IB Mathematics: Applications and Interpretation Quiz: Data Representation And Summary Stats
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Data Representation And Summary StatsQuestion 1 of 20

Two basketball players, Alex and Ben, record their points per game over a season. Alex's scores have a median of 24 points and an interquartile range of 7 points. Ben's scores have a median of 21 points and an interquartile range of 12 points.

Which statement is the most accurate interpretation of these statistics?

Alex typically scores more points and is a more consistent scorer than Ben.
Alex typically scores more points but is a less consistent scorer than Ben.
Ben typically scores more points and is a more consistent scorer than Alex.
Ben typically scores more points but is a less consistent scorer than Alex.
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Data Representation And Summary Stats

Practice Data Representation And Summary Stats in IB Mathematics: Applications and Interpretation 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: Applications and Interpretation.

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

Two basketball players, Alex and Ben, record their points per game over a season. Alex's scores have a median of 24 points and an interquartile range of 7 points. Ben's scores have a median of 21 points and an interquartile range of 12 points.

Which statement is the most accurate interpretation of these statistics?

  1. Alex typically scores more points and is a more consistent scorer than Ben. (correct answer)
  2. Alex typically scores more points but is a less consistent scorer than Ben.
  3. Ben typically scores more points and is a more consistent scorer than Alex.
  4. Ben typically scores more points but is a less consistent scorer than Alex.
Explanation: The median represents a typical value. Alex's median (24) is higher than Ben's (21), so Alex typically scores more points. The interquartile range (IQR) is a measure of spread or consistency. A smaller IQR indicates greater consistency. Alex's IQR (7) is smaller than Ben's (12), so Alex is the more consistent scorer.

Question 2

The following is a list of house prices in a neighbourhood, in thousands of dollars: {350, 375, 380, 400, 410, 420, 850}. The house priced at $850,000 is identified as an outlier and is removed from the dataset.

How will the mean and standard deviation of the house prices change after removing the outlier?

  1. The mean will decrease and the standard deviation will decrease. (correct answer)
  2. The mean will decrease and the standard deviation will increase.
  3. The mean will increase and the standard deviation will decrease.
  4. The mean will increase and the standard deviation will increase.
Explanation: The outlier (850) is a high value. Removing a value that is significantly higher than the rest of the data will cause the mean to decrease. The standard deviation is a measure of spread. Since the outlier is very far from the other data points, it contributes significantly to the spread. Removing it will make the data less spread out, causing the standard deviation to decrease.

Question 3

The journey times for a bus route have a mean of 40 minutes and a variance of 9 minutes2^2. Due to a new speed limit, all journey times are expected to increase by 10%.

What will be the new mean and new standard deviation of the journey times?

  1. Mean = 44 min, Standard Deviation = 3.0 min
  2. Mean = 44 min, Standard Deviation = 3.3 min (correct answer)
  3. Mean = 40 min, Standard Deviation = 3.3 min
  4. Mean = 44 min, Standard Deviation = 10.9 min
Explanation: An increase of 10% is equivalent to multiplying by 1.1. The new mean is 40×1.1=4440 \times 1.1 = 44 minutes. The original variance is 9 min2^2, so the original standard deviation is 9=3\sqrt{9} = 3 minutes. The standard deviation is also multiplied by 1.1: new standard deviation = 3×1.1=3.33 \times 1.1 = 3.3 minutes. Note that the new variance would be 9×1.12=10.899 \times 1.1^2 = 10.89 min2^2.

Question 4

A student's scores on 10 tests are: 65, 72, 75, 78, 80, 82, 85, 88, 91, 98. The student then scores 45 on an eleventh test.

Using the 1.5 × IQR rule on the new set of 11 scores, what is the lower boundary for outliers, and is the score of 45 an outlier?

  1. The lower boundary is 48, and the score is an outlier. (correct answer)
  2. The lower boundary is 48, and the score is not an outlier.
  3. The lower boundary is 55.5, and the score is an outlier.
  4. The lower boundary is 55.5, and the score is not an outlier.
Explanation: First, order the 11 scores: {45, 65, 72, 75, 78, 80, 82, 85, 88, 91, 98}. The lower quartile (Q1) is the 3rd value, which is 72. The upper quartile (Q3) is the 9th value, which is 88. The interquartile range (IQR) is 88−72=1688 - 72 = 16. The lower boundary for outliers is calculated as Q1−1.5×IQRQ1 - 1.5 \times IQR. This gives 72−1.5×16=72−24=4872 - 1.5 \times 16 = 72 - 24 = 48. Since the score of 45 is less than 48, it is an outlier.

