Takeaway: $u_i$ are the smallest numbers of all — step-deviation is the quickest here.

Example 3: Choosing the assumed mean

Using $a = 47.5$ for the same table, confirm the mean is still 62.

Solution:

1. $d_i = -30,-15,0,15,30,45$.
2. $f_id_i = -60,-45,0,90,180,270$; $\sum f_id_i = 435$; $\sum f_i=30$.
3. $\bar{x} = 47.5 + \dfrac{435}{30} = 47.5 + 14.5 = 62$.

Final Answer: Mean $= 62$.

Takeaway: Any assumed mean gives the same result — the method is self-correcting.

Example 4: Large numbers

Find the mean by step-deviation: class marks $150,250,350,450,550$; frequencies $2,3,5,6,4$; take $a=350$, $h=100$.

Solution:

1. $u_i = -2,-1,0,1,2$.
2. $f_iu_i = -4,-3,0,6,8$; $\sum f_iu_i = 7$; $\sum f_i = 20$.
3. $\bar{x} = 350 + 100\times\dfrac{7}{20} = 350 + 35 = 385$.

Final Answer: Mean $= 385$.

Takeaway: With big, equally spaced marks, step-deviation turns the work into tiny integers.