This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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To calculate the 3rd quartile, 5th decile, and 90th percentile for grouped data, we first need to determine the true class boundaries and cumulative frequencies. The given data is: | Score Interval | Number of Students (f) | | :------------- | :----------------------- | | 30-39 | 5 | | 40-49 | 8 | | 50-59 | 7 | | 60-69 | 10 | | 70-79 | 6 | | 80-89 | 4 | Step 1: Create a table with true class boundaries and cumulative frequencies. The class width (w) is 39.5 - 29.5 = 10. The total number of students (N) is the sum of frequencies: 5+8+7+10+6+4 = 40. | Score Interval | True Class Boundaries | Frequency (f) | Cumulative Frequency (cf) | | :------------- | :-------------------- | :-------------- | :-------------------------- | | 30-39 | 29.5 - 39.5 | 5 | 5 | | 40-49 | 39.5 - 49.5 | 8 | 13 | | 50-59 | 49.5 - 59.5 | 7 | 20 | | 60-69 | 59.5 - 69.5 | 10 | 30 | | 70-79 | 69.5 - 79.5 | 6 | 36 | | 80-89 | 79.5 - 89.5 | 4 | 40 | The formula for the k-th percentile (P_k) is: P_k = L + ((kN)/(100) - cf_bf_p) × w Where: L = lower class boundary of the percentile class N = total number of observations cf_b = cumulative frequency of the class before the percentile class f_p = frequency of the percentile class w = class width a) 3rd quartile (Q_3) Step 1: Determine the position of Q_3. The 3rd quartile is the 75th percentile (P_75). Position = (3N)/(4) = (3 × 40)/(4) = 30. Step 2: Identify the 3rd quartile class. The 30th observation falls in the class with a cumulative frequency of 30, which is 60-69 (true boundaries 59.5 - 69.5). L = 59.5 cf_b = 20 (cumulative frequency of the class before 60-69) f_p = 10 (frequency of the class 60-69) w = 10 Step 3: Calculate Q_3. Q_3 = 59.5 + ((30 - 20)/(10)) × 10 Q_3 = 59.5 + ((10)/(10)) × 10 Q_3 = 59.5 + 1 × 10 Q_3 = 59.5 + 10 Q_3 = 69.5 The 3rd quartile is 69.5. b) 5th decile (D_5) Step 1: Determine the position of D_5. The 5th decile is the 50th percentile (P_50). Position = (5N)/(10) = (5 × 40)/(10) = 20. Step 2: Identify the 5th decile class. The 20th observation falls in the class with a cumulative frequency of 20, which is 50-59 (true boundaries 49.5 - 59.5). L = 49.5 cf_b = 13 (cumulative frequency of the class before 50-59) f_p = 7 (frequency of the class 50-59) w = 10 Step 3: Calculate D_5. D_5 = 49.5 + ((20 - 13)/(7)) × 10 D_5 = 49.5 + ((7)/(7)) × 10 D_5 = 49.5 + 1 × 10 D_5 = 49.5 + 10 D_5 = 59.5 The 5th decile is 59.5. c) 90th percentile (P_90) Step 1: Determine the position of P_90. Position = (90N)/(100) = (90 × 40)/(100) = 36. Step 2: Identify the 90th percentile class. The 36th observation falls in the class with a cumulative frequency of 36, which is 70-79 (true boundaries 69.5 - 79.5). L = 69.5 cf_b = 30 (cumulative frequency of the class before 70-79) f_p = 6 (frequency of the class 70-79) w = 10 Step 3: Calculate P_90. P_90 = 69.5 + ((36 - 30)/(6)) × 10 P_90 = 69.5 + ((6)/(6)) × 10 P_90 = 69.5 + 1 × 10 P_90 = 69.5 + 10 P_90 = 79.5 The 90th percentile is 79.5.