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 — here's the solution for question 17. Question 17: The table below shows the wages in US dollars earned by technical developers in an ICT firm. | Amount (USD) | 10 - 14 | 15 - 24 | 25 - 29 | 30 - 44 | 45 - 64 | | :----------- | :------ | :------ | :------ | :------ | :------ | | Number of workers | 6 | 16 | a | 10 | 5 | (a) Given that the mean wage is 27 (11)/(16) USD, find the value of a. Step 1: Calculate the midpoint (x) for each class. For grouped data, the mean is given by the formula: Mean = ( (f × x))/( f). | Class (Amount USD) | Frequency (f) | Midpoint (x) | f × x | | :----------------- | :-------------- | :------------- | :----------- | | 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 (f × x) ( (f × x)). f = 6 + 16 + a + 10 + 5 = 37 + a (f × x) = 72 + 312 + 27a + 370 + 272.5 = 1026.5 + 27a Step 3: Set up the equation for the mean. The given mean wage is 27 (11)/(16) USD. 27 (11)/(16) = (27 × 16 + 11)/(16) = (432 + 11)/(16) = (443)/(16) Using 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 Subtract 432a from both sides: 16424 = 16391 + 443a - 432a 16424 = 16391 + 11a Subtract 16391 from both sides: 16424 - 16391 = 11a 33 = 11a a = (33)/(11) a = 3 The value of a is 3. (b) Find the median class and its frequency. Step 1: Use the value of a=3 to list the frequencies and calculate the total number of workers. Frequencies: 6, 16, 3, 10, 5. Total number of workers ( f) = 6 + 16 + 3 + 10 + 5 = 40. Step 2: Determine the position of the median. The median position is ( f)/(2) = (40)/(2) = 20^th value. Step 3: Calculate the cumulative frequencies to find the median class. | Class (Amount USD) | Frequency (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 value falls into the class where the cumulative frequency first exceeds or equals 20. This is the class 15 - 24, which has a cumulative frequency of 22. The median class is 15 - 24. The frequency of the median class is 16. (c) On the grid provided, draw a histogram to represent the information. To draw a histogram for this data, you would follow these steps: 1. Determine Class Boundaries: Adjust the class intervals to be continuous. For example, 10-14 becomes 9.5-14.5, 15-24 becomes 14.5-24.5, and so on. 2. Calculate Class Widths: The width of each class interval (e.g., 14.5 - 9.5 = 5). 3. Calculate Frequency Densities: Since the class widths are not uniform, you need to calculate frequency density for each class using the formula: Frequency Density = FrequencyClass Width. 10-14: (6)/(5) = 1.2 15-24: (16)/(10) = 1.6 25-29: (3)/(5) = 0.6 30-44: (10)/(15) ≈ 0.67 45-64: (5)/(20) = 0.25 4. Draw Axes: The x-axis represents the Amount (USD) using the class boundaries. The y-axis represents the Frequency Density. 5. Draw Bars: For each class, draw a rectangular bar with a width corresponding to its class width and a height corresponding to its frequency density. The bars should touch each other. What's next?