Question 5

The daily temperatures in Celsius for a week in a city have a mean of 15°C and a standard deviation of 3°C.

The temperatures are converted to Fahrenheit using the formula F=1.8C+32F = 1.8C + 32. What are the new mean and standard deviation in degrees Fahrenheit, correct to one decimal place?

  1. Mean = 59.0°F, Standard Deviation = 3.0°F
  2. Mean = 59.0°F, Standard Deviation = 5.4°F (correct answer)
  3. Mean = 59.0°F, Standard Deviation = 35.4°F
  4. Mean = 27.0°F, Standard Deviation = 5.4°F
Explanation: When data is transformed by multiplying by a constant aa and adding a constant bb, the new mean is a×old mean+ba \times \text{old mean} + b and the new standard deviation is ∣a∣×old standard deviation|a| \times \text{old standard deviation}. New mean = 1.8×15+32=27+32=59.01.8 \times 15 + 32 = 27 + 32 = 59.0. New standard deviation = 1.8×3=5.41.8 \times 3 = 5.4. Adding a constant does not affect measures of spread like standard deviation.

Question 6

Two basketball players, Alex and Ben, record their points per game over a season. Alex's scores have a median of 24 points and an interquartile range of 7 points. Ben's scores have a median of 21 points and an interquartile range of 12 points.

Which statement is the most accurate interpretation of these statistics?

  1. Alex typically scores more points and is a more consistent scorer than Ben. (correct answer)
  2. Alex typically scores more points but is a less consistent scorer than Ben.
  3. Ben typically scores more points and is a more consistent scorer than Alex.
  4. Ben typically scores more points but is a less consistent scorer than Alex.
Explanation: The median represents a typical value. Alex's median (24) is higher than Ben's (21), so Alex typically scores more points. The interquartile range (IQR) is a measure of spread or consistency. A smaller IQR indicates greater consistency. Alex's IQR (7) is smaller than Ben's (12), so Alex is the more consistent scorer.

Question 7

In an IB school, the 20 students in the Higher Level (HL) Math class have a mean score of 85 on a test. The 15 students in the Standard Level (SL) Math class have a mean score of 74 on the same test.

What is the combined mean score for all 35 students, correct to three significant figures?

  1. 79.5
  2. 80.0
  3. 80.3 (correct answer)
  4. 81.1
Explanation: This is a weighted average problem. The total score for the HL class is 20×85=170020 \times 85 = 1700. The total score for the SL class is 15×74=111015 \times 74 = 1110. The combined total score is 1700+1110=28101700 + 1110 = 2810. The total number of students is 20+15=3520 + 15 = 35. The combined mean is 2810÷35≈80.28572810 \div 35 \approx 80.2857. To three significant figures, this is 80.3.

Question 8

The following is a list of house prices in a neighbourhood, in thousands of dollars: {350, 375, 380, 400, 410, 420, 850}. The house priced at $850,000 is identified as an outlier and is removed from the dataset.

How will the mean and standard deviation of the house prices change after removing the outlier?

  1. The mean will decrease and the standard deviation will decrease. (correct answer)
  2. The mean will decrease and the standard deviation will increase.
  3. The mean will increase and the standard deviation will decrease.
  4. The mean will increase and the standard deviation will increase.
Explanation: The outlier (850) is a high value. Removing a value that is significantly higher than the rest of the data will cause the mean to decrease. The standard deviation is a measure of spread. Since the outlier is very far from the other data points, it contributes significantly to the spread. Removing it will make the data less spread out, causing the standard deviation to decrease.

Question 9

A data set consists of six ordered numbers: {10, 12, 15, 17, 20, 24}.

The number 10 was recorded by mistake. If it is corrected to 22, how do the mean and median of the data set change?

  1. The mean increases and the median stays the same.
  2. The mean increases and the median increases. (correct answer)
  3. The mean decreases and the median increases.
  4. Both the mean and the median stay the same.
Explanation: Original set: {10, 12, 15, 17, 20, 24}. Original mean = 98/6 ≈ 16.3. Original median = (15+17)/2 = 16. New set (ordered): {12, 15, 17, 20, 22, 24}. The sum increases from 98 to 110, so the mean increases to 110/6 ≈ 18.3. The new median is the average of the two new middle numbers, 17 and 20. New median = (17+20)/2 = 18.5. Therefore, both the mean and the median increase.

Question 10

A factory produces bolts with a mean length of 50.0 mm. The lengths of bolts from Machine A have a standard deviation of 0.1 mm, while bolts from Machine B have a standard deviation of 0.5 mm.

What can be concluded about the two machines based on this information?

