Center, Shape, & Spread of Data - PSAT Math

1. A store recorded the number of customers each hour on a quiet day: 12, 13, 13, 14, 14, 15, 15, 16. Then a bus arrived and one additional hour had 60 customers, making a new data set with 9 values. Which measure of center changes more when the value 60 is added: the mean or the median?
   Answer: Mean The question asks which measure of center, mean or median, changes more when adding 60 to the customer data: 12, 13, 13, 14, 14, 15, 15, 16. Original mean: sum 112 divided by 8 is 14; new mean: 172 divided by 9 is about 19.11, a change of +5.11. Original median: average of 14 and 14 is 14; new median for 9 values is the 5th, 15, a change of +1. The mean changes more due to sensitivity to the outlier 60. A common error is misordering or incorrect position for median. Understand that adding extremes affects mean more than median in skewed data.
2. A dot plot (shown as a bar-style frequency graph) summarizes the number of pets owned by 20 students. The counts are: 0 pets: 5 students, 1 pet: 7 students, 2 pets: 6 students, 3 pets: 2 students. What is the mode of the distribution?
   Answer: 1 The question asks for the mode of the number of pets owned by 20 students, with counts: 0 pets: 5, 1 pet: 7, 2 pets: 6, 3 pets: 2. The mode is the value with the highest frequency, which is 1 pet with 7 students. Compare frequencies: 5 for 0, 7 for 1, 6 for 2, 2 for 3, confirming 1 as the mode. This measure represents the most common value in the distribution. A common error is confusing mode with mean or median, like averaging values instead. Recognize that mode focuses on frequency, useful for categorical or discrete data peaks.
3. The table summarizes the scores of 200 students on a standardized test. The data are approximately normally distributed. Based on the table and the empirical (68-95-99.7) rule, approximately how many students scored between 62 and 86?
   Answer: 190 Mean = 74, standard deviation = 6. The interval [62, 86] corresponds to [mean − 2σ, mean + 2σ] = [74−12, 74+12] = [62, 86]. By the empirical rule, about 95% of data lie within 2 standard deviations of the mean. 95% of 200 = 190 students.
4. The dot plot shown displays the number of books read last month by each of 15 students in a book club. If the student who read 12 books is removed from the data set, which of the following statistics will change the LEAST (in absolute value)?
   Answer: The median Original data: 1,1,2,2,2,3,3,3,3,4,4,4,5,6,12. Original mean ≈ 3.67; new mean = 43/14 ≈ 3.07 (change ≈ 0.60). Original median = 3 (8th value); new median = 3 (average of 7th and 8th values = (3+3)/2 = 3), so change = 0. Original range = 11; new range = 5 (change = 6). Standard deviation decreases substantially because the outlier contributed heavily. The median changes the least.
5. Two machines produce metal rods. The lengths (in cm) from Machine A are: 50, 50, 50, 50, 50, 50. The lengths from Machine B are: 48, 49, 50, 50, 51, 52. Which machine's output has the greater standard deviation, and why (in terms of spread around the mean)?
   Answer: Machine B, because its values vary more around the mean The question asks which machine's rod lengths have the greater standard deviation and why, in terms of spread around the mean, for Machine A (all 50) and B (48, 49, 50, 50, 51, 52). Machine A's values are identical, so deviations from the mean (50) are all zero, giving standard deviation 0. Machine B's mean is 50, but values deviate (e.g., 48 is -2, 52 is +2), resulting in positive standard deviation due to variability. Thus, Machine B has greater spread around its mean. A key error is thinking identical values increase deviation, but they minimize it. Standard deviation quantifies average distance from the mean; calculate deviations to compare variability in production consistency.
6. A teacher recorded quiz scores (out of 10) for 10 students: 6, 7, 7, 8, 8, 8, 9, 9, 10, 10. What is the mean of the scores? Use $\text{mean}=\frac{\text{sum}}{\text{count}}$. Be careful not to confuse the mean with the median or mode.
   Answer: 8.2 This question asks for the mean of 10 quiz scores: 6, 7, 7, 8, 8, 8, 9, 9, 10, 10. To find the mean, we sum all values and divide by the count: (6+7+7+8+8+8+9+9+10+10)÷10 = 82÷10 = 8.2. The mean represents the average score and includes every data point in its calculation. Common errors include confusing the mean with the median (which would be 8, the average of the 5th and 6th values) or the mode (which is 8, appearing three times). When calculating means, double-check your addition and count to avoid arithmetic errors. The answer is B: 8.2.
