1. **Stating the problem:** We are given a grouped frequency distribution table with class intervals, frequencies, class marks, and other statistical values. We want to understand how to calculate the mean ($\bar{x}$) and median ($M_d$) using the given data and formulas. 2. **Mean calculation:** The mean of grouped data is calculated using the formula: $$\bar{x} = \frac{\sum f x}{\sum f}$$ where $f$ is the frequency and $x$ is the class mark. 3. **Given data:** - $\sum f = 50$ - $\sum f x = 1735$ 4. **Calculate the mean:** $$\bar{x} = \frac{1735}{50} = 34.7$$ This means the average score is 34.7. 5. **Median calculation:** The median for grouped data is calculated by: $$M_d = L + \left(\frac{\frac{n}{2} - F}{f_m}\right) \times c$$ where: - $L$ = lower boundary of median class - $n$ = total frequency - $F$ = cumulative frequency before median class - $f_m$ = frequency of median class - $c$ = class width 6. **From the problem:** - $L = 35.5$ (lower boundary of median class) - $n = 50$ - $F = 25$ (cumulative frequency before median class) - $f_m = 26$ (frequency of median class) - $c = 3$ (class interval width) 7. **Calculate median:** $$M_d = 35.5 + \left(\frac{25 - 26}{26}\right) \times 3 = 35.5 + \left(-\frac{1}{26}\right) \times 3 = 35.5 - 0.115 = 35.385$$ 8. **Explanation:** - The mean is the weighted average of class marks. - The median locates the middle value by finding the class where the cumulative frequency reaches half the total frequency. - The formula adjusts within the median class to estimate the exact median value. 9. **Additional notes:** - $\sum f(x - \bar{x})^2 = 2280.5$ is the sum of squared deviations used for variance and standard deviation calculations. **Final answers:** - Mean $\bar{x} = 34.7$ - Median $M_d \approx 35.39$