Question 4 of 17intermediate🔧 ApplyNumerical4 marks

A dart-throwing competition was organised in a school. The number of throws participants took to hit the bull's eye (the centre circle) is given in the table below. Describe the data using its minimum, maximum, mean and median.

Correct Answer

Minimum: 1 throw Maximum: 10 throws

Mean: Mean = Total throws / Total students = (1×11 \times 1 + 4×14 \times 1 + 5×45 \times 4 + 6×96 \times 9 + 7×127 \times 12 + 8×158 \times 15 + 9×109 \times 10 + 10×1010 \times 10) / 62 ≈\approx 7.6 throws

Median: Median = Average of 31st and 32nd values = 8 throws

Exercise: Figure it Out (Section 1) | Q: 11 | (Chapter: Page 14)
For More Understanding

Explanation

The chapter explains calculating mean and median from frequency tables. For mean, multiply each value by its frequency, sum these products, and divide by total frequency. For median, find the cumulative frequency to locate the middle position(s).

Solution Steps

  1. Step 1: Calculate total students: 1 + 0 + 0 + 1 + 4 + 9 + 12 + 15 + 10 + 10 = 62

  2. Step 2: Calculate weighted sum: (1×11 \times 1) + (4×14 \times 1) + (5×45 \times 4) + (6×96 \times 9) + (7×127 \times 12) + (8×158 \times 15) + (9×109 \times 10) + (10×1010 \times 10) = 1 + 4 + 20 + 54 + 84 + 120 + 90 + 100 = 473

  3. Step 3: Mean = 473/62 ≈\approx 7.6

  4. Step 4: For median, find cumulative frequencies: 1, 1, 1, 2, 6, 15, 27, 42, 52, 62

  5. Step 5: With 62 values, median is average of 31st and 32nd values

  6. Step 6: Both 31st and 32nd values fall in the '8 throws' category (cumulative 27 to 42)

  7. Step 7: Median = 8

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