Data Representation & Summary Stats - IB Mathematics Analysis and Approaches

For the dataset $\{5, 9, 11, x\}$ , the mean is equal to the median. Which of the following is a possible value for $x$ ?
Answer: 7 We must consider the ordering of the data. Case 1: $x \le 5$ . The ordered set is $\{x, 5, 9, 11\}$ . The median is $\frac{5+9}{2} = 7$ . The mean is $\frac{x+5+9+11}{4} = \frac{x+25}{4}$ . Setting them equal: $\frac{x+25}{4} = 7 \implies x+25=28 \implies x=3$ . This is consistent with $x \le 5$ . Case 2: $5 < x < 9$ . The ordered set is $\{5, x, 9, 11\}$ . The median is $\frac{x+9}{2}$ . The mean is $\frac{x+25}{4}$ . Setting them equal: $\frac{x+9}{2} = \frac{x+25}{4} \implies 2(x+9) = x+25 \implies 2x+18 = x+25 \implies x=7$ . This is consistent with $5 < x < 9$ . Case 3: $9 \le x \le 11$ . The ordered set is $\{5, 9, x, 11\}$ . Median is $(9+x)/2$ . Mean is $(x+25)/4$ . $(9+x)/2 = (x+25)/4 \implies 18+2x=x+25 \implies x=7$ . This contradicts $x \ge 9$ . Case 4: $x > 11$ . The ordered set is $\{5, 9, 11, x\}$ . The median is $\frac{9+11}{2} = 10$ . The mean is $\frac{x+25}{4}$ . Setting them equal: $\frac{x+25}{4} = 10 \implies x+25=40 \implies x=15$ . This is consistent with $x>11$ . The possible values for $x$ are 3, 7, and 15. Of the given choices, only 7 is a possible value.
A dataset has mean $\bar{x}$ , median $M$ , and interquartile range $I$ . Each value in the dataset is first multiplied by 3 and then increased by 4. What are the new mean, median, and interquartile range?
Answer: New mean $3\bar{x}+4$ , new median $3M+4$ , new IQR $3I$ When each data point $x_i$ is transformed to $ax_i+b$ , measures of central tendency (like mean and median) are transformed in the same way. The new mean is $3\bar{x}+4$ and the new median is $3M+4$ . Measures of spread (like range and IQR) are only affected by multiplication, not by addition. The new interquartile range is $I' = Q'_3 - Q'_1 = (3Q_3+4) - (3Q_1+4) = 3Q_3 - 3Q_1 = 3(Q_3 - Q_1) = 3I$ .
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?
Answer: 11 Let the five integers in ascending order be $a, b, c, d, e$ . The median is the middle value, so $c = 12$ . The mode is 15, and since the numbers are ordered and the median is 12, the two largest numbers must be 15. So, $d = 15$ and $e = 15$ . The set is $\{a, b, 12, 15, 15\}$ . The mean is 10, so the sum is $5 \times 10 = 50$ . The sum is $a+b+12+15+15 = a+b+42$ . So, $a+b+42 = 50$ , which means $a+b=8$ . Since the numbers are in ascending order, we have $a \le b \le c = 12$ . We want to find the smallest possible range, which is $e-a = 15-a$ . To minimize the range, we must maximize $a$ . Given $a+b=8$ and $a \le b$ , the maximum value of $a$ occurs when $a=b$ . Thus, $2a=8$ , so $a=4$ . This gives the set $\{4, 4, 12, 15, 15\}$ , which satisfies all conditions. The range is $15-4=11$ .
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)?
Answer: The IQR will increase. 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.
Find the value of $a$ , the frequency of scoring 3 goals.
