Setup: Standard deviation measures how far the values in a data set typically fall from the mean. All four tables use the same five values ($5, 6, 7, 8, 9$), each table is symmetric about $7$, and each has the same total frequency ($n = 75$). So every table has the same mean, $\bar{x} = 7$ — the only thing that separates them is how tightly the values cluster around that mean.

Compare the spread (sum of squared deviations from $7$):

• (A) $55$ of the $75$ values sit exactly at the mean; only $20$ sit one unit away: $10(1) + 10(1) = 20$
• (B) Values fan out to $5$ and $9$: $11(4) + 17(1) + 17(1) + 11(4) = 122$
• (C) Every value occurs equally often (flat): $15(4) + 15(1) + 15(1) + 15(4) = 150$
• (D) The heaviest frequencies sit at the extremes: $18(4) + 14(1) + 14(1) + 18(4) = 172$

Dividing each by $75$ and taking the square root: $\text{SD}_A \approx 0.52 < \text{SD}_B \approx 1.28 < \text{SD}_C \approx 1.41 < \text{SD}_D \approx 1.51$.

Answer: A

Common slips:

• Assuming the flat, uniform table (C) has the smallest spread. A uniform distribution is actually quite wide — it puts just as many values at $5$ and $9$ as at $7$.
• Choosing (D) because its bars look tall. Its peaks are at the extremes, which maximizes spread, not minimizes it.
• Computing full standard deviations for all four tables. Since the mean and $n$ are identical, you only need to compare how concentrated each set is around $7$.