IB Mathematics: Analysis and Approaches Quiz: Data Representation And Summary Stats
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Data Representation And Summary StatsQuestion 1 of 20

The five-number summary for a dataset is: Minimum = 10, Q1Q_1 = 18, Median = 25, Q3Q_3 = 30, Maximum = 50. Which of the following values in the dataset would be considered an outlier using the 1.5×IQR1.5 \times IQR rule?

10 only
Neither 10 nor 50
Both 10 and 50
50 only
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Data Representation And Summary Stats

Practice Data Representation And Summary Stats in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Data Representation And Summary Stats, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The five-number summary for a dataset is: Minimum = 10, Q1Q_1 = 18, Median = 25, Q3Q_3 = 30, Maximum = 50. Which of the following values in the dataset would be considered an outlier using the 1.5×IQR1.5 \times IQR rule?

  1. 10 only
  2. Neither 10 nor 50
  3. Both 10 and 50
  4. 50 only (correct answer)
Explanation: When you encounter outlier detection problems, you're applying the standard 1.5×IQR1.5 \times IQR rule to determine which data points fall unusually far from the central tendency of your dataset. To find outliers, first calculate the interquartile range: IQR=Q3−Q1=30−18=12IQR = Q_3 - Q_1 = 30 - 18 = 12. Then determine the outlier boundaries by extending 1.5×IQR=1.5×12=181.5 \times IQR = 1.5 \times 12 = 18 units beyond the quartiles. The lower boundary is Q1−18=18−18=0Q_1 - 18 = 18 - 18 = 0, and the upper boundary is Q3+18=30+18=48Q_3 + 18 = 30 + 18 = 48. Any value below 0 or above 48 is an outlier. Checking our extreme values: the minimum value of 10 falls between 0 and 48, so it's not an outlier. However, the maximum value of 50 exceeds the upper boundary of 48, making it an outlier. Looking at the wrong answers: Choice A suggests only 10 is an outlier, but 10 > 0, so it's within normal range. Choice B claims neither value is an outlier, missing that 50 exceeds the upper boundary. Choice C suggests both are outliers, but this incorrectly includes 10, which falls comfortably within the acceptable range. The correct answer is D: only 50 is an outlier. Remember this two-step process: calculate 1.5×IQR1.5 \times IQR, then check if extreme values fall outside the boundaries Q1−1.5×IQRQ_1 - 1.5 \times IQR and Q3+1.5×IQRQ_3 + 1.5 \times IQR. Don't assume minimum and maximum values are automatically outliers—they must exceed these calculated boundaries.

Question 2

For a dataset representing the annual incomes of employees at a large corporation, the mean income is found to be significantly higher than the median income. Which of the following is the most likely reason for this difference?

  1. The income distribution is symmetric.
  2. The income distribution is negatively skewed.
  3. A small number of executives have extremely high incomes. (correct answer)
  4. The interquartile range of the incomes is small.
Explanation: The mean is sensitive to extreme values (outliers), while the median is resistant. If the mean is significantly higher than the median, it indicates that there are some very large values pulling the mean upwards. This scenario corresponds to a positively (or right-skewed) distribution. A small number of executives with extremely high incomes would create such a skew. A symmetric distribution would have a mean approximately equal to the median. A negatively skewed distribution would have a mean lower than the median. A small IQR relates to the spread of the middle 50% of data, not directly to the relationship between the mean and median.

Question 3

A dataset is given by {10,15,17,19,20,23,25,45}\{10, 15, 17, 19, 20, 23, 25, 45\}. Using the criterion that an outlier is a value more than 1.5×IQR1.5 \times IQR below the lower quartile or above the upper quartile, which of the following statements is true?

