Worked Examples: GRE Data Analysis

Example 1: Missing Value from a Mean (Medium)

The five numbers $3, 7, 7, x, 15$ have an average (arithmetic mean) of 8. What is the value of $x$?

Solution. Turn the mean into a sum: the total is $5 \times 8 = 40$. The four known values sum to $3 + 7 + 7 + 15 = 32$, so $x = 40 - 32 = 8.$

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Example 2: Comparing Standard Deviations (Medium)

Quantity A: the standard deviation of $\{10, 20, 30, 40, 50\}$ Quantity B: the standard deviation of $\{28, 29, 30, 31, 32\}$

Solution. Both sets have mean 30, but the first set's values sit far from the mean (distances up to 20) while the second set's values are clustered tightly (distances at most 2). Greater spread means greater standard deviation, so Quantity A is greater — answer (A). No numeric computation is required; SD measures spread.

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Example 3: Probability Without Replacement (Medium–Hard)

A bag contains 3 red marbles and 5 blue marbles. If two marbles are drawn at random without replacement, what is the probability that both are red?

Solution. First draw: $P(\text{red}) = \frac{3}{8}$. After removing one red, 2 red remain of 7 total, so the second draw is $\frac{2}{7}$. Multiply: $\frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}.$ The trap $\left(\frac{3}{8}\right)^2 = \frac{9}{64}$ assumes with replacement.

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Example 4: Combinations (Medium)

A committee of 3 people is to be selected from a group of 8. How many different committees are possible?

Solution. A committee is unordered, so use a combination: $C(8, 3) = \frac{8!}{3!\,5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56.$ The trap answer 336 is $P(8, 3)$ — the permutation count, which would apply only if the three roles were distinct (e.g., president, VP, treasurer).

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Example 5: Percent Change from a Table (Medium)

A store recorded daily sales: Mon $200, Tue $250, Wed $180, Thu $300, Fri $270. By approximately what percent did sales increase from Wednesday to Thursday?

Solution. Percent change uses the original (Wednesday) value as the base: $\frac{300 - 180}{180} \times 100\% = \frac{120}{180} \times 100\% \approx 66.7\%.$ The trap 40% divides by 300 (the new value) instead of 180.

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Example 6: Empirical Rule (Medium)

Test scores are normally distributed with a mean of 500 and a standard deviation of 100. Approximately what percent of scores fall between 400 and 600?

Solution. The interval $400$ to $600$ is $\mu \pm 1\sigma$ (that is, $500 \pm 100$). By the empirical rule, about 68% of a normal distribution lies within one standard deviation of the mean.