Answer: 4 The total frequency is $N = 4+7+a+3+2 = 16+a$ . The median is 2. The cumulative frequency for $x=1$ is 4. The cumulative frequency for $x=2$ is $4+7=11$ . For the median to be 2, the median position $\frac{N+1}{2}$ must correspond to a value of 2. This means $4 < \frac{N+1}{2} \le 11$ . This condition holds for a wide range of $a$ . Now, use the mean. The sum of the scores is $\sum fx = 1(4)+2(7)+3(a)+4(3)+5(2) = 4+14+3a+12+10 = 40+3a$ . The mean is $\frac{\sum fx}{N} = \frac{40+3a}{16+a}$ . We are given that the mean is 2.6. So, $\frac{40+3a}{16+a} = 2.6$ . This gives $40+3a = 2.6(16+a)$ , which is $40+3a = 41.6+2.6a$ . Rearranging gives $0.4a = 1.6$ , so $a=4$ . We can check this is consistent with the median condition.
A dataset is given by $\{10, 15, 17, 19, 20, 23, 25, 45\}$ . Using the criterion that an outlier is a value more than $1.5 \times IQR$ below the lower quartile or above the upper quartile, which of the following statements is true?
Answer: 45 is the only outlier in the dataset. The dataset is ordered and has $n=8$ values. The lower quartile $Q_1$ is the median of the lower half $\{10, 15, 17, 19\}$ , so $Q_1 = \frac{15+17}{2} = 16$ . The upper quartile $Q_3$ is the median of the upper half $\{20, 23, 25, 45\}$ , so $Q_3 = \frac{23+25}{2} = 24$ . The interquartile range is $IQR = Q_3 - Q_1 = 24 - 16 = 8$ . The lower boundary for outliers is $Q_1 - 1.5 \times IQR = 16 - 1.5 \times 8 = 16 - 12 = 4$ . The upper boundary for outliers is $Q_3 + 1.5 \times IQR = 24 + 1.5 \times 8 = 24 + 12 = 36$ . 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.
The mean score of $n$ 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 $n$ .
Answer: 21 The sum of the boys' scores is $70n$ . The sum of the girls' scores is $15 \times 82 = 1230$ . The total sum of scores for all students is $70n + 1230$ . The total number of students is $n+15$ . The combined mean is the total sum divided by the total number of students. Therefore, $\frac{70n + 1230}{n+15} = 75$ . Multiplying both sides by $(n+15)$ gives $70n + 1230 = 75(n+15)$ , which simplifies to $70n + 1230 = 75n + 1125$ . Rearranging the terms, we get $5n = 1230 - 1125$ , so $5n = 105$ . Dividing by 5 gives $n=21$ .
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?
Answer: A small number of executives have extremely high incomes. 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.
Find the exact value of the mean of the set $\{\cos^2(0^\circ), \cos^2(30^\circ), \cos^2(45^\circ), \cos^2(60^\circ), \cos^2(90^\circ)\}$ .
Answer: $\frac{1}{2}$ When you encounter questions asking for the mean of trigonometric values at standard angles, you need to evaluate each trigonometric function at the given angles, then apply the basic mean formula. Let's calculate each cosine squared value at the standard angles: - $\cos^2(0°) = 1^2 = 1$ - $\cos^2(30°) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$ - $\cos^2(45°) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}$ - $\cos^2(60°) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$ - $\cos^2(90°) = 0^2 = 0$ Now find the mean by adding these values and dividing by 5: $\text{Mean} = \frac{1 + \frac{3}{4} + \frac{1}{2} + \frac{1}{4} + 0}{5} = \frac{\frac{4+3+2+1+0}{4}}{5} = \frac{\frac{10}{4}}{5} = \frac{10}{20} = \frac{1}{2}$ This confirms answer D is correct. Looking at the wrong answers: A) $\frac{2}{5}$ results from incorrectly adding the numerators without proper fraction conversion. B) $\frac{5}{8}$ comes from computational errors in the fraction arithmetic. C) $\frac{3}{5}$ typically arises from mistakes in evaluating the cosine values at standard angles or incorrect simplification. Study tip: Memorize the exact values of sine and cosine at 0°, 30°, 45°, 60°, and 90°. These appear frequently on IB exams, and having them memorized prevents calculation errors that lead to wrong answer choices.
