Word Length Stats 4Ba28E

1. **Problem Statement:** We are given frequency data for the number of letters in words and asked to find the mean and a probability for part (b), and then find the median and range for a list of word lengths in part (c). --- ### Part (b) 2. **Calculate the mean number of letters:** The mean is given by the formula: $$\text{mean} = \frac{\sum (\text{number of letters} \times \text{frequency})}{\sum \text{frequency}}$$ 3. **Calculate total frequency:** $$10 + 18 + 9 + 6 + 5 + 2 = 50$$ 4. **Calculate sum of products:** $$1 \times 10 + 2 \times 18 + 3 \times 9 + 4 \times 6 + 5 \times 5 + 6 \times 2$$ $$= 10 + 36 + 27 + 24 + 25 + 12 = 134$$ 5. **Calculate mean:** $$\text{mean} = \frac{134}{50}$$ Show cancellation: $$\frac{\cancel{134}}{\cancel{50}} = 2.68$$ 6. **Answer for (b)(i):** The mean number of letters is **2.68**. --- 7. **Find the probability that a randomly chosen word has more than 3 letters:** Words with more than 3 letters have lengths 4, 5, or 6. 8. **Sum frequencies for letters > 3:** $$6 + 5 + 2 = 13$$ 9. **Calculate probability:** $$P(>3) = \frac{13}{50} = 0.26$$ 10. **Answer for (b)(ii):** The probability is **0.26**. --- ### Part (c) 11. **Given data:** $$3, 4, 1, 7, 9, 2, 6, 5, 4, 2, 3, 2$$ 12. **Sort the data:** $$1, 2, 2, 2, 3, 3, 4, 4, 5, 6, 7, 9$$ 13. **Find the median:** Since there are 12 numbers (even), median is average of 6th and 7th values. 14. **Identify 6th and 7th values:** 6th value = 3, 7th value = 4 15. **Calculate median:** $$\text{median} = \frac{3 + 4}{2} = 3.5$$ 16. **Answer for (c)(i):** The median is **3.5**. --- 17. **Find the range:** $$\text{range} = \text{max} - \text{min} = 9 - 1 = 8$$ 18. **Answer for (c)(ii):** The range is **8**.