1. **Problem Statement:** Given the age distribution of patients admitted to a hospital and their frequencies, find the mode and mean and then compare these measures of central tendency. 2. **Data Table:** Age intervals: 5-15, 15-25, 25-35, 35-45, 45-55, 55-65 Number of patients (frequency): 6, 11, 21, 23, 14, 5 3. **Find the Mode:** The mode is the class interval with the highest frequency. Here, max frequency = 23 in the interval 35-45. So, mode class is 35-45. 4. **Calculate Mean:** Step 1: Find midpoints (x) of each class: - 5 - 15: $\frac{5+15}{2}=10$ - 15 - 25: $\frac{15+25}{2}=20$ - 25 - 35: $\frac{25+35}{2}=30$ - 35 - 45: $\frac{35+45}{2}=40$ - 45 - 55: $\frac{45+55}{2}=50$ - 55 - 65: $\frac{55+65}{2}=60$ Step 2: Multiply midpoints by frequencies to find $f \times x$: - $6 \times 10 = 60$ - $11 \times 20 = 220$ - $21 \times 30 = 630$ - $23 \times 40 = 920$ - $14 \times 50 = 700$ - $5 \times 60 = 300$ Step 3: Find sum of frequencies, $\sum f = 6+11+21+23+14+5=80$ Step 4: Find sum of $f \times x$, $\sum fx = 60+220+630+920+700+300=2830$ Step 5: Calculate mean $$\text{Mean} = \frac{\sum fx}{\sum f} = \frac{2830}{80} = 35.375$$ 5. **Interpretation and Comparison:** - The mode class is 35-45 years, meaning most patients fall within this age group. - The mean age is approximately 35.38 years, which also lies in the 35-45 interval. - Since the mean and mode are close, the age distribution is roughly symmetric or moderately skewed. - Mode gives the most frequent age group, while mean gives the average age of patients. **Final answers:** - Mode class: 35-45 years - Mean age: 35.375 years