1. **Problem Statement:** Calculate the geometric mean of the monthly wages given the frequency distribution of workers. 2. **Formula for Geometric Mean:** The geometric mean $G$ for grouped data is given by: $$G = \exp\left(\frac{\sum f_i \ln x_i}{\sum f_i}\right)$$ where $f_i$ is the frequency of the $i$th class and $x_i$ is the class midpoint. 3. **Step 1: Find the midpoints $x_i$ of each class interval:** - For 13-17: $x_1 = \frac{13 + 17}{2} = 15$ - For 18-22: $x_2 = \frac{18 + 22}{2} = 20$ - For 23-27: $x_3 = \frac{23 + 27}{2} = 25$ - For 28-32: $x_4 = \frac{28 + 32}{2} = 30$ - For 33-37: $x_5 = \frac{33 + 37}{2} = 35$ - For 38-42: $x_6 = \frac{38 + 42}{2} = 40$ - For 43-47: $x_7 = \frac{43 + 47}{2} = 45$ - For 48-52: $x_8 = \frac{48 + 52}{2} = 50$ - For 53-57: $x_9 = \frac{53 + 57}{2} = 55$ 4. **Step 2: List frequencies $f_i$:** - $f_1 = 2$, $f_2 = 22$, $f_3 = 19$, $f_4 = 14$, $f_5 = 3$, $f_6 = 4$, $f_7 = 6$, $f_8 = 1$, $f_9 = 1$ 5. **Step 3: Calculate $f_i \ln x_i$ for each class:** - $2 \times \ln 15 = 2 \times 2.70805 = 5.4161$ - $22 \times \ln 20 = 22 \times 2.99573 = 65.9061$ - $19 \times \ln 25 = 19 \times 3.21888 = 61.1587$ - $14 \times \ln 30 = 14 \times 3.40120 = 47.6168$ - $3 \times \ln 35 = 3 \times 3.55535 = 10.6660$ - $4 \times \ln 40 = 4 \times 3.68888 = 14.7555$ - $6 \times \ln 45 = 6 \times 3.80666 = 22.8399$ - $1 \times \ln 50 = 1 \times 3.91202 = 3.9120$ - $1 \times \ln 55 = 1 \times 4.00733 = 4.0073$ 6. **Step 4: Sum frequencies and weighted logs:** - Total frequency $N = 2 + 22 + 19 + 14 + 3 + 4 + 6 + 1 + 1 = 72$ - Sum $\sum f_i \ln x_i = 5.4161 + 65.9061 + 61.1587 + 47.6168 + 10.6660 + 14.7555 + 22.8399 + 3.9120 + 4.0073 = 236.2784$ 7. **Step 5: Calculate geometric mean:** $$G = \exp\left(\frac{236.2784}{72}\right) = \exp(3.2816) = 26.65$$ **Final answer:** The geometric mean of the monthly wages is approximately **26.65**.