The five-number summary for a dataset is: Minimum = 10, $Q_1$ = 18, Median = 25, $Q_3$ = 30, Maximum = 50. Which of the following values in the dataset would be considered an outlier using the $1.5 \times IQR$ rule?
Answer: 50 only When you encounter outlier detection problems, you're applying the standard $1.5 \times IQR$ rule to determine which data points fall unusually far from the central tendency of your dataset. To find outliers, first calculate the interquartile range: $IQR = Q_3 - Q_1 = 30 - 18 = 12$ . Then determine the outlier boundaries by extending $1.5 \times IQR = 1.5 \times 12 = 18$ units beyond the quartiles. The lower boundary is $Q_1 - 18 = 18 - 18 = 0$ , and the upper boundary is $Q_3 + 18 = 30 + 18 = 48$ . 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 $1.5 \times IQR$ , then check if extreme values fall outside the boundaries $Q_1 - 1.5 \times IQR$ and $Q_3 + 1.5 \times IQR$ . Don't assume minimum and maximum values are automatically outliers—they must exceed these calculated boundaries.
The mean of a set of $n$ numbers is $M$ . A new number, $X$ , is added to the set. What is the new mean of the $n+1$ numbers?
Answer: $\frac{nM+X}{n+1}$ The sum of the original $n$ numbers is given by $S_n = n \times M$ . When the new number $X$ is added, the new sum is $S_{n+1} = S_n + X = nM + X$ . The new set has $n+1$ numbers. Therefore, the new mean is the new sum divided by the new count of numbers: $\text{New Mean} = \frac{nM+X}{n+1}$ .
In a dataset, the first quartile $Q_1$ is 50 and the third quartile $Q_3$ is 80. Which of the following statements must be true?
Answer: Approximately 50% of the data lies between 50 and 80. 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.
The mean of a list of 8 numbers is 12.5. One of the numbers, 20, is removed and replaced by a new number, $x$ . The new mean is 11.0. Find the value of $x$ .
Answer: 8 The original sum of the 8 numbers is $S_{old} = 8 \times 12.5 = 100$ . When the number 20 is removed, the sum of the remaining 7 numbers is $100 - 20 = 80$ . The new number $x$ is added to these 7 numbers, so the new sum of 8 numbers is $S_{new} = 80 + x$ . The new mean is 11.0, so $\frac{S_{new}}{8} = 11.0$ . This means $\frac{80+x}{8} = 11$ . Multiplying by 8 gives $80+x = 88$ . Therefore, $x = 8$ .
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?
Answer: The mean and the median only 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 $10 \times 15 = 150$ 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: $(Q_3 - 10) - (Q_1 - 10) = Q_3 - Q_1$ . 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.
Find the mean of the dataset $\{\log_2(k), \log_2(2k), \log_2(4k), \log_2(8k)\}$ for $k>0$ .
Answer: $\log_2(k) + \frac{3}{2}$ First, simplify the terms using the logarithm property $\log(ab) = \log(a) + \log(b)$ : $\log_2(k)$ , $\log_2(2) + \log_2(k) = 1 + \log_2(k)$ , $\log_2(4) + \log_2(k) = 2 + \log_2(k)$ , $\log_2(8) + \log_2(k) = 3 + \log_2(k)$ . The sum of these four terms is $S = \log_2(k) + (1+\log_2(k)) + (2+\log_2(k)) + (3+\log_2(k)) = 4\log_2(k) + (1+2+3) = 4\log_2(k) + 6$ . Since there are 4 terms, the mean is $\frac{S}{4} = \frac{4\log_2(k) + 6}{4} = \log_2(k) + \frac{6}{4} = \log_2(k) + \frac{3}{2}$ .

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