1. **Problem Statement:** We want to find the mean weekly earning of employees using the step-deviation method from the given grouped data. 2. **Understanding the Table:** The table shows weekly earning intervals, their mid-values ($x_i$), number of employees ($f_i$), and calculated columns $f_i \times x_i$, deviations ($u_i$), and $f_i \times u_i$. 3. **Key Formulas:** - Mean using direct method: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ - Mean using step-deviation method: $$\bar{x} = A + h \times \frac{\sum f_i u_i}{\sum f_i}$$ Where: - $A$ is the assumed mean (a mid-value chosen from the data, here 25), - $h$ is the class width (here 2), - $u_i = \frac{x_i - A}{h}$ is the deviation of each mid-value from $A$ divided by $h$. 4. **Calculations:** - Total number of employees: $$\sum f_i = 509$$ - Sum of products of frequency and mid-values: $$\sum f_i x_i = 13315$$ - Sum of products of frequency and deviations: $$\sum f_i u_i = 295$$ 5. **Direct Mean Calculation:** $$\bar{x} = \frac{13315}{509} = 26.16$$ 6. **Step-Deviation Mean Calculation:** - Calculate $u_i$ for each class: $$u_i = \frac{x_i - 25}{2}$$ - Multiply each $u_i$ by corresponding $f_i$ and sum: $$\sum f_i u_i = 295$$ - Apply formula: $$\bar{x} = 25 + 2 \times \frac{295}{509} = 25 + 2 \times 0.579 = 25 + 1.16 = 26.16$$ 7. **Interpretation:** Both methods give the same mean, but the step-deviation method simplifies calculations by using deviations from an assumed mean and smaller numbers. This method is especially useful for large data sets or when mid-values are large, making direct multiplication cumbersome.