GRE Data Analysis & Statistics Practice & Questions

1. Set $S = \{4, 8, 11, 15, 22\}$ has $5$ elements. The median is the middle element: $11$. 2. Set $T = \{4, 8, 11, 11, 15, 22\}$ has $6$ elements. The median is the average of the two middle numbers: $\frac{11 + 11}{2} = 11$. 3. Therefore, the median remains exactly $11$, so the median of $T$ is equal to the median of $S$. (Note: The mean of $S$ is $\frac{60}{5} = 12$. Adding $11$ lowers the mean to $\frac{71}{6} \approx 11.83$, so statement A is false. The range remains $22 - 4 = 18$, so C is false.)

When an added element equals the existing median of an odd-count sorted list, the new median of the even-count list is guaranteed to be unchanged.

Use the complement principle (Total $-$ Unwanted): 1. Total ways to select any $4$ members from the total pool of $5 + 4 = 9$ scientists: $$\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$$ 2. Unwanted scenario: committees with ZERO biologists (all $4$ members chosen exclusively from the $5$ chemists): $$\binom{5}{4} = 5$$ 3. Subtract the unwanted from the total: $$\text{Committees with at least 1 biologist} = 126 - 5 = 121$$

Examinees who calculate by cases (1 bio + 2 bio + 3 bio + 4 bio) often make arithmetic errors or inadvertently double-count committees by selecting "1 biologist first, then 3 remaining". The complement method is airtight.

For "at least one" counting problems, always use: $\text{Total} - \text{None}$.

According to the Empirical Rule (68–95–99.7 rule) for normal distributions: 1. Approximately $68\%$ of data falls within $1$ standard deviation of the mean ($\mu \pm 1\sigma$): $$500 - 100 = 400 \quad \text{to} \quad 500 + 100 = 600$$ 2. Compute $68\%$ of the total $10{,}000$ scores: $$0.68 \times 10{,}000 = 6{,}800$$ Therefore, approximately $6{,}800$ scores fall in this interval.

Examinees often confuse 1 standard deviation (68%) with 2 standard deviations (95%) or confuse the full range with one side (34%).

In a normal distribution, $\mu \pm 1\sigma \approx 68\%$, $\mu \pm 2\sigma \approx 95\%$, and $\mu \pm 3\sigma \approx 99.7\%$.

The committee has 5 members, requiring at least 2 biologists and at least 2 chemists. Only two mutually exclusive team compositions satisfy this constraint: **Case 1: 2 Biologists and 3 Chemists** $$\binom{6}{2} \times \binom{5}{3} = \frac{6 \times 5}{2 \times 1} \times \frac{5 \times 4}{2 \times 1} = 15 \times 10 = 150$$ **Case 2: 3 Biologists and 2 Chemists** $$\binom{6}{3} \times \binom{5}{2} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times 10 = 20 \times 10 = 200$$ **Total Possible Committees:** $$\text{Total} = 150 + 200 = 350$$

Examinees often overcount by selecting 2 biologists, 2 chemists, and then 1 remaining member from the pool: $\binom{6}{2}\binom{5}{2}\binom{7}{1} = 15 \times 10 \times 7 = 1050$. This overcounts because the 5th member duplicates combinations.

For "at least" combinatorics problems with multiple categories, break into mutually exclusive cases to eliminate double-counting.

To determine the probability that the chosen token is not blue, first find the total number of tokens in the container: $5 + 7 + 8 = 20$. The number of tokens that are not blue is the sum of the red and green tokens: $5 + 8 = 13$. Alternatively, calculate the complement of choosing a blue token: $1 - P(\text{blue}) = 1 - \frac{7}{20} = \frac{13}{20}$.

The most common trap is selecting $\frac{7}{20}$, which represents the probability of selecting a blue token rather than one that is NOT blue.

Always verify whether the prompt asks for the direct event or its complement ($P(E') = 1 - P(E)$).