  1. There are no outliers in the dataset.
  2. 45 is the only outlier in the dataset. (correct answer)
  3. 10 and 45 are both outliers in the dataset.
  4. 10 is the only outlier in the dataset.
Explanation: The dataset is ordered and has n=8n=8 values. The lower quartile Q1Q_1 is the median of the lower half {10,15,17,19}\{10, 15, 17, 19\}, so Q1=15+172=16Q_1 = \frac{15+17}{2} = 16. The upper quartile Q3Q_3 is the median of the upper half {20,23,25,45}\{20, 23, 25, 45\}, so Q3=23+252=24Q_3 = \frac{23+25}{2} = 24. The interquartile range is IQR=Q3−Q1=24−16=8IQR = Q_3 - Q_1 = 24 - 16 = 8. The lower boundary for outliers is Q1−1.5×IQR=16−1.5×8=16−12=4Q_1 - 1.5 \times IQR = 16 - 1.5 \times 8 = 16 - 12 = 4. The upper boundary for outliers is Q3+1.5×IQR=24+1.5×8=24+12=36Q_3 + 1.5 \times IQR = 24 + 1.5 \times 8 = 24 + 12 = 36. The value 10 is greater than 4, so it is not an outlier. The value 45 is greater than 36, so it is an outlier. Thus, 45 is the only outlier.

Question 4

A dataset has mean xˉ\bar{x}, median MM, and interquartile range II. Each value in the dataset is first multiplied by 3 and then increased by 4. What are the new mean, median, and interquartile range?

  1. New mean 3xˉ+43\bar{x}+4, new median 3M+43M+4, new IQR 3I+43I+4
  2. New mean 3xˉ+43\bar{x}+4, new median 3M+43M+4, new IQR 3I3I (correct answer)
  3. New mean 3xˉ3\bar{x}, new median (3M), new IQR 3I3I
  4. New mean xˉ+4\bar{x}+4, new median M+4M+4, new IQR II
Explanation: When each data point xix_i is transformed to axi+bax_i+b, measures of central tendency (like mean and median) are transformed in the same way. The new mean is 3xˉ+43\bar{x}+4 and the new median is 3M+43M+4. Measures of spread (like range and IQR) are only affected by multiplication, not by addition. The new interquartile range is I′=Q3′−Q1′=(3Q3+4)−(3Q1+4)=3Q3−3Q1=3(Q3−Q1)=3II' = Q'_3 - Q'_1 = (3Q_3+4) - (3Q_1+4) = 3Q_3 - 3Q_1 = 3(Q_3 - Q_1) = 3I.

Question 5

For a dataset representing the annual incomes of employees at a large corporation, the mean income is found to be significantly higher than the median income. Which of the following is the most likely reason for this difference?

  1. The income distribution is symmetric.
  2. The income distribution is negatively skewed.
  3. A small number of executives have extremely high incomes. (correct answer)
  4. The interquartile range of the incomes is small.
Explanation: The mean is sensitive to extreme values (outliers), while the median is resistant. If the mean is significantly higher than the median, it indicates that there are some very large values pulling the mean upwards. This scenario corresponds to a positively (or right-skewed) distribution. A small number of executives with extremely high incomes would create such a skew. A symmetric distribution would have a mean approximately equal to the median. A negatively skewed distribution would have a mean lower than the median. A small IQR relates to the spread of the middle 50% of data, not directly to the relationship between the mean and median.

Question 6

A dataset is given by {10,15,17,19,20,23,25,45}\{10, 15, 17, 19, 20, 23, 25, 45\}. Using the criterion that an outlier is a value more than 1.5×IQR1.5 \times IQR below the lower quartile or above the upper quartile, which of the following statements is true?

  1. There are no outliers in the dataset.
  2. 45 is the only outlier in the dataset. (correct answer)
  3. 10 and 45 are both outliers in the dataset.
  4. 10 is the only outlier in the dataset.
Explanation: The dataset is ordered and has n=8n=8 values. The lower quartile Q1Q_1 is the median of the lower half {10,15,17,19}\{10, 15, 17, 19\}, so Q1=15+172=16Q_1 = \frac{15+17}{2} = 16. The upper quartile Q3Q_3 is the median of the upper half {20,23,25,45}\{20, 23, 25, 45\}, so Q3=23+252=24Q_3 = \frac{23+25}{2} = 24. The interquartile range is IQR=Q3−Q1=24−16=8IQR = Q_3 - Q_1 = 24 - 16 = 8. The lower boundary for outliers is Q1−1.5×IQR=16−1.5×8=16−12=4Q_1 - 1.5 \times IQR = 16 - 1.5 \times 8 = 16 - 12 = 4. The upper boundary for outliers is Q3+1.5×IQR=24+1.5×8=24+12=36Q_3 + 1.5 \times IQR = 24 + 1.5 \times 8 = 24 + 12 = 36. The value 10 is greater than 4, so it is not an outlier. The value 45 is greater than 36, so it is an outlier. Thus, 45 is the only outlier.

