Hard GRE Data Analysis Set 2 Practice Questions

Concept Explanation

GRE Data Analysis involves interpreting, modeling, and analyzing data using statistical methods such as mean, median, standard deviation, and probability distributions.

At the harder levels of the GRE Prep curriculum, data analysis questions often combine multiple concepts, such as requiring you to calculate a weighted average before determining the probability of a specific outcome. You will frequently encounter complex graphical representations, including box plots, scatterplots with lines of best fit, and multi-layered bar charts. Success on these problems depends on your ability to distinguish between absolute values and percentage changes, as well as understanding how outliers affect measures of central tendency. For those seeking structured support, using an Adaptive GRE Practice Test can help identify which specific data subsets—like combinatorics or normal distributions—require more focus.

Solved Examples

1. Standard Deviation and Mean: A set of 5 distinct integers has a mean of 20 and a standard deviation of $\sigma$. If every number in the set is increased by 5, what happens to the new mean and the new standard deviation?
   1. The original sum was $5 \times 20 = 100$.
   2. After adding 5 to each of the 5 integers, 25 is added to the total sum ($100 + 25 = 125$).
   3. The new mean is $\frac{125}{5} = 25$.
   4. Standard deviation measures the spread or distance of data points from the mean. Since every point shifted by the same amount, the relative distances between points remain unchanged.
   5. Therefore, the new mean is 25 and the new standard deviation remains $\sigma$.
2. Probability of Independent Events: In a bag of 10 marbles, 4 are red and 6 are blue. If two marbles are picked at random without replacement, what is the probability that both are red?
   1. The probability of the first marble being red is $\frac{4}{10}$.
   2. Since the marble is not replaced, there are now 9 marbles left, 3 of which are red.
   3. The probability of the second marble being red is $\frac{3}{9}$.
   4. Multiply the probabilities: $\frac{4}{10} \times \frac{3}{9} = \frac{12}{90}$.
   5. Simplify the fraction: $\frac{12}{90} = \frac{2}{15}$.
3. Weighted Averages: Class A has 20 students with an average score of 80. Class B has 30 students with an average score of 90. What is the average score for all 50 students combined?
   1. Calculate the total points for Class A: $20 \times 80 = 1,600$.
   2. Calculate the total points for Class B: $30 \times 90 = 2,700$.
   3. Add the total points: $1,600 + 2,700 = 4,300$.
   4. Divide by the total number of students: $\frac{4,300}{50} = 86$.
   5. The combined average is 86.

Practice Questions

1. A dataset consists of 11 unique integers. If the median of the set is 15 and the largest possible integer is 40, what is the maximum possible value for the arithmetic mean of the set?

2. In a group of 100 people, 60 people own a car, 45 people own a bicycle, and 20 people own both. If a person is chosen at random, what is the probability that the person owns neither a car nor a bicycle?

3. The standard deviation of a distribution of scores is 8.2. If every score in the distribution is multiplied by 3, what is the standard deviation of the new distribution?

4. A box contains 5 red, 4 green, and 3 blue balls. If 3 balls are selected at random without replacement, what is the probability that at least one ball is green?

5. A set of numbers has a mean of 50 and a standard deviation of 0. If one more number, 50, is added to the set, what is the new standard deviation?

6. In a normal distribution, approximately what percentage of the data falls within two standard deviations of the mean? (Refer to Normal Distribution for empirical rule details).

7. If the ratio of the number of men to women in a room is 3:4 and the average age of the men is 30 while the average age of the women is 37, what is the average age of all people in the room?

8. A list of 5 numbers has a median of 12, a mode of 8, and a range of 15. If the smallest number is 8, what is the largest possible sum of these 5 numbers?

