Percentiles & Quartiles - College Statistics

1. A dataset of 200 observations has the following five-number summary: {Min: 10, Q1: 40, Median: 60, Q3: 90, Max: 150}. Which of the following statements is necessarily false based on this information?
   Answer: The interquartile range is exactly half of the range from the median to the maximum value. When analyzing five-number summaries, you need to carefully evaluate each statement using the specific values provided and your knowledge of statistical measures and distribution properties. Let's examine each option systematically. For option A, calculate the interquartile range: $IQR = Q3 - Q1 = 90 - 40 = 50$. The range from median to maximum is $150 - 60 = 90$. Half of this range is $90 ÷ 2 = 45$. Since $50 ≠ 45$, this statement is false. Option B requires checking the 1.5 × IQR rule for outliers. The boundaries are $Q1 - 1.5(50) = 40 - 75 = -35$ and $Q3 + 1.5(50) = 90 + 75 = 165$. Since both the minimum (10) and maximum (150) fall within these boundaries, there are no outliers by this rule, making this statement potentially true. Option C suggests right skewness. The median (60) is closer to Q1 (40) than to Q3 (90), with distances of 20 and 30 respectively. This pattern, along with a longer tail toward the maximum, indicates right skewness, making this statement reasonable. Option D involves quartile positioning. In any dataset, exactly 25% of observations fall between the minimum and Q1. With 200 observations, this equals $200 × 0.25 = 50$ observations, making this statement true. Remember that five-number summaries provide exact quartile positions, so always calculate precise values rather than estimating. Pay special attention to IQR calculations and outlier rules, as these frequently appear on statistics exams.
2. The 20th percentile of a large dataset of measurements is 60 and the 80th percentile is 120. Which of the following statements must be true about the interquartile range (IQR)?
   Answer: The IQR is at most 60. Let Pk be the k-th percentile. We are given P20 = 60 and P80 = 120. The first quartile (Q1) is the 25th percentile (P25), and the third quartile (Q3) is the 75th percentile (P75). By the definition of percentiles, a lower percentile value must be less than or equal to a higher percentile value. Therefore, we have the following relationships: P20 ≤ P25 (Q1) and P75 (Q3) ≤ P80. Substituting the given values, we get: 60 ≤ Q1 and Q3 ≤ 120. The interquartile range is IQR = Q3 - Q1. To find the maximum possible value of the IQR, we would need the largest possible Q3 and the smallest possible Q1. From our inequalities, the highest Q3 can be is 120 and the lowest Q1 can be is 60. Therefore, IQR = Q3 - Q1 ≤ 120 - 60 = 60. The IQR must be less than or equal to 60. It could be equal to 60 if, for example, all data between the 20th and 80th percentiles were either 60 or 120.
3. A data set consists of the following 11 measurements: {3, 10, 12, 15, 18, 20, 21, 24, 28, 30, 45}. Using the 1.5 × IQR rule, what is the upper boundary for identifying potential outliers?
   Answer: 52 To find the upper boundary for outliers, we first need to find the quartiles. The data is already sorted: {3, 10, 12, 15, 18, 20, 21, 24, 28, 30, 45}. The sample size is n=11. 1. Find the median (Q2): The median is the (11+1)/2 = 6th value, which is 20. 2. Find the first quartile (Q1): Q1 is the median of the lower half of the data (excluding Q2): {3, 10, 12, 15, 18}. The median of these 5 values is the 3rd value, so Q1 = 12. 3. Find the third quartile (Q3): Q3 is the median of the upper half of the data (excluding Q2): {21, 24, 28, 30, 45}. The median of these 5 values is the 3rd value, so Q3 = 28. 4. Calculate the Interquartile Range (IQR): IQR = Q3 - Q1 = 28 - 12 = 16. 5. Calculate the upper outlier boundary: The formula is Q3 + 1.5 × IQR. So, the boundary is 28 + 1.5 × 16 = 28 + 24 = 52.