Question 7

The mean of a list of 8 numbers is 12.5. One of the numbers, 20, is removed and replaced by a new number, xx. The new mean is 11.0. Find the value of xx.

  1. -3
  2. 8 (correct answer)
  3. 12
  4. 14
Explanation: The original sum of the 8 numbers is Sold=8×12.5=100S_{old} = 8 \times 12.5 = 100. When the number 20 is removed, the sum of the remaining 7 numbers is 100−20=80100 - 20 = 80. The new number xx is added to these 7 numbers, so the new sum of 8 numbers is Snew=80+xS_{new} = 80 + x. The new mean is 11.0, so Snew8=11.0\frac{S_{new}}{8} = 11.0. This means 80+x8=11\frac{80+x}{8} = 11. Multiplying by 8 gives 80+x=8880+x = 88. Therefore, x=8x = 8.

Question 8

The first quartile of a dataset of 21 unique student scores is 60 and the third quartile is 80. Three new students join the class with scores 55, 85, and 90. What is the most likely effect on the interquartile range (IQR)?

  1. The IQR will increase. (correct answer)
  2. The IQR will remain the same.
  3. The IQR will decrease.
  4. The effect on the IQR cannot be determined.
Explanation: When analyzing how new data points affect quartiles, you need to understand that quartiles are positional statistics that depend on the location of values in an ordered dataset, not just the values themselves. With 21 students, the first quartile (Q1) sits at position 5.5 (between the 5th and 6th values) and the third quartile (Q3) sits at position 16.5 (between the 16th and 17th values). Currently, Q1 = 60 and Q3 = 80, giving an IQR of 20. Adding three new students creates a dataset of 24 values. Now Q1 will be at position 6.25 (approximately the 6th value) and Q3 at position 18.75 (approximately the 19th value). The key insight is determining where the new scores (55, 85, 90) fall relative to the existing quartiles. The score of 55 falls below the current Q1 of 60, which will push more values above the new Q1 position. The scores of 85 and 90 both fall above the current Q3 of 80, pushing the new Q3 higher. Since Q1 will likely remain near 60 (or possibly decrease slightly) while Q3 will definitely increase beyond 80, the IQR must increase. Answer A is correct because the new Q3 will be higher while Q1 remains similar or decreases. Answer B is wrong because adding values outside the current IQR range affects the quartile positions. Answer C is incorrect since Q3 increases more than any potential decrease in Q1. Answer D is wrong because we can determine the direction of change by analyzing where new values fall relative to existing quartiles. Study tip: When new data points are added, focus on whether they fall above Q3, below Q1, or between the quartiles to predict IQR changes.

Question 9

A dataset has mean xˉ\bar{x}, median MM, and interquartile range II. Each value in the dataset is first multiplied by 3 and then increased by 4. What are the new mean, median, and interquartile range?

  1. New mean 3xˉ+43\bar{x}+4, new median 3M+43M+4, new IQR 3I+43I+4
  2. New mean 3xˉ+43\bar{x}+4, new median 3M+43M+4, new IQR 3I3I (correct answer)
  3. New mean 3xˉ3\bar{x}, new median (3M), new IQR 3I3I
  4. New mean xˉ+4\bar{x}+4, new median M+4M+4, new IQR II
Explanation: When each data point xix_i is transformed to axi+bax_i+b, measures of central tendency (like mean and median) are transformed in the same way. The new mean is 3xˉ+43\bar{x}+4 and the new median is 3M+43M+4. Measures of spread (like range and IQR) are only affected by multiplication, not by addition. The new interquartile range is I′=Q3′−Q1′=(3Q3+4)−(3Q1+4)=3Q3−3Q1=3(Q3−Q1)=3II' = Q'_3 - Q'_1 = (3Q_3+4) - (3Q_1+4) = 3Q_3 - 3Q_1 = 3(Q_3 - Q_1) = 3I.

