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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You're on a roll — To solve this problem, we first need to set up a frequency distribution table with additional columns for calculations using the assumed mean method. Given assumed mean A = 59.5. The class width c = 10 (e.g., 34.5 - 24.5 = 10). Step 1: Create a table for calculations. | Length (Class) | Frequency (f) | Midpoint (x) | d = x - A | u = (d)/(c) | fu | fu^2 | |:---------------|:----------------|:---------------|:------------|:----------------|:-----|:-------| | 25-34 | 9 | 29.5 | -30 | -3 | -27 | 81 | | 35-44 | 12 | 39.5 | -20 | -2 | -24 | 48 | | 45-54 | 55 | 49.5 | -10 | -1 | -55 | 55 | | 55-64 | 68 | 59.5 | 0 | 0 | 0 | 0 | | 65-74 | 32 | 69.5 | 10 | 1 | 32 | 32 | | 75-84 | 18 | 79.5 | 20 | 2 | 36 | 72 | | 85-94 | 6 | 89.5 | 30 | 3 | 18 | 54 | | Total | f = 200 | | | | fu = -20 | fu^2 = 342 | a) (i) The mean length The formula for the mean using the assumed mean method with coding is: x = A + c ( fu)/( f) Substitute the values from the table: x = 59.5 + 10 × (-20)/(200) x = 59.5 + 10 × (-0.1) x = 59.5 - 1 x = 58.5 The mean length is 58.5 mm. a) (ii) The variance The formula for the variance using the assumed mean method with coding is: ^2 = c^2 [ ( fu^2)/( f) - (( fu)/( f))^2 ] Substitute the values from the table: ^2 = 10^2 [ (342)/(200) - ((-20)/(200))^2 ] ^2 = 100 [ 1.71 - (-0.1)^2 ] ^2 = 100 [ 1.71 - 0.01 ] ^2 = 100 [ 1.70 ] ^2 = 170 The variance is 170 mm^2. a) (iii) The standard deviation The standard deviation is the square root of the variance: = sqrt(^2) = sqrt(170) ≈ 13.0384 Rounding to two decimal places: ≈ 13.04 The standard deviation is 13.04 mm. b) Find the percentage of leaves that has a length of 69.5 mm and above. First, determine the class boundaries: 25-34 becomes 24.5 - 34.5 35-44 becomes 34.5 - 44.5 45-54 becomes 44.5 - 54.5 55-64 becomes 54.5 - 64.5 65-74 becomes 64.5 - 74.5 75-84 becomes 74.5 - 84.5 85-94 becomes 84.5 - 94.5 We need to find the number of leaves with a length of 69.5 mm and above. • Leaves in the class 75-84 (74.5 - 84.5): 18 leaves • Leaves in the class 85-94 (84.5 - 94.5): 6 leaves • For the class 65-74 (64.5 - 74.5), which has 32 leaves, we need the portion from 69.5 mm to 74.5 mm. The class width is 74.5 - 64.5 = 10 mm. The portion of the interval we are interested in is $74.5 - 69.5 =