Q.In Worked Example 3, the mean of a grouped marks distribution (classes 0–10 to 40–50) was found to be 27 by the direct method. If the marks are coded using $u = \dfrac{x-25}{10}$, find the mean of the coded values $u$, and use the linear-transformation rule for the mean to confirm that the actual mean $\bar{x}$ is indeed 27.

The coding $u=(x-25)/10$ is a linear transformation of $x$; once $\bar{u}$ is found, reversing the transformation should recover the same mean as the direct method.

The coding $u = \dfrac{x-25}{10}$ can be rearranged as $x = 10u + 25$, which is a linear relation $x = au+b$ with $a=10, b=25$. By the linear-transformation rule, $\bar{x} = 10\bar{u} + 25$. We first find $\bar{u}$ from the coded data.

Step 1 — Recall the grouped data and compute $u$ for each class. Using the class marks from Worked Example 3 ($5,15,25,35,45$) and $A=25,h=10$:

$\sum fu = -8-6+0+10+12 = 8$.

Step 2 — Find $\bar{u}$.

Step 3 — Apply the linear-transformation rule to recover $\bar{x}$. Since $x = 10u+25$, the rule $\bar{y}=a\bar{x}+b$ (here with the roles of $x$ and $u$ swapped: $\bar{x} = 10\bar{u}+25$) gives

$\bar{x} = 10(0.2) + 25 = 2 + 25 = 27$ …

This same idea, run in reverse, is exactly what the step-deviation method of Worked Example 5 does: it computes $\bar{u}$ (there called the step-deviation mean) from small coded numbers, then applies $\bar{x}=A+h\bar{u}$ to re …

Forgetting to reverse the transformation — reporting $\bar{u}=0.2$ itself as if it were the answer to "find the mean marks" — is the key error; a coded mean must always be converted back using the origina …