Question 10

The number of goals scored by a football team in its matches is recorded. For a season of games, the data is partially summarized in the frequency table below. Goals (x): 1, 2, 3, 4, 5 Frequency (f): 4, 7, a, 3, 2 It is known that the median number of goals is 2 and the mean number of goals is 2.6.

Find the value of aa, the frequency of scoring 3 goals.

  1. 2
  2. 3
  3. 4 (correct answer)
  4. 5
Explanation: The total frequency is N=4+7+a+3+2=16+aN = 4+7+a+3+2 = 16+a. The median is 2. The cumulative frequency for x=1x=1 is 4. The cumulative frequency for x=2x=2 is 4+7=114+7=11. For the median to be 2, the median position N+12\frac{N+1}{2} must correspond to a value of 2. This means 4<N+12≤114 < \frac{N+1}{2} \le 11. This condition holds for a wide range of aa. Now, use the mean. The sum of the scores is ∑fx=1(4)+2(7)+3(a)+4(3)+5(2)=4+14+3a+12+10=40+3a\sum fx = 1(4)+2(7)+3(a)+4(3)+5(2) = 4+14+3a+12+10 = 40+3a. The mean is ∑fxN=40+3a16+a\frac{\sum fx}{N} = \frac{40+3a}{16+a}. We are given that the mean is 2.6. So, 40+3a16+a=2.6\frac{40+3a}{16+a} = 2.6. This gives 40+3a=2.6(16+a)40+3a = 2.6(16+a), which is 40+3a=41.6+2.6a40+3a = 41.6+2.6a. Rearranging gives 0.4a=1.60.4a = 1.6, so a=4a=4. We can check this is consistent with the median condition.

Question 11

The first quartile of a dataset of 21 unique student scores is 60 and the third quartile is 80. Three new students join the class with scores 55, 85, and 90. What is the most likely effect on the interquartile range (IQR)?

  1. The IQR will increase. (correct answer)
  2. The IQR will remain the same.
  3. The IQR will decrease.
  4. The effect on the IQR cannot be determined.
Explanation: When analyzing how new data points affect quartiles, you need to understand that quartiles are positional statistics that depend on the location of values in an ordered dataset, not just the values themselves. With 21 students, the first quartile (Q1) sits at position 5.5 (between the 5th and 6th values) and the third quartile (Q3) sits at position 16.5 (between the 16th and 17th values). Currently, Q1 = 60 and Q3 = 80, giving an IQR of 20. Adding three new students creates a dataset of 24 values. Now Q1 will be at position 6.25 (approximately the 6th value) and Q3 at position 18.75 (approximately the 19th value). The key insight is determining where the new scores (55, 85, 90) fall relative to the existing quartiles. The score of 55 falls below the current Q1 of 60, which will push more values above the new Q1 position. The scores of 85 and 90 both fall above the current Q3 of 80, pushing the new Q3 higher. Since Q1 will likely remain near 60 (or possibly decrease slightly) while Q3 will definitely increase beyond 80, the IQR must increase. Answer A is correct because the new Q3 will be higher while Q1 remains similar or decreases. Answer B is wrong because adding values outside the current IQR range affects the quartile positions. Answer C is incorrect since Q3 increases more than any potential decrease in Q1. Answer D is wrong because we can determine the direction of change by analyzing where new values fall relative to existing quartiles. Study tip: When new data points are added, focus on whether they fall above Q3, below Q1, or between the quartiles to predict IQR changes.

Question 12

The five-number summary for a dataset is: Minimum = 10, Q1Q_1 = 18, Median = 25, Q3Q_3 = 30, Maximum = 50. Which of the following values in the dataset would be considered an outlier using the 1.5×IQR1.5 \times IQR rule?

