Mean Height 2Bbc6C

1. **State the problem:** We have a grouped frequency table of heights and frequencies. We want to estimate the mean height using the grouped data (part a), then explain the difference from the actual mean (part b). 2. **Formula for estimated mean from grouped data:** $$\text{Estimated mean} = \frac{\sum (f \times x)}{\sum f}$$ where $f$ is the frequency of each class and $x$ is the midpoint of each class interval. 3. **Calculate midpoints for each class:** - For $150 < h \leq 160$, midpoint $= \frac{150 + 160}{2} = 155$ - For $160 < h \leq 170$, midpoint $= \frac{160 + 170}{2} = 165$ - For $170 < h \leq 180$, midpoint $= \frac{170 + 180}{2} = 175$ - For $180 < h \leq 190$, midpoint $= \frac{180 + 190}{2} = 185$ 4. **Multiply each midpoint by its frequency:** - $155 \times 3 = 465$ - $165 \times 9 = 1485$ - $175 \times 14 = 2450$ - $185 \times 5 = 925$ 5. **Sum the products and frequencies:** - Sum of $f \times x = 465 + 1485 + 2450 + 925 = 5325$ - Sum of frequencies $= 3 + 9 + 14 + 5 = 31$ 6. **Calculate the estimated mean:** $$\text{Estimated mean} = \frac{5325}{31} \approx 171.7742$$ Rounded to 1 decimal place: $$171.8$$ 7. **Answer for part a:** The estimated mean height is $171.8$ cm. 8. **Explain difference for part b:** The estimated mean uses midpoints of grouped intervals, assuming all data in a class is at the midpoint. The actual mean $176.1$ cm is calculated from the original ungrouped data, which is more precise. Grouping causes loss of detail and can lead to differences between estimated and actual means. **Final answers:** - a) Estimated mean height $= 171.8$ cm - b) Difference arises because the estimate uses grouped data midpoints, which approximate the actual data distribution, causing some error compared to the actual mean calculated from raw data.