1. **State the problem:** We have grouped data with class intervals and their frequencies. We want to estimate the mean and variance using the coding method. 2. **Identify the class with highest frequency:** The class 360-369 has the highest frequency, 39, so we choose its midpoint, 364.5, as the assumed mean (a). 3. **Calculate midpoints (x) for each class:** Given already as 344.5, 354.5, 364.5, 374.5, and 384.5. 4. **Calculate deviations from assumed mean:** $$ t = x - a $$ So, 340-349: $$ 344.5 - 364.5 = -20 $$ 350-359: $$ 354.5 - 364.5 = -10 $$ 360-369: $$ 364.5 - 364.5 = 0 $$ 370-379: $$ 374.5 - 364.5 = 10 $$ 380-389: $$ 384.5 - 364.5 = 20 $$ 5. **Fill t-values and calculate ft and ft^2:** Calculate $$ ft = f \times t $$ and $$ ft^2 = f \times t^2 $$ | Class | f | t | ft | ft^2 | |-------|----|-----|------|---------| |340-349| 3 | -20 | 3\times(-20) = -60 | 3\times(400)=1200 | |350-359| 12 | -10 | 12\times(-10)=-120 | 12\times(100)=1200 | |360-369| 39 | 0 | 0 | 0 | |370-379| 23 | 10 | 23\times 10=230 | 23\times 100=2300 | |380-389| 7 | 20 | 7\times 20=140 | 7\times 400=2800 | 6. **Calculate totals:** $$ \Sigma f = 3+12+39+23+7 = 84 $$ $$ \Sigma ft = -60 -120 + 0 + 230 + 140 = 190 $$ $$ \Sigma ft^2 = 1200 + 1200 + 0 + 2300 + 2800 = 7500 $$ 7. **Mean using coding method:** $$ \bar{x} = a + \frac{\Sigma ft}{\Sigma f} = 364.5 + \frac{190}{84} = 364.5 + 2.26 = 366.76 $$ 8. **Variance using coding method:** $$ s^2 = \frac{\Sigma ft^2}{\Sigma f} - \left(\frac{\Sigma ft}{\Sigma f}\right)^2 = \frac{7500}{84} - \left(\frac{190}{84}\right)^2 = 89.29 - 5.11 = 84.18 $$ **Final answers:** - Mean = $366.76$ (in '000s) - Variance = $84.18$ (square of '000s)