4. For the dataset {15, 20, 22, 25, 28, 30, 34, 40}, what is the percentile rank of the value 34?
   Answer: 87.5th percentile There are different definitions for percentile rank, but a common one defines the percentile rank of a score as the percentage of scores in the distribution that are less than or equal to that score. 1. Count the total number of values in the dataset: n = 8. 2. Identify the score for which we want the percentile rank: X = 34. 3. Count the number of values in the dataset that are less than or equal to 34. These values are {15, 20, 22, 25, 28, 30, 34}. There are 7 such values. 4. Calculate the percentile rank: (Number of values ≤ X) / n * 100%. Percentile Rank = (7 / 8) 100% = 0.875 100% = 87.5%. Therefore, 34 is at the 87.5th percentile. The distractor 75.0th percentile comes from using the count of values strictly less than 34, which is 6, leading to (6/8)*100 = 75%.
5. A dataset of 100 test scores has an interquartile range (IQR) of 20 points. The teacher first adds 5 points to every student's score, and then increases each new score by 10%. What is the IQR of the final scores?
   Answer: 22 Let the original scores be denoted by X, the first transformation by Y, and the final scores by Z. 1. Original IQR: IQR_X = Q3_X - Q1_X = 20. 2. Add 5 points: Y = X + 5. Adding a constant to all data points shifts the entire distribution but does not change its spread. Therefore, the IQR is unaffected. IQR_Y = IQR_X = 20. 3. Increase by 10%: Z = Y 1.10. Multiplying all data points by a positive constant scales the measures of spread, including the IQR. The new IQR will be the old IQR multiplied by the constant. IQR_Z = 1.10 IQR_Y = 1.10 * 20 = 22. Therefore, the IQR of the final scores is 22. Distractors arise from incorrectly applying the additive constant to the IQR or applying the transformations in the wrong conceptual order.
6. A student scores at the 60th percentile on a national standardized test. Which of the following is the correct interpretation of this score?
   Answer: The student's score was higher than or equal to the scores of approximately 60% of the students who took the test. When you encounter percentile questions, remember that percentiles tell you about relative position within a group, not absolute performance or raw scores. The 60th percentile means the student's score was at or above 60% of all test-takers' scores. Think of it this way: if you lined up all students from lowest to highest score, this student would be positioned where 60% of students are at or below their score level. This makes choice A correct. Let's examine why the other options are wrong. Choice B reverses the logic—it incorrectly suggests the student scored lower than 60% of test-takers, which would actually describe the 40th percentile. Choice C confuses percentiles with percentage correct. A student could theoretically answer only 30% of questions correctly but still score at the 60th percentile if most students performed poorly. Percentiles are about relative ranking, not the proportion of correct answers. Choice D makes the common error of thinking percentiles equal raw scores. The 60th percentile score might be 450, 1200, or any other number depending on the test's scoring system. Here's your key takeaway: percentiles always answer the question "What percentage of people scored at or below this level?" When you see percentile problems, focus on relative position within the group, not raw performance measures. This distinction frequently appears on statistics exams, so practice converting between percentile language and its proper interpretation.
7. The five-number summary for a set of exam scores is {10, 25, 40, 60, 90}. If the professor decides to scale the scores by first doubling every score and then adding 5 points, what will be the interquartile range (IQR) of the new, scaled scores?
   Answer: 70 The original five-number summary is {Min, Q1, Median, Q3, Max} = {10, 25, 40, 60, 90}. The original interquartile range (IQR) is Q3 - Q1 = 60 - 25 = 35. Let X be an original score. The transformation is Y = 2X + 5. Measures of position (like Q1 and Q3) are affected by both multiplication and addition. New Q1 = 2 (Old Q1) + 5 = 2 25 + 5 = 55. New Q3 = 2 (Old Q3) + 5 = 2 60 + 5 = 125. Measures of spread (like the IQR) are affected by multiplication but not by addition. The new IQR is New Q3 - New Q1 = 125 - 55 = 70. Alternatively, New IQR = |2| (Old IQR) = 2 35 = 70.