  1. 10 only
  2. Neither 10 nor 50
  3. Both 10 and 50
  4. 50 only (correct answer)
Explanation: When you encounter outlier detection problems, you're applying the standard 1.5×IQR1.5 \times IQR rule to determine which data points fall unusually far from the central tendency of your dataset. To find outliers, first calculate the interquartile range: IQR=Q3−Q1=30−18=12IQR = Q_3 - Q_1 = 30 - 18 = 12. Then determine the outlier boundaries by extending 1.5×IQR=1.5×12=181.5 \times IQR = 1.5 \times 12 = 18 units beyond the quartiles. The lower boundary is Q1−18=18−18=0Q_1 - 18 = 18 - 18 = 0, and the upper boundary is Q3+18=30+18=48Q_3 + 18 = 30 + 18 = 48. Any value below 0 or above 48 is an outlier. Checking our extreme values: the minimum value of 10 falls between 0 and 48, so it's not an outlier. However, the maximum value of 50 exceeds the upper boundary of 48, making it an outlier. Looking at the wrong answers: Choice A suggests only 10 is an outlier, but 10 > 0, so it's within normal range. Choice B claims neither value is an outlier, missing that 50 exceeds the upper boundary. Choice C suggests both are outliers, but this incorrectly includes 10, which falls comfortably within the acceptable range. The correct answer is D: only 50 is an outlier. Remember this two-step process: calculate 1.5×IQR1.5 \times IQR, then check if extreme values fall outside the boundaries Q1−1.5×IQRQ_1 - 1.5 \times IQR and Q3+1.5×IQRQ_3 + 1.5 \times IQR. Don't assume minimum and maximum values are automatically outliers—they must exceed these calculated boundaries.

Question 13

The number of goals scored by a football team in its matches is recorded. For a season of games, the data is partially summarized in the frequency table below. Goals (x): 1, 2, 3, 4, 5 Frequency (f): 4, 7, a, 3, 2 It is known that the median number of goals is 2 and the mean number of goals is 2.6.

Find the value of aa, the frequency of scoring 3 goals.

  1. 2
  2. 3
  3. 4 (correct answer)
  4. 5
Explanation: The total frequency is N=4+7+a+3+2=16+aN = 4+7+a+3+2 = 16+a. The median is 2. The cumulative frequency for x=1x=1 is 4. The cumulative frequency for x=2x=2 is 4+7=114+7=11. For the median to be 2, the median position N+12\frac{N+1}{2} must correspond to a value of 2. This means 4<N+12≤114 < \frac{N+1}{2} \le 11. This condition holds for a wide range of aa. Now, use the mean. The sum of the scores is ∑fx=1(4)+2(7)+3(a)+4(3)+5(2)=4+14+3a+12+10=40+3a\sum fx = 1(4)+2(7)+3(a)+4(3)+5(2) = 4+14+3a+12+10 = 40+3a. The mean is ∑fxN=40+3a16+a\frac{\sum fx}{N} = \frac{40+3a}{16+a}. We are given that the mean is 2.6. So, 40+3a16+a=2.6\frac{40+3a}{16+a} = 2.6. This gives 40+3a=2.6(16+a)40+3a = 2.6(16+a), which is 40+3a=41.6+2.6a40+3a = 41.6+2.6a. Rearranging gives 0.4a=1.60.4a = 1.6, so a=4a=4. We can check this is consistent with the median condition.

Question 14

In a dataset, the first quartile Q1Q_1 is 50 and the third quartile Q3Q_3 is 80. Which of the following statements must be true?

  1. Approximately 50% of the data lies between 50 and 80. (correct answer)
  2. The median of the dataset is exactly 65.
  3. The mean of the dataset must be between 50 and 80.
  4. The range of the dataset is 30.
Explanation: When you encounter quartile problems, remember that quartiles divide your dataset into four equal parts, each containing exactly 25% of the data. The first quartile (Q₁) marks the point below which 25% of data falls, and the third quartile (Q₃) marks the point below which 75% of data falls. Given Q₁ = 50 and Q₃ = 80, you know that 25% of data lies below 50, and 75% lies below 80. This means the middle 50% of your data (from the 25th percentile to the 75th percentile) must lie between 50 and 80. This makes option A correct. Option B assumes the median equals the midpoint of Q₁ and Q₃, but the median (Q₂) can be anywhere between these quartiles depending on data distribution. It's not necessarily their arithmetic average. Option C suggests the mean must fall between the quartiles, but means are sensitive to outliers. Extreme values beyond Q₃ or below Q₁ can pull the mean outside the interquartile range entirely. Option D confuses the interquartile range (IQR = Q₃ - Q₁ = 30) with the total range. The range spans from minimum to maximum values, which could extend far beyond the quartiles. The IQR only captures the middle 50% spread. Study tip: Remember that quartiles give you precise information about data distribution percentages, but they don't determine the median's exact value, the mean's location, or the dataset's total spread. Focus on what quartiles definitively tell you versus what they leave uncertain.