  1. Machine A produces longer bolts on average than Machine B.
  2. Machine B produces bolts with more consistent lengths than Machine A.
  3. Machine A produces bolts with more consistent lengths than Machine B. (correct answer)
  4. The total output of Machine A is higher than Machine B.
Explanation: Standard deviation is a measure of the consistency or precision of a process. A smaller standard deviation indicates that the data points tend to be very close to the mean. Since Machine A has a smaller standard deviation (0.1 mm) than Machine B (0.5 mm), the bolts produced by Machine A have lengths that are more consistent and closer to the mean length of 50.0 mm.

Question 11

The following data shows the time, in minutes, that 100 people spent in a supermarket. Time (t minutes): 0 < t ≤ 10, Frequency: 15 Time (t minutes): 10 < t ≤ 20, Frequency: 35 Time (t minutes): 20 < t ≤ 30, Frequency: 40 Time (t minutes): 30 < t ≤ 40, Frequency: 10

What is the best estimate for the mean time spent in the supermarket?

  1. 14.5 minutes
  2. 19.5 minutes (correct answer)
  3. 24.5 minutes
  4. 25.0 minutes
Explanation: To estimate the mean from grouped data, use the midpoint of each class interval. The midpoints are 5, 15, 25, and 35. The estimated mean is ∑(f×x)∑f\frac{\sum(f \times x)}{\sum f}, where ff is the frequency and xx is the midpoint. Mean =(15×5)+(35×15)+(40×25)+(10×35)100=75+525+1000+350100=1950100=19.5= \frac{(15 \times 5) + (35 \times 15) + (40 \times 25) + (10 \times 35)}{100} = \frac{75 + 525 + 1000 + 350}{100} = \frac{1950}{100} = 19.5.

Question 12

In an IB school, the 20 students in the Higher Level (HL) Math class have a mean score of 85 on a test. The 15 students in the Standard Level (SL) Math class have a mean score of 74 on the same test.

What is the combined mean score for all 35 students, correct to three significant figures?

  1. 79.5
  2. 80.0
  3. 80.3 (correct answer)
  4. 81.1
Explanation: This is a weighted average problem. The total score for the HL class is 20×85=170020 \times 85 = 1700. The total score for the SL class is 15×74=111015 \times 74 = 1110. The combined total score is 1700+1110=28101700 + 1110 = 2810. The total number of students is 20+15=3520 + 15 = 35. The combined mean is 2810÷35≈80.28572810 \div 35 \approx 80.2857. To three significant figures, this is 80.3.

Question 13

A survey records the annual income of 9 employees at a small company: {$45k, $48k, $50k, $52k, $55k, $58k, $60k, $65k, $150k}. The CEO's income of $150k is significantly higher than the others.

Which statement best describes the effect of the CEO's high income on the measures of spread?

  1. The range is heavily influenced by the high income, while the interquartile range is not. (correct answer)
  2. The interquartile range is heavily influenced by the high income, while the range is not.
  3. Both the range and the interquartile range are heavily influenced by the high income.
  4. The high income does not significantly affect either the range or the interquartile range.
Explanation: The range is calculated as the maximum value minus the minimum value. The high income of $150k is the maximum, so it directly and significantly increases the range. The interquartile range (IQR) measures the spread of the middle 50% of the data and is resistant to outliers. The high income value will not affect Q1 and is unlikely to significantly affect Q3, thus leaving the IQR relatively unchanged compared to the range.

Question 14

The heights of 80 saplings were measured. The results are summarized in the frequency table: Height (h cm): 0 < h ≤ 20, Frequency: 12 Height (h cm): 20 < h ≤ 40, Frequency: 25 Height (h cm): 40 < h ≤ 60, Frequency: 30 Height (h cm): 60 < h ≤ 80, Frequency: 13

In which class interval does the median height lie?

  1. 0 < h ≤ 20
  2. 20 < h ≤ 40
  3. 40 < h ≤ 60 (correct answer)
  4. 60 < h ≤ 80
Explanation: There are 80 data points, so the median is the average of the 40th and 41st values. We need to find which class interval contains these values. We can use cumulative frequency: The first class (0-20 cm) contains the first 12 saplings. The second class (20-40 cm) contains the next 25, so it ends at the 12 + 25 = 37th sapling. The third class (40-60 cm) contains the next 30 saplings, from the 38th to the 67th. Since the 40th and 41st values fall within this range, the median lies in the 40 < h ≤ 60 interval.

Question 15

The mean salary at a company is $70,000, while the median salary is $55,000. Which of the following statements is the most likely explanation for this difference?

