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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Here are the solutions to the problems. Problem 5: To find the mean weight using the assumed mean method, we first determine the midpoint (x) for each weight class, then calculate the deviation (d) from the assumed mean (A=36), and finally apply the formula x = A + ( fd)/( f). Step 1: Calculate the midpoints (x), deviations (d = x - 36), and fd values for each class. | Weight Range | Midpoint (x) | No. of Livestock (f) | d = x - 36 | fd | | :----------- | :------------- | :--------------------- | :----------- | :--- | | 20 - 24 | 22 | 5 | -14 | -70 | | 25 - 29 | 27 | 15 | -9 | -135 | | 30 - 34 | 32 | 12 | -4 | -48 | | 35 - 39 | 37 | 9 | 1 | 9 | | 40 - 44 | 42 | 6 | 6 | 36 | | 45 - 49 | 47 | 3 | 11 | 33 | Step 2: Sum the frequencies ( f) and the fd values ( fd). f = 5 + 15 + 12 + 9 + 6 + 3 = 50 fd = -70 - 135 - 48 + 9 + 36 + 33 = -175 Step 3: Apply the assumed mean formula. x = A + ( fd)/( f) x = 36 + (-175)/(50) x = 36 - 3.5 x = 32.5 The mean weight of the livestock is 32.5. Problem 6: We need to form committees of 4 persons from a group of 9 persons. This is a combination problem, as the order of selection does not matter. The formula for combinations is C(n, k) = nk = (n!)/(k!(n-k)!). a) there are no restrictions? Step 1: Calculate the number of ways to choose 4 persons from 9 without any restrictions. C(9, 4) = (9!)/(4!(9-4)!) = (9!)/(4!5!) C(9, 4) = (9 × 8 × 7 × 6)/(4 × 3 × 2 × 1) C(9, 4) = 9 × 2 × 7 = 126 There are 126 possible committees. b) both Kwame and Rita must be on the committee? Step 1: If Kwame and Rita must be on the committee, 2 spots are already filled. We need to choose 4 - 2 = 2 more persons. Step 2: The remaining pool of people to choose from is 9 - 2 = 7 persons (excluding Kwame and Rita). Step 3: Calculate the number of ways to choose 2 persons from these 7. C(7, 2) = (7!)/(2!(7-2)!) = (7!)/(2!5!) C(7, 2) = (7 × 6)/(2 × 1) C(7, 2) = 21 There are 21 possible committees. c) either Kwame or Rita, but not both, must be on the committee? This means we consider two mutually exclusive cases: Case 1: Kwame is on the committee, and Rita is not. Case 2: Rita is on the committee, and Kwame is not. Step 1: Calculate Case 1 (Kwame is on, Rita is not). If Kwame is on, 1 spot is filled. We need to choose 4 - 1 = 3 more persons. If Rita is not on, she is excluded from the selection pool. The remaining pool of people to choose from is 9 - 1 (Kwame) - 1 (Rita) = 7 persons. Number of ways for Case 1: Choose 3 persons from these 7. C(7, 3) = (7!)/(3!(7-3)!) = (7!)/(3!4!) C(7, 3) = (7 × 6 × 5)/(3 × 2 × 1) C(7, 3) = 35 Step 2: Calculate Case 2 (Rita is on, Kwame is not). This is symmetrical to Case 1. If Rita is on, 1 spot is filled. We need to choose 4 - 1 = 3 more persons. If Kwame is not on, he is excluded from the selection pool. The remaining pool of people to choose from is 9 - 1 (Rita) - 1 (Kwame) = 7 persons. Number of ways for Case 2: Choose 3 persons from these 7. C(7, 3) = 35 Step 3: Add the possibilities from both cases. Total ways = 35 + 35 = 70. There are 70 possible committees. Send me the next one 📸