1. **Problem Statement:** We are given a frequency distribution table and asked to find the assumed mean, median interval, estimate the mode, calculate the semi-interquartile range, the 2nd decile, and the 75th percentile. 2. **Given Frequency Distribution:** Class Interval | Frequency (f) 0 - 9 | 10 10 - 19 | 15 20 - 29 | 23 30 - 39 | 27 40 - 49 | 44 50 - 59 | 21 60 - 69 | 10 3. **Step 1: Calculate Midpoints (x) for each class:** Midpoint $x = \frac{\text{Lower limit} + \text{Upper limit}}{2}$ | Class Interval | Midpoint (x) | |---|---| | 0 - 9 | $\frac{0+9}{2} = 4.5$ | | 10 - 19 | $\frac{10+19}{2} = 14.5$ | | 20 - 29 | $\frac{20+29}{2} = 24.5$ | | 30 - 39 | $\frac{30+39}{2} = 34.5$ | | 40 - 49 | $\frac{40+49}{2} = 44.5$ | | 50 - 59 | $\frac{50+59}{2} = 54.5$ | | 60 - 69 | $\frac{60+69}{2} = 64.5$ | 4. **Step 2: Calculate Assumed Mean (A):** Choose an assumed mean, usually a midpoint near the center. Here, $A = 44.5$ (midpoint of 40-49 class). 5. **Step 3: Calculate deviations $d = x - A$ and $fd = f \times d$:** | Class | f | x | d = x - 44.5 | fd | |---|---|---|---|---| | 0-9 | 10 | 4.5 | $4.5 - 44.5 = -40$ | $10 \times (-40) = -400$ | | 10-19 | 15 | 14.5 | $14.5 - 44.5 = -30$ | $15 \times (-30) = -450$ | | 20-29 | 23 | 24.5 | $24.5 - 44.5 = -20$ | $23 \times (-20) = -460$ | | 30-39 | 27 | 34.5 | $34.5 - 44.5 = -10$ | $27 \times (-10) = -270$ | | 40-49 | 44 | 44.5 | $44.5 - 44.5 = 0$ | $44 \times 0 = 0$ | | 50-59 | 21 | 54.5 | $54.5 - 44.5 = 10$ | $21 \times 10 = 210$ | | 60-69 | 10 | 64.5 | $64.5 - 44.5 = 20$ | $10 \times 20 = 200$ | 6. **Step 4: Calculate total frequency $N$ and sum of $fd$:** $$N = 10 + 15 + 23 + 27 + 44 + 21 + 10 = 150$$ $$\sum fd = -400 - 450 - 460 - 270 + 0 + 210 + 200 = -1170$$ 7. **Step 5: Calculate Mean:** $$\bar{x} = A + \frac{\sum fd}{N} = 44.5 + \frac{-1170}{150} = 44.5 - 7.8 = 36.7$$ 8. **Step 6: Find Median Interval:** Median position = $\frac{N}{2} = \frac{150}{2} = 75$ Cumulative frequencies: | Class | f | Cumulative Frequency (CF) | |---|---|---| | 0-9 | 10 | 10 | | 10-19 | 15 | 25 | | 20-29 | 23 | 48 | | 30-39 | 27 | 75 | | 40-49 | 44 | 119 | | 50-59 | 21 | 140 | | 60-69 | 10 | 150 | Median class is 30-39 because CF reaches 75 here. 9. **Step 7: Estimate Mode:** Mode formula for grouped data: $$\text{Mode} = L + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times h$$ Where: - $L$ = lower class boundary of modal class - $f_1$ = frequency of modal class - $f_0$ = frequency of class before modal class - $f_2$ = frequency of class after modal class - $h$ = class width Modal class is 40-49 (highest frequency 44). $L = 39.5$ (true lower limit), $f_1 = 44$, $f_0 = 27$, $f_2 = 21$, $h = 10$ Calculate: $$\text{Mode} = 39.5 + \frac{44 - 27}{2 \times 44 - 27 - 21} \times 10 = 39.5 + \frac{17}{88 - 48} \times 10 = 39.5 + \frac{17}{40} \times 10 = 39.5 + 4.25 = 43.75$$ 10. **Step 8: Calculate Semi-Interquartile Range (SIQR):** SIQR = $\frac{Q_3 - Q_1}{2}$ Find $Q_1$ (25th percentile) and $Q_3$ (75th percentile) class intervals using cumulative frequency. - $Q_1$ position = $\frac{N}{4} = 37.5$ - $Q_3$ position = $\frac{3N}{4} = 112.5$ From CF table: - $Q_1$ class: 20-29 (CF before = 25, f = 23) - $Q_3$ class: 40-49 (CF before = 75, f = 44) Calculate $Q_1$: $$Q_1 = L + \frac{\frac{N}{4} - F}{f} \times h = 19.5 + \frac{37.5 - 25}{23} \times 10 = 19.5 + \frac{12.5}{23} \times 10 = 19.5 + 5.43 = 24.93$$ Calculate $Q_3$: $$Q_3 = L + \frac{\frac{3N}{4} - F}{f} \times h = 39.5 + \frac{112.5 - 75}{44} \times 10 = 39.5 + \frac{37.5}{44} \times 10 = 39.5 + 8.52 = 48.02$$ SIQR: $$\frac{48.02 - 24.93}{2} = \frac{23.09}{2} = 11.545$$ 11. **Step 9: Calculate 2nd Decile (D2):** Position of $D_2 = \frac{2N}{10} = 30$ $D_2$ class: 20-29 (CF before = 25, f = 23) $$D_2 = L + \frac{D_2\text{ position} - F}{f} \times h = 19.5 + \frac{30 - 25}{23} \times 10 = 19.5 + \frac{5}{23} \times 10 = 19.5 + 2.17 = 21.67$$ 12. **Step 10: Calculate 75th Percentile (P75):** Position of $P_{75} = \frac{75N}{100} = 112.5$ $P_{75}$ class: 40-49 (CF before = 75, f = 44) $$P_{75} = L + \frac{P_{75}\text{ position} - F}{f} \times h = 39.5 + \frac{112.5 - 75}{44} \times 10 = 39.5 + \frac{37.5}{44} \times 10 = 39.5 + 8.52 = 48.02$$ **Final Answers:** - Assumed Mean $\approx 36.7$ - Median Interval: 30-39 - Estimated Mode $\approx 43.75$ - Semi-Interquartile Range $\approx 11.545$ - 2nd Decile $\approx 21.67$ - 75th Percentile $\approx 48.02$