Q.The wages (in ₹) of 7 workers in a small unit are: 8000, 9000, 9000, 10000, 11000, 9000, 12000. Find the mean, median and mode of the wages.

Sum and divide by $n$ for the mean; sort and pick the middle value for the median; count frequencies for the mode.

Mean $=68000/7\approx9714.3$. Sorted: 8000,9000,9000,9000,10000,11000,12000 — median (4th value) $=9000$. Mode $=9000$ (appears 3 times).

Step 1 — Find the mean. Sum $=8000+9000+9000+10000+11000+9000+12000=68{,}000$. $n=7$. $\bar x = 68{,}000/7 \approx 9714.3$.

Step 2 — Find the median. Sorted ascending: $8000, 9000, 9000, 9000, 10000, 11000, 12000$. With $n=7$ (odd), the median is the $\left(\dfrac{7+1}{2}\right)^{\text{th}}=4^{\text{th}}$ value, which is $9000$.

Step 3 — Find the mode. Frequencies: $8000$ (once), $9000$ (three times), $10000$ (once), $11000$ (once), $12000$ (once). Mode $=9000$.

Independent check. All three measures — mean ($\approx9714$), median ($9000$) and mode ($9000$) — sit close together, with the mean pulled slightly higher than the other two by the single larger wage of ₹12,000; this is exactly the expected pattern for a mildly right-skewed (positively skewed) small data set with one high outlying value, confirming the calculations are internally consistent.

Reporting the median as the average of the 3rd and 4th sorted values (a rule that only applies when $n$ is even) is a common error here, since $n=7$ is odd and has a single, uniquely defined middle value.