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's the solution to question 17: Part (a): Find the value of a. Step 1: Calculate the midpoint (x) for each class interval and the product of frequency and midpoint (fx). The mean of grouped data is given by the formula: Mean = ( fx)/( f). | Amount (USD) | Number of workers (f) | Midpoint (x) | fx | |:--------------|:------------------------|:---------------|:-----| | 10 - 14 | 6 | (10+14)/(2) = 12 | 6 × 12 = 72 | | 15 - 24 | 16 | (15+24)/(2) = 19.5 | 16 × 19.5 = 312 | | 25 - 29 | a | (25+29)/(2) = 27 | a × 27 = 27a | | 30 - 44 | 10 | (30+44)/(2) = 37 | 10 × 37 = 370 | | 45 - 64 | 5 | (45+64)/(2) = 54.5 | 5 × 54.5 = 272.5 | Step 2: Calculate the sum of frequencies ( f) and the sum of fx ( fx). f = 6 + 16 + a + 10 + 5 = 37 + a fx = 72 + 312 + 27a + 370 + 272.5 = 1026.5 + 27a Step 3: Use the given mean wage to set up an equation. The mean wage is 27(11)/(16) USD. Convert this to an improper fraction: 27(11)/(16) = (27 × 16 + 11)/(16) = (432 + 11)/(16) = (443)/(16) Now, apply the mean formula: (1026.5 + 27a)/(37 + a) = (443)/(16) Step 4: Solve for a. Cross-multiply: 16(1026.5 + 27a) = 443(37 + a) 16424 + 432a = 16391 + 443a Rearrange the terms to isolate a: 16424 - 16391 = 443a - 432a 33 = 11a a = (33)/(11) a = 3 The value of a is 3. Part (b): Find the median class and its frequency. Step 1: Calculate the total number of workers ( f). Using a=3 from part (a): f = 37 + a = 37 + 3 = 40 Step 2: Determine the position of the median. The median position is ( f)/(2) = (40)/(2) = 20^th position. Step 3: Find the median class using cumulative frequency. | Amount (USD) | Number of workers (f) | Cumulative Frequency (CF) | |:--------------|:------------------------|:--------------------------| | 10 - 14 | 6 | 6 | | 15 - 24 | 16 | 6 + 16 = 22 | | 25 - 29 | 3 | 22 + 3 = 25 | | 30 - 44 | 10 | 25 + 10 = 35 | | 45 - 64 | 5 | 35 + 5 = 40 | The 20^th worker falls into the class where the cumulative frequency first exceeds 20. This is the 15 - 24 class, which has a cumulative frequency of 22. The median class is 15 - 24. The frequency of the median class is 16. Part (c): On the grid provided, draw a histogram to represent the information. To draw a histogram for data with unequal class widths, we need to calculate the frequency density for each class. Frequency Density = FrequencyClass Width Step 1: Determine class boundaries and class widths. | Amount (USD) | Number of workers (f) | Class Boundaries | Class Width | Frequency Density | |:--------------|:------------------------|:-----------------|:------------|:------------------| | 10 - 14 | 6 | 9.5 - 14.5 | 14.5 - 9.5 = 5 | (6)/(5) = 1.2 | | 15 - 24 | 16 | 14.5 - 24.5 | 24.5 - 14.5 = 10 | (16)/(10) = 1.6 | | 25 - 29 | 3 | 24.5 - 29.5 | 29.5 - 24.5 = 5 | (3)/(5) = 0.6 | | 30 - 44 | 10 | 29.5 - 44.5 | 44.5 - 29.5 = 15 | (10)/(15) ≈ 0.67 | | 45 - 64 | 5 | 44.5 - 64.5 | 64.5 - 44.5 = 20 | (5)/(20) = 0.25 | Step 2: Draw the histogram. • X-axis: Label "Amount (USD)". Mark the class boundaries: 9.5, 14.5, 24.5, 29.5, 44.5, 64.5. • Y-axis: Label "Frequency Density". Choose a suitable scale, for example, 1 unit on the grid represents 0.2 frequency density, going up to at least 1.8. Draw rectangles for each class: 1. 10 - 14 (9.5 - 14.5): Draw a bar from 9.5 to 14.5 on the x-axis, with a height of 1.2 on the y-axis. 2. 15 - 24 (14.5 - 24.5): Draw a bar from 14.5 to 24.5 on the x-axis, with a height of 1.6 on the y-axis. 3. **25 -