8. Side-by-side box plots for the daily sales of two stores, Store A and Store B, are shown. Based on these plots, which statement provides an accurate comparison of the two stores? Refer to the box plots below.
   Answer: The middle 50% of daily sales for Store B is more variable than for Store A, and Store B's median sales are higher. Let's analyze the box plots: - Median: Store B's median line is higher than Store A's, indicating higher median sales. - Variability (IQR): The length of the box represents the interquartile range (IQR), which measures the variability of the middle 50% of the data. Store B's box is visibly longer than Store A's box, so its IQR is larger. This means the middle 50% of sales for Store B is more variable, not more consistent. - Range: The total length from whisker tip to whisker tip represents the range. Store B's range is larger than Store A's. - Symmetry: Store A's plot is roughly symmetric. Store B's plot is right-skewed (median closer to Q1, longer right whisker). Evaluating the options: A is false because Store B has the larger range and IQR. B is false because larger variability (larger IQR) means less consistency. C is true because Store B has a larger IQR (more variable middle 50%) and a higher median. D is false because Store B's distribution is not symmetric.
9. The table shows the frequency distribution of ages for a sample of 100 employees. In which age group is the 80th percentile located? Refer to the table.
   Answer: 50-59 To find the interval containing the 80th percentile, we first need to find the position of the 80th percentile in the ordered data. The total number of employees is n=100. The position is calculated as (80/100) * 100 = 80. We are looking for the 80th value in the dataset. Next, we calculate the cumulative frequency for each age group: - 20-29: 15 - 30-39: 15 + 25 = 40 - 40-49: 40 + 30 = 70 - 50-59: 70 + 20 = 90 - 60-69: 90 + 10 = 100 The cumulative frequency tells us the position of the last observation in each group. The first 70 employees are in the age groups up to 40-49. The employees from the 71st position to the 90th position are in the 50-59 age group. Since we are looking for the 80th value, it falls within the 50-59 age group.
10. The salaries for a small company of 10 employees are analyzed. The interquartile range (IQR) is $20,000 and the standard deviation is $18,000. The CEO's salary, an extreme outlier, is removed from the dataset and replaced by a new salary equal to the original median. Which of the following describes the most likely impact on the IQR and standard deviation?
    Answer: The standard deviation will decrease substantially, while the IQR will likely experience a much smaller change. The standard deviation is a measure of spread that is highly sensitive to outliers. Removing an extreme outlier and replacing it with a central value (the median) will drastically reduce the overall variation, so the standard deviation will decrease substantially. The interquartile range (IQR = Q3 - Q1) is a resistant measure of spread, meaning it is not strongly affected by outliers. Since the change involves removing an extreme value, it will not affect the values that determine Q1 and Q3 unless the dataset is very small and the outlier was close to Q3. Even if Q1 or Q3 change, the effect on the IQR is expected to be much smaller than the effect on the standard deviation. Therefore, the most accurate description is that the standard deviation will decrease substantially, while the IQR will see a much smaller change (it could slightly increase, decrease, or stay the same, but a large change is unlikely).
11. The histogram displays exam scores for a class of 200 students. Based on the histogram, the 70th percentile of the scores is located in which score interval? Use the histogram shown to answer.
    Answer: 70-80 First, find the position of the 70th percentile in the ordered dataset of 200 scores. Position = (70/100) * 200 = 140. We are looking for the 140th score. Next, calculate the cumulative frequencies from the histogram: - Interval 40-50: Frequency = 10. Cumulative Frequency = 10. - Interval 50-60: Frequency = 30. Cumulative Frequency = 10 + 30 = 40. - Interval 60-70: Frequency = 60. Cumulative Frequency = 40 + 60 = 100. - Interval 70-80: Frequency = 50. Cumulative Frequency = 100 + 50 = 150. - Interval 80-90: Frequency = 40. Cumulative Frequency = 150 + 40 = 190. - Interval 90-100: Frequency = 10. Cumulative Frequency = 190 + 10 = 200. The cumulative frequency up to the 60-70 interval is 100. The cumulative frequency up to the 70-80 interval is 150. This means the scores from the 101st to the 150th position fall within the 70-80 interval. Since we are looking for the 140th score, it must be in the 70-80 interval.
