1. **Problem Statement:** We have the ages of 20 teachers: 23, 20, 24, 26, 29, 39, 35, 37, 39, 25, 24, 20, 28, 30, 34, 39, 45, 54, 58. We need to create a frequency distribution table with class intervals of 5 years, then find the mean and median age. 2. **Step 1: Create Frequency Distribution Table** - Class intervals of 5 years mean intervals like 20-24, 25-29, 30-34, etc. - Count how many ages fall into each interval. | Class Interval | Frequency | |----------------|-----------| | 20 - 24 | 5 | | 25 - 29 | 3 | | 30 - 34 | 3 | | 35 - 39 | 4 | | 40 - 44 | 0 | | 45 - 49 | 1 | | 50 - 54 | 1 | | 55 - 59 | 1 | 3. **Step 2: Calculate the Mean** - Use the formula for mean from grouped data: $$\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}$$ where $f_i$ is frequency and $x_i$ is the class midpoint. - Calculate midpoints $x_i$: - 20-24 midpoint = $\frac{20+24}{2} = 22$ - 25-29 midpoint = $\frac{25+29}{2} = 27$ - 30-34 midpoint = $\frac{30+34}{2} = 32$ - 35-39 midpoint = $\frac{35+39}{2} = 37$ - 40-44 midpoint = $\frac{40+44}{2} = 42$ - 45-49 midpoint = $\frac{45+49}{2} = 47$ - 50-54 midpoint = $\frac{50+54}{2} = 52$ - 55-59 midpoint = $\frac{55+59}{2} = 57$ - Calculate $f_i x_i$: - $5 \times 22 = 110$ - $3 \times 27 = 81$ - $3 \times 32 = 96$ - $4 \times 37 = 148$ - $0 \times 42 = 0$ - $1 \times 47 = 47$ - $1 \times 52 = 52$ - $1 \times 57 = 57$ - Sum frequencies: $\sum f_i = 5 + 3 + 3 + 4 + 0 + 1 + 1 + 1 = 18$ (Note: We have 20 teachers, so recount frequencies carefully.) - Recount frequencies: - 20-24: 23,20,24,24,20 = 5 - 25-29: 26,29,25,28 = 4 - 30-34: 30,34 = 2 - 35-39: 35,37,39,39,39 = 5 - 40-44: none = 0 - 45-49: 45 = 1 - 50-54: 54 = 1 - 55-59: 58 = 1 - Correct frequencies: - 20-24: 5 - 25-29: 4 - 30-34: 2 - 35-39: 5 - 40-44: 0 - 45-49: 1 - 50-54: 1 - 55-59: 1 - Recalculate $f_i x_i$: - $5 \times 22 = 110$ - $4 \times 27 = 108$ - $2 \times 32 = 64$ - $5 \times 37 = 185$ - $0 \times 42 = 0$ - $1 \times 47 = 47$ - $1 \times 52 = 52$ - $1 \times 57 = 57$ - Sum $f_i x_i = 110 + 108 + 64 + 185 + 0 + 47 + 52 + 57 = 623$ - Sum frequencies $\sum f_i = 5 + 4 + 2 + 5 + 0 + 1 + 1 + 1 = 19$ (Still 19, but we have 20 teachers, check again.) - Check original data count: 23,20,24,26,29,39,35,37,39,25,24,20,28,30,34,39,45,54,58 (19 numbers listed, user gave 20 teachers but only 19 ages? Recount original list carefully.) - Original ages count: 23, 20, 24, 26, 29, 39, 35, 37, 39, 25, 24, 20, 28, 30, 34, 39, 45, 54, 58 - Count: 19 ages only. - So total teachers = 19. - Sum frequencies = 19 correct. - Calculate mean: $$\text{Mean} = \frac{623}{19} \approx 32.79$$ 4. **Step 3: Calculate the Median** - Median is the middle value when data is ordered. - Since $n=19$ (odd), median position is $\frac{19+1}{2} = 10^{th}$ value. - Order data: 20,20,23,24,24,25,26,28,29,30,34,35,37,39,39,39,39,45,54,58 - The 10th value is 30. - So, median age = 30. **Final answers:** - Frequency distribution table as above. - Mean age $\approx 32.79$ - Median age = 30