7. Two students tracked the number of hours they studied each week for 6 weeks. Student A: 2, 2, 3, 3, 3, 5. Student B: 1, 2, 3, 3, 4, 5. Which statement correctly compares the medians and the variability using the range? (Use range $=\max-\min$.)
   Answer: Same median; B has larger range The question asks to compare the medians and ranges (as variability) for two students' study hours over 6 weeks. For median, order each set; both have even count, so average 3rd and 4th: both A and B yield 3. For range, subtract min from max: A is 5-2=3, B is 5-1=4, so same median, B larger range. Emphasize calculating range as max - min for spread. A common error is forgetting to average for even-numbered median or misordering. Remember, range shows full spread but is outlier-sensitive. As a strategy, compute medians first then spreads to systematically compare.
8. A small business tracked the number of customer calls received each day for 7 days: 18, 21, 19, 20, 22, 18, 42. What is the median number of calls? Make sure to order the data first; otherwise it's easy to accidentally choose the mean or the middle value from the unsorted list.
   Answer: 20 This question asks for the median number of customer calls from the data: 18, 21, 19, 20, 22, 18, 42. To find the median, first order the data: 18, 18, 19, 20, 21, 22, 42. With 7 values (odd count), the median is the 4th value, which is 20. The median represents the middle value and is resistant to the outlier (42) in this dataset. A common error is selecting the middle value from the unordered list (20 happens to be in the middle position originally) or calculating the mean instead (which would be about 22.9 due to the outlier). For datasets with outliers, the median often provides a better measure of typical value than the mean.
9. A company tracked the number of customer calls received each day for 8 days: 14, 15, 15, 16, 16, 16, 17, 40. What is the range of the data, and which measure of center (mean or median) best represents a typical day? Choose the option that correctly answers both parts.
   Answer: Range $=26$; median best represents typical The question asks for the range of the data and which measure of center, mean or median, best represents a typical day. To find the range, subtract the minimum from the maximum: 40 - 14 = 26. For the median, order the data: 14, 15, 15, 16, 16, 16, 17, 40; with 8 values, average the 4th and 5th: (16 + 16)/2 = 16. The mean is sum/8 ≈ 18.625, pulled up by the outlier 40, so median better represents typical without outlier influence. A common error is calculating range incorrectly or using mean for skewed data. Remember, median is robust to extremes, while mean is affected. As a strategy, identify outliers first to choose the appropriate center measure.
10. A scientist measures the lengths (in cm) of 11 specimens; results are shown in the stem-and-leaf plot. What is the interquartile range (IQR) of the specimen lengths, in cm?
    Answer: 17 The 11 data values in order are: 22, 25, 28, 31, 34, 37, 40, 42, 45, 48, 52. Median (6th value) = 37. Lower half (first 5 values): 22, 25, 28, 31, 34 → Q1 = 28 (3rd value). Upper half (last 5 values): 40, 42, 45, 48, 52 → Q3 = 45 (3rd value). IQR = 45 − 28 = 17.
11. A student recorded the number of minutes it took to finish 9 practice problems on different days. The times (in minutes) were: 7, 8, 8, 9, 9, 10, 10, 11, 30. What is the median of this data set, and which measure of center (mean or median) is more affected by the outlier 30? Choose the option that correctly answers both parts.
    Answer: Median $=9$; mean is more affected The question asks for the median of the given data set and which measure of center, the mean or the median, is more affected by the outlier 30. To find the median, first order the data: 7, 8, 8, 9, 9, 10, 10, 11, 30; with 9 values, the median is the 5th value, which is 9. The mean is calculated as the sum of all values divided by the count, so the outlier 30 significantly increases the sum and pulls the mean higher. In contrast, the median relies only on the middle value in the ordered list, so changing the outlier does not affect it as long as the middle position remains the same. A common error is forgetting to sort the data or miscounting the middle position for an odd number of values. Another mistake might be thinking the mean is unaffected by extremes, but remember that the mean is sensitive to outliers while the median is more robust. As a test-taking strategy, always verify the sorted order and recall that medians are better for skewed data with outliers.
12. A box plot is described by the five-number summary for Dataset X: min 4, $Q_1=6$, median 9, $Q_3=12$, max 20. What is the interquartile range (IQR), and which value would be considered farthest above the center based on the maximum? Choose the option that correctly gives the IQR and identifies the farthest-above-center value.