12. For the ordered dataset {5, 8, 10, 15, 18, 22, 25, 30}, what is the value of the first quartile (Q1)?
    Answer: 9 There are several methods for calculating quartiles. A common method, often called Tukey's hinges, is as follows: 1. The dataset is already ordered: {5, 8, 10, 15, 18, 22, 25, 30}. The size is n=8. 2. Find the median of the entire dataset. Since n is even, the median is the average of the two middle values: (15 + 18) / 2 = 16.5. 3. Divide the dataset into a lower half and an upper half. The lower half consists of all values below the median: {5, 8, 10, 15}. 4. The first quartile (Q1) is the median of the lower half. Since the lower half has an even number of values (4), its median is the average of its two middle values: (8 + 10) / 2 = 9. Distractor B (8.5) can be obtained using an interpolation method (e.g., the one used by Excel's QUARTILE.INC or TI calculators), which calculates the position as 0.25*(n+1) = 2.25, and interpolates between the 2nd and 3rd values.
13. Summary statistics for two datasets are provided: Dataset A (n=50, median=100, IQR=20) and Dataset B (n=50, median=120, IQR=20). Which of the following statements is best supported by this information?
    Answer: The middle 50% of scores in both datasets have a similar level of dispersion. The Interquartile Range (IQR) is a measure of statistical dispersion, representing the spread of the middle 50% of the data. Since both Dataset A and Dataset B have an IQR of 20, the middle 50% of their scores have a similar level of dispersion. A. We cannot determine the relationship between the maximum values from the given information. B. The median and IQR do not provide enough information to determine the shape (e.g., skewness, modality) of the distributions. C. Q1 marks the 25th percentile. 25% of the 50 scores in Dataset A would be 0.25 * 50 = 12.5, so about 12 or 13 scores are less than or equal to Q1. The statement that 25 scores are less than or equal to 90 (a possible value for Q1) is incorrect.
14. A box plot summarizes a dataset with 500 observations. Approximately how many observations lie outside the box (i.e., in the whiskers)?
    Answer: 250 A box plot is constructed based on the five-number summary. The 'box' represents the interquartile range (IQR), spanning from the first quartile (Q1, the 25th percentile) to the third quartile (Q3, the 75th percentile). This means the box contains the middle 50% of the data. The whiskers represent the data outside the box. If the box contains the middle 50%, then the remaining 50% of the data must lie in the whiskers (25% in the lower whisker, from the minimum to Q1, and 25% in the upper whisker, from Q3 to the maximum). Given n=500 observations, the number of observations outside the box is 50% of 500, which is 0.50 * 500 = 250.
15. A statistician defines a 'strong outlier' as any data point below Q1 − 2.5×IQR or above Q3 + 2.5×IQR. Given the dataset {5, 40, 42, 45, 48, 50, 55, 58, 60, 105}, which of the following is the complete set of strong outliers?
    Answer: {105} First, we must calculate the necessary statistics from the sorted dataset: {5, 40, 42, 45, 48, 50, 55, 58, 60, 105}. n=10. 1. Find Q1: Q1 is the median of the lower half {5, 40, 42, 45, 48}. The median is the 3rd value, so Q1 = 42. 2. Find Q3: Q3 is the median of the upper half {50, 55, 58, 60, 105}. The median is the 3rd value, so Q3 = 58. 3. Calculate IQR: IQR = Q3 - Q1 = 58 - 42 = 16. 4. Calculate the fences for strong outliers: The rule is 2.5 × IQR. - Lower Fence = Q1 - 2.5 × IQR = 42 - 2.5 × 16 = 42 - 40 = 2. - Upper Fence = Q3 + 2.5 × IQR = 58 + 2.5 × 16 = 58 + 40 = 98. 5. Identify outliers: We look for values below the lower fence or above the upper fence. - Is any value < 2? No. - Is any value > 98? Yes, the value 105. Therefore, the only strong outlier is 105. The distractor {5, 105} would be correct if using the standard 1.5 × IQR rule (Lower Fence = 42 - 1.5*16 = 18), which is a common trap.
