Hard GRE Statistics Set 1 Practice Questions

Concept Explanation

GRE Statistics involves the collection, analysis, interpretation, and presentation of data, focusing on measures of central tendency, dispersion, and probability distributions. To excel in GRE Prep, you must understand how to manipulate data sets to find the mean, median, mode, and range, as well as more complex concepts like standard deviation and normal distribution. At the hard difficulty level, the GRE often tests the relationships between these measures—for instance, how adding a constant to every value in a set affects the standard deviation or how the median changes when outliers are introduced. Understanding the properties of normal distributions and the 68-95-99.7 rule is essential for solving high-level quantitative comparison and data interpretation questions.

Solved Examples

Reviewing these worked examples will help you identify patterns in hard GRE statistics set 1 practice questions.

1. Example 1: Impact of Outliers
   A set of 7 distinct integers has a median of 20 and a mean of 25. If the largest number is increased by 50, what are the new median and the change in the mean?
   1. The median is the middle value. Since the order of the numbers does not change when the largest value is increased, the median remains 20.
   2. The mean is the sum divided by the number of elements $n$. The original sum was $25 \times 7 = 175$.
   3. The new sum is $175 + 50 = 225$.
   4. The new mean is $\frac{225}{7} \approx 32.14$. The mean increased by $\frac{50}{7} \approx 7.14$.
2. Example 2: Standard Deviation Properties
   Set $S$ consists of the numbers $\{x, y, z\}$ with a standard deviation of $\sigma$. If a new set $T$ is created such that $T = \{2x+5, 2y+5, 2z+5\}$, what is the standard deviation of set $T$ in terms of $\sigma$?
   1. Adding a constant to every term in a set (like the +5 here) does not change the standard deviation.
   2. Multiplying every term in a set by a constant $k$ multiplies the standard deviation by $|k|$.
   3. Since every term was multiplied by 2, the new standard deviation is $2\sigma$.
3. Example 3: Combined Means
   Class A has 20 students with an average score of 80. Class B has 30 students. If the combined average of both classes is 86, what is the average score of Class B?
   1. Calculate total points for Class A: $20 \times 80 = 1600$.
   2. Calculate total points for both classes: $(20 + 30) \times 86 = 50 \times 86 = 4300$.
   3. Subtract Class A total from the combined total: $4300 - 1600 = 2700$.
   4. Divide by the number of students in Class B: $\frac{2700}{30} = 90$.

Practice Questions

Test your skills with these hard GRE statistics set 1 practice questions. For more targeted practice, you can use an AI Question Generator to create similar sets.

1. Set $X$ contains 5 consecutive even integers. If the sum of these integers is 120, what is the standard deviation of Set $X$?
2. In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean. If a test has a mean of 75 and a standard deviation of 8, what percentage of scores are greater than 91?
3. A list of 11 numbers has a median of 15. If the 4 smallest numbers are each decreased by 5 and the 4 largest numbers are each increased by 10, what is the new median?

4. The average (arithmetic mean) of 10 numbers is $A$. If one number, 45, is removed, the average of the remaining numbers is $A - 3$. What is the value of $A$?
5. Quantity A: The standard deviation of $\{10, 20, 30, 40, 50\}$. Quantity B: The standard deviation of $\{110, 120, 130, 140, 150\}$. Compare the two quantities.
6. A set of 5 positive integers has a mean of 12, a median of 10, and a mode of 8. What is the maximum possible value for the largest integer in the set?
7. If the range of a set of 8 numbers is 25 and the smallest number is increased by 5 while the largest number remains the same, what is the new range?
8. In a group of 100 people, the mean height is 170 cm with a standard deviation of 10 cm. If the distribution is normal, how many people are expected to be between 160 cm and 180 cm?

