1. **Problem Statement:** (a) Explain why organizing raw data is necessary and define frequency and frequency distribution. (b) Given tea production data (in 000 kg) per day, construct a frequency distribution using the exclusive method, draw an ogive curve, find the first and third quartiles, and calculate the quartile deviation. 2. **Why organize raw data?** Organizing raw data helps to summarize and simplify large data sets, making it easier to analyze and interpret patterns or trends. 3. **Definitions:** - **Frequency:** The number of times a particular value or range of values occurs in the data. - **Frequency Distribution:** A table that shows the frequency of various outcomes in a sample. 4. **Data given:** 3.2, 6.2, 6.1, 6.4, 3.5, 6.8, 6.9, 6.8, 6.8, 6.5, 8.2, 3.8, 9.3, 3.9, 9.8, 9.6, 4.5, 7.6, 4.7, 7.6, 8.1, 4.0, 7.1, 4.2, 4.9, 4.7, 8.7, 8.5, 4.4, 5.1 5. **Step (i): Construct frequency distribution using exclusive method** - Find minimum and maximum values: min = 3.2, max = 9.8 - Choose class width: approx 1 unit - Classes (exclusive upper limit): 3.0-4.0, 4.0-5.0, 5.0-6.0, 6.0-7.0, 7.0-8.0, 8.0-9.0, 9.0-10.0 6. **Count frequencies:** - 3.0-4.0: 3.2, 3.5, 3.8, 3.9 → 4 - 4.0-5.0: 4.0, 4.2, 4.4, 4.5, 4.7, 4.7, 4.9 → 7 - 5.0-6.0: 5.1 → 1 - 6.0-7.0: 6.1, 6.2, 6.4, 6.5, 6.8, 6.8, 6.8, 6.9 → 8 - 7.0-8.0: 7.1, 7.6, 7.6 → 3 - 8.0-9.0: 8.1, 8.2, 8.5, 8.7 → 4 - 9.0-10.0: 9.3, 9.6, 9.8 → 3 7. **Frequency distribution table:** | Class Interval | Frequency | | 3.0 - 4.0 | 4 | | 4.0 - 5.0 | 7 | | 5.0 - 6.0 | 1 | | 6.0 - 7.0 | 8 | | 7.0 - 8.0 | 3 | | 8.0 - 9.0 | 4 | | 9.0 - 10.0 | 3 | 8. **Step (ii): Draw ogive and find quartiles** - Calculate cumulative frequencies: - 3.0-4.0: 4 - 4.0-5.0: 4+7=11 - 5.0-6.0: 11+1=12 - 6.0-7.0: 12+8=20 - 7.0-8.0: 20+3=23 - 8.0-9.0: 23+4=27 - 9.0-10.0: 27+3=30 - Total observations $n=30$ - First quartile $Q_1$ position: $\frac{n}{4} = \frac{30}{4} = 7.5$ - Third quartile $Q_3$ position: $\frac{3n}{4} = \frac{90}{4} = 22.5$ 9. **Find $Q_1$:** - $Q_1$ lies in cumulative frequency 11 (class 4.0-5.0) - Class lower boundary $l=4.0$, cumulative frequency before class $F=4$, class frequency $f=7$, class width $h=1$ - Use formula: $$Q_1 = l + \left(\frac{\frac{n}{4} - F}{f}\right) \times h = 4.0 + \left(\frac{7.5 - 4}{7}\right) \times 1 = 4.0 + 0.5 = 4.5$$ 10. **Find $Q_3$:** - $Q_3$ lies in cumulative frequency 23 (class 7.0-8.0) - Class lower boundary $l=7.0$, cumulative frequency before class $F=20$, class frequency $f=3$, class width $h=1$ - Use formula: $$Q_3 = l + \left(\frac{\frac{3n}{4} - F}{f}\right) \times h = 7.0 + \left(\frac{22.5 - 20}{3}\right) \times 1 = 7.0 + 0.8333 = 7.8333$$ 11. **Calculate quartile deviation:** $$QD = \frac{Q_3 - Q_1}{2} = \frac{7.8333 - 4.5}{2} = \frac{3.3333}{2} = 1.6667$$ **Final answers:** - Frequency distribution as above - $Q_1 = 4.5$ - $Q_3 = 7.8333$ - Quartile deviation $= 1.6667$