Question 15

The mean score of nn boys in a class is 70. The mean score of the 15 girls in the same class is 82. The combined mean score for all students is 75. Find the value of nn.

  1. 10
  2. 15
  3. 21 (correct answer)
  4. 25
Explanation: The sum of the boys' scores is 70n70n. The sum of the girls' scores is 15×82=123015 \times 82 = 1230. The total sum of scores for all students is 70n+123070n + 1230. The total number of students is n+15n+15. The combined mean is the total sum divided by the total number of students. Therefore, 70n+1230n+15=75\frac{70n + 1230}{n+15} = 75. Multiplying both sides by (n+15)(n+15) gives 70n+1230=75(n+15)70n + 1230 = 75(n+15), which simplifies to 70n+1230=75n+112570n + 1230 = 75n + 1125. Rearranging the terms, we get 5n=1230−11255n = 1230 - 1125, so 5n=1055n = 105. Dividing by 5 gives n=21n=21.

Question 16

A dataset contains 15 distinct numbers. If each number in the dataset is decreased by 10, which of the following summary statistics will also decrease by 10?

  1. The interquartile range only
  2. The mean only
  3. The mean, median, and interquartile range
  4. The mean and the median only (correct answer)
Explanation: When you encounter questions about transformations of data, think about whether the operation affects measures of center versus measures of spread. Adding or subtracting a constant to every data point shifts the entire distribution without changing its shape or variability. Let's examine what happens when we subtract 10 from each number. The mean is the sum of all values divided by the count. If we subtract 10 from each of the 15 numbers, we're subtracting 10×15=15010 \times 15 = 150 from the total sum, so the new mean equals the old mean minus 10. Similarly, the median is the middle value when data is ordered. Since every number decreases by 10, the middle position still contains the same relative value, just shifted down by 10. However, the interquartile range (IQR) measures spread—the difference between the third and first quartiles. When you subtract 10 from every value, both quartiles decrease by 10, so their difference remains unchanged: (Q3−10)−(Q1−10)=Q3−Q1(Q_3 - 10) - (Q_1 - 10) = Q_3 - Q_1. Option A is wrong because the IQR doesn't change, only measures of center do. Option B misses that the median also shifts by 10, not just the mean. Option C incorrectly includes the IQR, which measures spread and remains constant under linear shifts. Study tip: Remember that adding or subtracting constants affects measures of center (mean, median, mode) but leaves measures of spread (range, IQR, standard deviation) unchanged. Only multiplication/division affects spread measures.

Question 17

In a dataset, the first quartile Q1Q_1 is 50 and the third quartile Q3Q_3 is 80. Which of the following statements must be true?

  1. Approximately 50% of the data lies between 50 and 80. (correct answer)
  2. The median of the dataset is exactly 65.
  3. The mean of the dataset must be between 50 and 80.
  4. The range of the dataset is 30.
Explanation: When you encounter quartile problems, remember that quartiles divide your dataset into four equal parts, each containing exactly 25% of the data. The first quartile (Q₁) marks the point below which 25% of data falls, and the third quartile (Q₃) marks the point below which 75% of data falls. Given Q₁ = 50 and Q₃ = 80, you know that 25% of data lies below 50, and 75% lies below 80. This means the middle 50% of your data (from the 25th percentile to the 75th percentile) must lie between 50 and 80. This makes option A correct. Option B assumes the median equals the midpoint of Q₁ and Q₃, but the median (Q₂) can be anywhere between these quartiles depending on data distribution. It's not necessarily their arithmetic average. Option C suggests the mean must fall between the quartiles, but means are sensitive to outliers. Extreme values beyond Q₃ or below Q₁ can pull the mean outside the interquartile range entirely. Option D confuses the interquartile range (IQR = Q₃ - Q₁ = 30) with the total range. The range spans from minimum to maximum values, which could extend far beyond the quartiles. The IQR only captures the middle 50% spread. Study tip: Remember that quartiles give you precise information about data distribution percentages, but they don't determine the median's exact value, the mean's location, or the dataset's total spread. Focus on what quartiles definitively tell you versus what they leave uncertain.

