The following table shows the number of working hours and the number of employees employed in a small scale industry. No. of working hours | No. of employees $3 - 5$ | $7$ $5 - 7$ | $10$ $7 - 9$ | $18$ $9 - 11$ | $57$ $11 - 13$ | $14$ $13 - 15$ | $8$ The average number of working hours of an employee is

### Concept & Formula When data is provided in class intervals (like $3-5$, $5-7$), we must first find the midpoint (class mark) of each interval to represent that group. The average of grouped data is then found using the weighted mean formula. $$ \text{Average} = \frac{\sum (f_i \times x_i)}{\sum f_i} $$ *(Where $f_i$ is the frequency and $x_i$ is the midpoint of the interval)* ### Step-by-Step Solution * **Given:** We need the midpoints ($x$) for each working hour interval and the corresponding number of employees ($f$). * **Calculation:** Step 1: Calculate the midpoints ($x$) for each interval. $3-5 \rightarrow 4$ $5-7 \rightarrow 6$ $7-9 \rightarrow 8$ $9-11 \rightarrow 10$ $11-13 \rightarrow 12$ $13-15 \rightarrow 14$ Step 2: Multiply each midpoint by its frequency ($f \times x$). $7 \times 4 = 28$ $10 \times 6 = 60$ $18 \times 8 = 144$ $57 \times 10 = 570$ $14 \times 12 = 168$ $8 \times 14 = 112$ Step 3: Sum the frequencies and the products. Total Employees ($\sum f$) = $7 + 10 + 18 + 57 + 14 + 8 = 114$ Total Hours ($\sum f \times x$) = $28 + 60 + 144 + 570 + 168 + 112 = 1082$ Step 4: Calculate the average. $\text{Average} = \frac{1082}{114} \approx 9.49$ Rounding to one decimal place gives $9.5$. ### Exam Strategy & Shortcut **Assumed Mean (Deviation) Method:** To avoid multiplying and adding large numbers, guess a central mean, say $10$. Calculate how far each midpoint is from $10$: $-6, -4, -2, 0, +2, +4$. Multiply these deviations by frequencies: $7(-6) = -42$, $10(-4) = -40$, $18(-2) = -36$, $57(0) = 0$, $14(2) = 28$, $8(4) = 32$. Sum of deviations = $-118 + 60 = -58$. Average deviation = $\frac{-58}{114} \approx -0.5$. Actual Average = Assumed Mean ($10$) - $0.5 = 9.5$. This involves much smaller numbers! ### Common Pitfall A very common mistake is attempting to calculate the average using the boundaries of the class intervals directly (like adding $3+5+7...$) instead of converting them into representative midpoints first. Always find the midpoint for grouped frequency tables. ### Final Answer **Therefore, the correct answer is 9.5.**