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The histogram below represents the score...

The histogram below represents the scores obtained by 25 students in a Mathematics mental test. Use the data to,
(i) Frame a frequency distribution table
(ii) To calculate mean

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To solve the problem step by step, we will first create a frequency distribution table based on the histogram data provided, and then we will calculate the mean of the scores. ### Step 1: Frame a Frequency Distribution Table 1. **Identify the Class Intervals**: From the histogram, we can identify the class intervals as follows: - 0 to 10 - 10 to 20 - 20 to 30 - 30 to 40 - 40 to 50 2. **Count the Frequencies**: We will count the number of students (frequency) for each class interval: - For 0 to 10: 2 students - For 10 to 20: 5 students - For 20 to 30: 8 students - For 30 to 40: 4 students - For 40 to 50: 6 students 3. **Create the Table**: | Class Interval | Frequency (f) | |----------------|----------------| | 0 - 10 | 2 | | 10 - 20 | 5 | | 20 - 30 | 8 | | 30 - 40 | 4 | | 40 - 50 | 6 | | **Total** | **25** | ### Step 2: Calculate the Mean 1. **Find the Midpoint (x) for Each Class Interval**: - For 0 to 10: Midpoint = (0 + 10)/2 = 5 - For 10 to 20: Midpoint = (10 + 20)/2 = 15 - For 20 to 30: Midpoint = (20 + 30)/2 = 25 - For 30 to 40: Midpoint = (30 + 40)/2 = 35 - For 40 to 50: Midpoint = (40 + 50)/2 = 45 2. **Calculate the Product of Midpoint and Frequency (fx)**: - For 0 to 10: fx = 5 * 2 = 10 - For 10 to 20: fx = 15 * 5 = 75 - For 20 to 30: fx = 25 * 8 = 200 - For 30 to 40: fx = 35 * 4 = 140 - For 40 to 50: fx = 45 * 6 = 270 3. **Create a New Table with fx**: | Class Interval | Frequency (f) | Midpoint (x) | fx | |----------------|----------------|---------------|------| | 0 - 10 | 2 | 5 | 10 | | 10 - 20 | 5 | 15 | 75 | | 20 - 30 | 8 | 25 | 200 | | 30 - 40 | 4 | 35 | 140 | | 40 - 50 | 6 | 45 | 270 | | **Total** | **25** | | **695** | 4. **Calculate the Mean**: - Use the formula for mean: \[ \text{Mean} = \frac{\sum (fx)}{\sum (f)} = \frac{695}{25} = 27.8 \] ### Final Result The mean of the scores obtained by the students is **27.8**. ---
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