Answers & Explanations

1. Answer: $\sqrt{8}$ or $2\sqrt{2}$
   Let the integers be $n-4, n-2, n, n+2, n+4$. Their sum is $5n = 120$, so $n = 24$. The set is $\{20, 22, 24, 26, 28\}$. The mean is 24. Deviations are $-4, -2, 0, 2, 4$. Squared deviations: $16, 4, 0, 4, 16$. Sum = 40. Variance = $\frac{40}{5} = 8$. Standard deviation = $\sqrt{8}$.
2. Answer: 2.5%
   A score of 91 is exactly 2 standard deviations above the mean $(75 + 2 \times 8 = 91)$. According to the 68-95-99.7 rule, 95% of data is within 2 standard deviations. This leaves 5% in the tails. Since the distribution is symmetric, 2.5% is in the upper tail (greater than 91).
3. Answer: 15
   The median is the 6th number in a sorted list of 11. Changing the 4 smallest and 4 largest numbers does not change the position or value of the 6th number, as long as they stay on their respective sides of the median.
4. Answer: 18
   Original sum: $10A$. New sum: $9(A-3)$. Since 45 was removed: $10A - 45 = 9A - 27$. Solving for $A$: $A = 18$.
5. Answer: The two quantities are equal
   Adding 100 to every term in Quantity A results in the set in Quantity B. Adding a constant to every term in a set does not change the standard deviation.
6. Answer: 26
   Let the set be $\{a, b, c, d, e\}$ in ascending order. Sum = $12 \times 5 = 60$. Median $c = 10$. Mode is 8, so $a = b = 8$. The set is $\{8, 8, 10, d, e\}$. To maximize $e$, minimize $d$. Since it is ascending, the smallest $d$ can be is 10. $8+8+10+10+e = 60 \rightarrow 36+e = 60 \rightarrow e = 24$. Wait, if $d=8$, the mode is 8. If $d=10$, there are two modes (8 and 10). If the mode is uniquely 8, $d$ must be at least 11. $36 + d + e = 60$. If $d=11$, $e = 25$. If $d=10$, the set is $\{8, 8, 10, 10, 24\}$. Check constraints: If the question implies a single mode, the answer is 25. If it allows multiple modes, it is 26 (using $d=8$, but $d$ must be $\geq c$). Thus, $d \geq 10$. If $d=10$, $e=24$. If $d=11$, $e=23$. Max $e = 24$. (Correction: If $a=b=8$, $c=10$, then $d$ could be 10, but the maximum $e$ occurs when $d$ is as small as possible. Since $d \geq c$, $d=10$. $8+8+10+10+24 = 60$. Max is 24).
7. Answer: 20
   Range = Max - Min. Initially, $Max - Min = 25$. New range = $Max - (Min + 5) = (Max - Min) - 5 = 25 - 5 = 20$.
8. Answer: 68
   160 is one standard deviation below the mean, and 180 is one standard deviation above. In a normal distribution, 68% of the population falls within this range. $68\% \text{ of } 100 = 68$.

Frequently Asked Questions

How is the weighted average different from the standard arithmetic mean?

A weighted average accounts for the relative importance or frequency of each value in a data set, whereas a standard mean treats all values as having equal weight. You calculate it by multiplying each value by its weight, summing those products, and dividing by the total weight.

What happens to the standard deviation if I add 10 to every data point?

The standard deviation remains exactly the same because the "spread" or distance between the data points does not change when the entire set is shifted. Standard deviation only changes when values are scaled by multiplication or when specific points are moved relative to others.

Can the median of a set be higher than the mean?

Yes, the median can be higher than the mean in a negatively skewed distribution where there are low-value outliers pulling the average down. In such cases, the "middle" value remains high while the arithmetic average is reduced by the small numbers.

What is the difference between range and interquartile range (IQR)?

The range is the difference between the absolute maximum and minimum values in a set, which makes it very sensitive to outliers. The IQR is the difference between the third quartile and the first quartile, measuring the spread of the middle 50% of the data.

How does the GRE test normal distribution without a calculator?

The GRE typically tests your knowledge of the 68-95-99.7 rule rather than complex calculations. You should memorize that 68% of data is within 1 standard deviation, 95% is within 2, and 99.7% is within 3, allowing you to estimate percentiles quickly.