Data Representation & Summary Stats - IB Mathematics Applications and Interpretation

The ages of 12 people on a committee are: 28, 35, 38, 42, 45, 45, 51, 53, 56, 60, 62, 68. Find the interquartile range (IQR) of the ages.
Answer: 18 The data is already ordered. With n=12, the lower half of the data is the first 6 values: \{28, 35, 38, 42, 45, 45\}. The lower quartile (Q1) is the median of this half, which is the average of the 3rd and 4th values: $(38+42)/2 = 40$ . The upper half is \{51, 53, 56, 60, 62, 68\}. The upper quartile (Q3) is the median of this half, which is the average of its 3rd and 4th values: $(56+60)/2 = 58$ . The IQR is $Q3 - Q1 = 58 - 40 = 18$ .
How will the mean and standard deviation of the house prices change after removing the outlier?
Answer: The mean will decrease and the standard deviation will decrease. 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.
Which statement is the most accurate interpretation of these statistics?
Answer: Alex typically scores more points and is a more consistent scorer than Ben. 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.
The mean height of 10 students is 165 cm. One student leaves the group, and the mean height of the remaining 9 students becomes 164 cm. What was the height of the student who left?
Answer: 174.0 cm The original total height of the 10 students was $10 \times 165 = 1650 \text{ cm}$ . The new total height of the remaining 9 students is $9 \times 164 = 1476 \text{ cm}$ . The height of the student who left is the difference between the original total height and the new total height: $1650 - 1476 = 174 \text{ cm}$ .
A dataset contains the scores of 5 students on a quiz: \{8, 8, 8, 8, 8\}. Which of the following statements correctly describes the measures of central tendency and dispersion for this dataset?
Answer: The mean is 8 and the standard deviation is 0. The mean is the sum of the values divided by the count: $(8+8+8+8+8)/5 = 40/5 = 8$ . The standard deviation measures the average distance of data points from the mean. Since all data points are exactly equal to the mean, there is no variation or spread. Therefore, the standard deviation is 0.
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?
Answer: The lower boundary is 48, and the score is an outlier. 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 $88 - 72 = 16$ . The lower boundary for outliers is calculated as $Q1 - 1.5 \times IQR$ . This gives $72 - 1.5 \times 16 = 72 - 24 = 48$ . Since the score of 45 is less than 48, it is an outlier.
What is the best estimate for the mean time spent in the supermarket?
Answer: 19.5 minutes 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 $\frac{\sum(f \times x)}{\sum f}$ , where $f$ is the frequency and $x$ is the midpoint. Mean $= \frac{(15 \times 5) + (35 \times 15) + (40 \times 25) + (10 \times 35)}{100} = \frac{75 + 525 + 1000 + 350}{100} = \frac{1950}{100} = 19.5$ .
What can be concluded about the two machines based on this information?
Answer: Machine A produces bolts with more consistent lengths than Machine B. 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.
The number 10 was recorded by mistake. If it is corrected to 22, how do the mean and median of the data set change?
Answer: The mean increases and the median increases. 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.
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?
Answer: The interquartile range 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.
Consider the ordered data set of 9 numbers: \{5, 8, 10, 12, 15, 18, 20, 21, 25\}. If the maximum value, 25, is removed from the set, what happens to the third quartile (Q3)?
Answer: It decreases from 20.5 to 19. In the original set (n=9), Q3 is the median of the upper half \{18, 20, 21, 25\}, which is $(20+21)/2 = 20.5$ . After removing 25, the new set is \{5, 8, 10, 12, 15, 18, 20, 21\} (n=8). The new upper half is \{15, 18, 20, 21\}. The new Q3 is the median of this new upper half, which is $(18+20)/2 = 19$ . Thus, the third quartile decreases from 20.5 to 19.
What is the combined mean score for all 35 students, correct to three significant figures?
Answer: 80.3 This is a weighted average problem. The total score for the HL class is $20 \times 85 = 1700$ . The total score for the SL class is $15 \times 74 = 1110$ . The combined total score is $1700 + 1110 = 2810$ . The total number of students is $20 + 15 = 35$ . The combined mean is $2810 \div 35 \approx 80.2857$ . To three significant figures, this is 80.3.
What will be the new mean and new standard deviation of the journey times?
Answer: Mean = 44 min, Standard Deviation = 3.3 min An increase of 10% is equivalent to multiplying by 1.1. The new mean is $40 \times 1.1 = 44$ minutes. The original variance is 9 min $^2$ , so the original standard deviation is $\sqrt{9} = 3$ minutes. The standard deviation is also multiplied by 1.1: new standard deviation = $3 \times 1.1 = 3.3$ minutes. Note that the new variance would be $9 \times 1.1^2 = 10.89$ min $^2$ .
Which statement best describes the effect of the CEO's high income on the measures of spread?
Answer: The range is heavily influenced by the high income, while the interquartile range is not. 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.
The temperatures are converted to Fahrenheit using the formula $F = 1.8C + 32$ . What are the new mean and standard deviation in degrees Fahrenheit, correct to one decimal place?
Answer: Mean = 59.0°F, Standard Deviation = 5.4°F When data is transformed by multiplying by a constant $a$ and adding a constant $b$ , the new mean is $a \times \text{old mean} + b$ and the new standard deviation is $|a| \times \text{old standard deviation}$ . New mean = $1.8 \times 15 + 32 = 27 + 32 = 59.0$ . New standard deviation = $1.8 \times 3 = 5.4$ . Adding a constant does not affect measures of spread like standard deviation.
In which class interval does the median height lie?
Answer: 40 < h ≤ 60 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.
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?
Answer: A few high-earning executives are pulling the mean salary upwards. 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.

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