  1. The salaries are symmetrically distributed around the mean.
  2. The data is skewed to the left by a few very low salaries.
  3. A few high-earning executives are pulling the mean salary upwards. (correct answer)
  4. There is a data entry error, as the mean and median should be closer.
Explanation: When the mean is significantly greater than the median, the data distribution is positively skewed (skewed to the right). This is typically caused by a few very high values (outliers) that have a large effect on the mean but little effect on the median. The high salaries of a few executives would pull the average up without changing the middle value (median) as much.

Question 16

A data set consists of six ordered numbers: {10, 12, 15, 17, 20, 24}.

The number 10 was recorded by mistake. If it is corrected to 22, how do the mean and median of the data set change?

  1. The mean increases and the median stays the same.
  2. The mean increases and the median increases. (correct answer)
  3. The mean decreases and the median increases.
  4. Both the mean and the median stay the same.
Explanation: Original set: {10, 12, 15, 17, 20, 24}. Original mean = 98/6 ≈ 16.3. Original median = (15+17)/2 = 16. New set (ordered): {12, 15, 17, 20, 22, 24}. The sum increases from 98 to 110, so the mean increases to 110/6 ≈ 18.3. The new median is the average of the two new middle numbers, 17 and 20. New median = (17+20)/2 = 18.5. Therefore, both the mean and the median increase.

Question 17

A factory produces bolts with a mean length of 50.0 mm. The lengths of bolts from Machine A have a standard deviation of 0.1 mm, while bolts from Machine B have a standard deviation of 0.5 mm.

What can be concluded about the two machines based on this information?

  1. Machine A produces longer bolts on average than Machine B.
  2. Machine B produces bolts with more consistent lengths than Machine A.
  3. Machine A produces bolts with more consistent lengths than Machine B. (correct answer)
  4. The total output of Machine A is higher than Machine B.
Explanation: Standard deviation is a measure of the consistency or precision of a process. A smaller standard deviation indicates that the data points tend to be very close to the mean. Since Machine A has a smaller standard deviation (0.1 mm) than Machine B (0.5 mm), the bolts produced by Machine A have lengths that are more consistent and closer to the mean length of 50.0 mm.

Question 18

A city council wants to understand income disparity. They are particularly interested in the spread of incomes for the middle half of the residents, ignoring the very poor and the very wealthy. Which statistic would be the most appropriate for this purpose?

  1. The mean
  2. The range
  3. The standard deviation
  4. The interquartile range (correct answer)
Explanation: The interquartile range (IQR) measures the spread of the middle 50% (the 'middle half') of the data, from the 25th percentile (Q1) to the 75th percentile (Q3). It is specifically designed to be resistant to the influence of extreme values (outliers), such as the incomes of the very poor and very wealthy, making it the most appropriate choice.

Question 19

The heights of 80 saplings were measured. The results are summarized in the frequency table: Height (h cm): 0 < h ≤ 20, Frequency: 12 Height (h cm): 20 < h ≤ 40, Frequency: 25 Height (h cm): 40 < h ≤ 60, Frequency: 30 Height (h cm): 60 < h ≤ 80, Frequency: 13

In which class interval does the median height lie?

  1. 0 < h ≤ 20
  2. 20 < h ≤ 40
  3. 40 < h ≤ 60 (correct answer)
  4. 60 < h ≤ 80
Explanation: There are 80 data points, so the median is the average of the 40th and 41st values. We need to find which class interval contains these values. We can use cumulative frequency: The first class (0-20 cm) contains the first 12 saplings. The second class (20-40 cm) contains the next 25, so it ends at the 12 + 25 = 37th sapling. The third class (40-60 cm) contains the next 30 saplings, from the 38th to the 67th. Since the 40th and 41st values fall within this range, the median lies in the 40 < h ≤ 60 interval.

Question 20

A survey records the annual income of 9 employees at a small company: {$45k, $48k, $50k, $52k, $55k, $58k, $60k, $65k, $150k}. The CEO's income of $150k is significantly higher than the others.

Which statement best describes the effect of the CEO's high income on the measures of spread?

  1. The range is heavily influenced by the high income, while the interquartile range is not. (correct answer)
  2. The interquartile range is heavily influenced by the high income, while the range is not.
  3. Both the range and the interquartile range are heavily influenced by the high income.
  4. The high income does not significantly affect either the range or the interquartile range.
Explanation: The range is calculated as the maximum value minus the minimum value. The high income of $150k is the maximum, so it directly and significantly increases the range. The interquartile range (IQR) measures the spread of the middle 50% of the data and is resistant to outliers. The high income value will not affect Q1 and is unlikely to significantly affect Q3, thus leaving the IQR relatively unchanged compared to the range.