1. **Problem statement:** We analyze the conversion rate (TC) data from various graphical representations to answer four questions. 2. **Maximum value in dot plot:** The highest point in the dot plot is the outlier near 75. 3. **Tenth smallest value (TC10):** Using the stem-and-leaf plot, list values in ascending order: - Stem 0: 1.4, 2.7, 3.9, 5, 6.2, 7.4, 8.6, 9.8 (8 values) - Stem 1: 1.1, 2.4 (2 values) The 10th value is the 2nd value in stem 1: 2.4 4. **Frequency of third class:** From the histogram description, the largest bar is the first interval, so the third class frequency is the third bar's height. Without exact numbers, we count the number of points in the third class from the stem-and-leaf plot or histogram data. Assuming classes correspond to stems: - Stem 0 (class 1): 8 values - Stem 1 (class 2): 6 values - Stem 2 (class 3): 5 values Frequency of third class = 5 5. **Third quartile Q3:** To find Q3, order all values and find the 75th percentile. Total values: 8 + 6 + 5 + 3 + 1 + 1 + 1 = 25 Position of Q3 = 0.75 * (25 + 1) = 19.5th value List values in order: 0: 1.4, 2.7, 3.9, 5, 6.2, 7.4, 8.6, 9.8 (1-8) 1: 1.1, 2.4, 3.8, 5.2, 6.7, 8.3 (9-14) 2: 0, 1.8, 3.8, 5.9, 8.3 (15-19) 3: 1, 4.2, 8 (20-22) 4: 2.9 (23) 5: 0.4 (24) 7: 4.6 (25) The 19th value is 8.3 (last in stem 2), 20th is 1 (stem 3). Interpolating between 19th and 20th: $$Q3 = 8.3 + 0.5 \times (1 - 8.3) = 8.3 - 3.65 = 4.65$$ But since 20th value is smaller, Q3 is better taken as 8.3 (the 19th value) or average of 19th and 20th values. Usually, Q3 is the value at or above 75%, so Q3 = 8.3 **Final answers:** 1. Maximum value $\approx 75$ 2. $TC_{10} = 2.4$ 3. Frequency of third class = 5 4. $Q_3(TC) = 8.3$