Question 18

A dataset contains 15 distinct numbers. If each number in the dataset is decreased by 10, which of the following summary statistics will also decrease by 10?

  1. The interquartile range only
  2. The mean only
  3. The mean, median, and interquartile range
  4. The mean and the median only (correct answer)
Explanation: When you encounter questions about transformations of data, think about whether the operation affects measures of center versus measures of spread. Adding or subtracting a constant to every data point shifts the entire distribution without changing its shape or variability. Let's examine what happens when we subtract 10 from each number. The mean is the sum of all values divided by the count. If we subtract 10 from each of the 15 numbers, we're subtracting 10×15=15010 \times 15 = 150 from the total sum, so the new mean equals the old mean minus 10. Similarly, the median is the middle value when data is ordered. Since every number decreases by 10, the middle position still contains the same relative value, just shifted down by 10. However, the interquartile range (IQR) measures spread—the difference between the third and first quartiles. When you subtract 10 from every value, both quartiles decrease by 10, so their difference remains unchanged: (Q3−10)−(Q1−10)=Q3−Q1(Q_3 - 10) - (Q_1 - 10) = Q_3 - Q_1. Option A is wrong because the IQR doesn't change, only measures of center do. Option B misses that the median also shifts by 10, not just the mean. Option C incorrectly includes the IQR, which measures spread and remains constant under linear shifts. Study tip: Remember that adding or subtracting constants affects measures of center (mean, median, mode) but leaves measures of spread (range, IQR, standard deviation) unchanged. Only multiplication/division affects spread measures.

Question 19

The mean of a set of nn numbers is MM. A new number, XX, is added to the set. What is the new mean of the n+1n+1 numbers?

  1. M+X2\frac{M+X}{2}
  2. M+Xn+1\frac{M+X}{n+1}
  3. nM+Xn+1\frac{nM+X}{n+1} (correct answer)
  4. nM+Xn\frac{nM+X}{n}
Explanation: The sum of the original nn numbers is given by Sn=n×MS_n = n \times M. When the new number XX is added, the new sum is Sn+1=Sn+X=nM+XS_{n+1} = S_n + X = nM + X. The new set has n+1n+1 numbers. Therefore, the new mean is the new sum divided by the new count of numbers: New Mean=nM+Xn+1\text{New Mean} = \frac{nM+X}{n+1}.

Question 20

The mean of five integers is 10, the median is 12, and the mode is 15. What is the smallest possible range for this set of five integers?

  1. 4
  2. 11 (correct answer)
  3. 15
  4. 19
Explanation: Let the five integers in ascending order be a,b,c,d,ea, b, c, d, e. The median is the middle value, so c=12c = 12. The mode is 15, and since the numbers are ordered and the median is 12, the two largest numbers must be 15. So, d=15d = 15 and e=15e = 15. The set is {a,b,12,15,15}\{a, b, 12, 15, 15\}. The mean is 10, so the sum is 5×10=505 \times 10 = 50. The sum is a+b+12+15+15=a+b+42a+b+12+15+15 = a+b+42. So, a+b+42=50a+b+42 = 50, which means a+b=8a+b=8. Since the numbers are in ascending order, we have a≤b≤c=12a \le b \le c = 12. We want to find the smallest possible range, which is e−a=15−ae-a = 15-a. To minimize the range, we must maximize aa. Given a+b=8a+b=8 and a≤ba \le b, the maximum value of aa occurs when a=ba=b. Thus, 2a=82a=8, so a=4a=4. This gives the set {4,4,12,15,15}\{4, 4, 12, 15, 15\}, which satisfies all conditions. The range is 15−4=1115-4=11.