16. The box plot shown summarizes the distribution of reaction times for a particular experiment. Which of the following statements about the distribution of reaction times is most likely to be true? Use the box plot shown to answer the question.
    Answer: The mean reaction time is greater than the median reaction time. The box plot shows a distribution that is skewed to the right. This is indicated by the median line being closer to the first quartile (Q1) than the third quartile (Q3) and the right whisker being substantially longer than the left whisker. In a right-skewed distribution, the mean is pulled towards the long tail, meaning the mean is typically greater than the median. Choice A describes a left-skewed distribution. Choice C describes a symmetric distribution. Choice D is incorrect because a box plot does not provide information about the mode of a distribution.
17. The box plot summarizes the heights of 400 professional basketball players. Approximately how many players have a height between 78 inches and 82 inches? Refer to the box plot provided.
    Answer: 200 The box in a box plot represents the middle 50% of the data. The bottom of the box is the first quartile (Q1) and the top of the box is the third quartile (Q3). From the provided box plot: - The first quartile (Q1) is at 78 inches. - The third quartile (Q3) is at 82 inches. This means that approximately 50% of the players have heights between Q1 (78 inches) and Q3 (82 inches). The total number of players is given as 400. To find the approximate number of players in this range, we calculate 50% of 400: Number of players = 0.50 × 400 = 200. Therefore, approximately 200 players have a height between 78 and 82 inches.
18. The following data set represents the ages of 9 employees, sorted in increasing order: {24, 28, 30, 32, 35, 38, 40, x, 50}. If the third quartile (Q3) of this dataset is 44, what is the value of x?
    Answer: 48 The dataset has n=9 employees. The data is already sorted. 1. Find the median: The median is the (9+1)/2 = 5th value, which is 35. 2. Identify the upper half: Using the method that excludes the median from the halves, the upper half of the data is {38, 40, x, 50}. 3. Find Q3: The third quartile (Q3) is the median of this upper half. Since there are 4 values, the median is the average of the middle two values: Q3 = (40 + x) / 2. 4. Solve for x: We are given that Q3 = 44. So, we set up the equation: (40 + x) / 2 = 44. Multiplying both sides by 2 gives 40 + x = 88. Subtracting 40 from both sides gives x = 48. 5. Check for consistency: The value x=48 must fit the sorted order, which requires 40 ≤ x ≤ 50. Our result x=48 satisfies this condition.
19. A statistics course has two sections. Section A has 29 students and a median exam score of 80. Section B has 39 students and a median exam score of 70. The scores for both sections are combined into a single dataset of 68 students. Which of the following statements MUST be true about the median of the combined dataset?
    Answer: The median must be between 70 and 80, inclusive. Let's analyze what we know. In Section A (n=29), at least 15 students scored ≤ 80 and at least 15 scored ≥ 80. In Section B (n=39), at least 20 students scored ≤ 70 and at least 20 scored ≥ 70. When combined (n=68), the median will be the average of the 34th and 35th scores in the sorted list. Consider a scenario to find the lowest possible combined median. Let scores in Section A be {79,...,79 (14 times), 80,...,80 (15 times)}. Let scores in Section B be {70,...,70 (20 times), 60,...,60 (19 times)}. The combined sorted list starts with 19 scores of 60, followed by 20 scores of 70. The 34th and 35th scores are both 70, so the median is 70. Consider a scenario to find the highest possible combined median. Let scores in Section A be {80,...,80 (15 times), 90,...,90 (14 times)}. Let scores in Section B be {70,...,70 (20 times), 80,...,80 (19 times)}. The combined list ends with (14+19)=33 scores of 80 or 90. The 35th highest score (and 34th) are both 80, making the median 80. Since the median can be 70, 80, or any value in between depending on the specific score distributions, the only statement that must be true is that the median is between 70 and 80, inclusive.

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