Q.The daily sales (in ₹ thousand) of a small retail shop over 6 days were: 10, 12, 8, 15, 14, 13. Find the mean deviation about the mean.

This is a set of 6 individual observations, so we first find the mean, then average the absolute distance of each value from it.

Mean $= 72/6 = 12$. Absolute deviations: $2,0,4,3,2,1$; sum $=12$.

$\text{MD}(\bar{x}) = 12/6 = 2$

Step 1 — Find the mean. $\sum x = 10+12+8+15+14+13 = 72$, $n=6$, so $\bar{x} = 72/6 = 12$.

Step 2 — Find the absolute deviations from the mean.

Step 3 — Average the absolute deviations. $\sum|x-\bar{x}| = 2+0+4+3+2+1 = 12$.

Independent check. Re-adding the deviations sorted by value ($8,10,12,13,14,15$, deviations $4,2,0,1,2,3$): $4+2+0+1+2+3 = 12$ — the same total, confirming the sum.

Forgetting to take the absolute value (letting negative and positive deviations cancel to give a mean deviation of 0) is the single most common error — always take $|x-\text{mean}|$ before averaging, never the plain signed deviation. Also recount $n$ correctly (6 days here) before dividing.