Answers & Explanations

1. Answer: 28.18 (approximately). To maximize the mean, we maximize all individual values. The 11 integers are $x_1, x_2, ... x_{11}$. The median $x_6 = 15$. To keep the set unique and maximize the sum: the first 5 values should be 10, 11, 12, 13, 14 (the largest possible integers less than 15). The values above the median should be 36, 37, 38, 39, 40. Sum = $10+11+12+13+14+15+36+37+38+39+40 = 265$. Mean = $265 / 11 \approx 24.09$. Correction: To truly maximize, the values below the median can be 10, 11, 12, 13, 14 only if we assume positive integers; however, the question says unique integers. If we use the highest possible values below 15 (which are 10, 11, 12, 13, 14) and the highest above 15 (36-40), the sum is 265.
2. Answer: 0.15. Using the formula for union: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Here, $60 + 45 - 20 = 85$. 85 people own at least one. Thus, $100 - 85 = 15$ people own neither. Probability = $15/100 = 0.15$.
3. Answer: 24.6. When every value in a set is multiplied by a constant $k$, the standard deviation is also multiplied by $|k|$. So, $8.2 \times 3 = 24.6$.
4. Answer: $\frac{37}{55}$. It is easier to find the probability of the complement (no green balls). Total balls = 12. Non-green = 8. $P( \text{no green}) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} = \frac{336}{1320} = \frac{14}{55}$. $P( \text{at least one}) = 1 - \frac{14}{55} = \frac{41}{55}$. (Recalculation: $\frac{2}{3} \times \frac{7}{11} \times \frac{3}{5} = \frac{42}{165} = \frac{14}{55}$). Corrected probability is $\frac{41}{55}$.
5. Answer: 0. If the standard deviation is 0, all numbers in the set are identical (all are 50). Adding another 50 does not change the spread; all numbers are still identical.
6. Answer: 95%. According to the empirical rule for normal distributions, approximately 68% falls within 1 SD, 95% within 2 SD, and 99.7% within 3 SD.
7. Answer: 34. Use weighted averages. Let there be 3 men and 4 women. Total age = $(3 \times 30) + (4 \times 37) = 90 + 148 = 238$. Average = $238 / 7 = 34$.
8. Answer: 68. Smallest is 8. Since the mode is 8, at least two numbers are 8. List: 8, 8, 12 (median), $x, y$. Range is 15, so $\text{Max} - 8 = 15 \Rightarrow \text{Max} = 23$. To maximize the sum, $x$ should be as large as possible but less than 23 (since 23 is the unique max to keep range 15). If integers: 8, 8, 12, 22, 23. Sum = 73. If not unique: 8, 8, 12, 23, 23. Sum = 74. (Note: The question asks for largest possible sum).

1. If a value that is exactly equal to the mean is added to a data set with a non-zero standard deviation, what happens to the standard deviation?

• AIt increases
• BIt decreases
• CIt remains the same
• DIt becomes zero

Frequently Asked Questions

How does the GRE test standard deviation?

The GRE rarely asks you to calculate standard deviation manually; instead, it tests your understanding of how the spread of data changes when values are added, removed, or transformed. You should know that adding a constant to all values doesn't change it, while multiplying all values by a constant scales the deviation by that same constant.

What is the difference between mutually exclusive and independent events?

Mutually exclusive events cannot happen at the same time, meaning their intersection is zero. Independent events are those where the occurrence of one does not affect the probability of the other, and their intersection is the product of their individual probabilities.

When should I use the median instead of the mean?

The median is a better measure of central tendency when a dataset contains significant outliers or is heavily skewed, as it is not pulled toward extreme values like the mean is. In GRE Data Analysis, you may be asked to compare these two reflecting on the skewness of a graph.

How do I calculate the probability of "at least one"?

The most efficient way to solve "at least one" problems is to calculate the probability of the event never happening and subtracting that value from 1. This avoids the need to sum multiple different successful scenarios.

What is a normal distribution on the GRE?

A normal distribution is a bell-shaped curve where the mean, median, and mode are all equal. On the GRE, you are expected to know the 68-95-99.7 rule regarding data distribution across standard deviations.