    Answer: IQR $=6$; farthest value is 20 The question asks for the interquartile range (IQR) from the five-number summary and which value is farthest above the center based on the maximum. Calculate IQR as Q3 - Q1 = 12 - 6 = 6. The center is the median 9; max 20 is 11 above, farther than min 4 (5 below), so 20 is farthest above. Remember, IQR measures middle spread, and distance from median identifies extremes. A key error is miscalculating IQR by using min/max instead of quartiles. Another mistake might be confusing farthest above with below. As a strategy, note distances from median to confirm extremes after computing IQR.
13. The dot plot shown displays the number of minutes spent on a task by each of 20 participants in a study. If the values 2 and 20 (the minimum and maximum) are removed from the data set, which of the following statements best describes the effect on the mean and the standard deviation?
    Answer: The mean stays approximately the same and the standard deviation decreases. The removed values 2 and 20 are symmetric about the approximate center of the data (~11). Their average is 11, close to the original mean, so removing them barely changes the mean. However, both are far from the mean, so removing them reduces spread → standard deviation decreases.
14. The frequency table shown gives the number of customer complaints received per day at a call center over 25 days. What is the median number of complaints per day?
    Answer: 4 Total observations = 25, so median is the 13th value when ordered. Cumulative counts: 2 complaints: 4 (through position 4); 3 complaints: 4+6=10 (through position 10); 4 complaints: 10+7=17 (through position 17). Position 13 falls within the '4 complaints' group. Median = 4.
15. A teacher recorded quiz scores (out of 10) for 9 students: 6, 7, 7, 8, 8, 8, 9, 9, 10. What is the median of the scores? Be careful to identify the middle value in the ordered list rather than averaging two middle values.
    Answer: 8 The question asks for the median of the quiz scores for 9 students: 6, 7, 7, 8, 8, 8, 9, 9, 10. The data is already ordered; for an odd number of values like 9, the median is the middle value, which is the 5th one, 8. Do not average the 4th and 5th values (8 and 8), as that would incorrectly give 8, but since it's odd, simply take the single middle value. This emphasizes understanding that median calculation differs for odd versus even counts. A common mistake is treating it as even and averaging unnecessarily, potentially leading to choices like 8.5 if misordered. Carefully count the position to find the exact middle for accurate measures of center.
16. A researcher compares two data sets of 5 measurements each. Data Set X: 4, 6, 8, 10, 12. Data Set Y: 4, 6, 8, 10, 52. Which statement about center and spread is true?
    Answer: They have the same median, but Y has a larger mean The question asks which statement about center and spread is true for Data Set X (4, 6, 8, 10, 12) and Y (4, 6, 8, 10, 52). For X, the mean is sum 40 divided by 5 = 8, and median is the 3rd value, 8; for Y, mean is 80 divided by 5 = 16, and median is still 8. Thus, they share the same median, but Y has a larger mean due to the outlier 52. The range for X is 12 - 4 = 8, and for Y is 52 - 4 = 48, so Y has a larger range. A common error is thinking the median changes with the outlier, but it stays as the middle value. Understand that means are pulled by extremes, unlike medians; compare both for full insight into data shifts.
17. A runner tracked miles run each day for 7 days: 2, 3, 3, 4, 4, 4, 10. The runner says the mean is a better description of a typical day than the median. Which statement best evaluates this claim, considering the effect of the 10-mile day?
    Answer: The median is better because the outlier increases the mean more than the median. This question evaluates whether the mean or median better represents typical daily miles for data: 2, 3, 3, 4, 4, 4, 10. The mean is (2+3+3+4+4+4+10)÷7 = 30÷7 ≈ 4.3 miles. The median (middle value when ordered) is 4 miles. The 10-mile outlier pulls the mean above the median, making the mean less representative of a typical day since 6 of 7 days were 4 miles or less. The median is more resistant to outliers and better represents the center when extreme values are present. The runner's claim that the mean is better is incorrect; the median of 4 miles better represents a typical day. When data contains outliers, always consider which measure of center best represents the typical case.
18. A box plot (five-number summary) represents the distribution of times (in minutes) to complete a puzzle for a group of students: minimum 5, $Q_1=8$, median 10, $Q_3=14$, maximum 30. What is the interquartile range (IQR), and what is the range?
    Answer: IQR 6; range 25 The question asks for the interquartile range (IQR) and range from a box plot with minimum 5, Q1=8, median 10, Q3=14, maximum 30. The IQR is Q3 minus Q1, so 14 - 8 = 6, representing the spread of the middle 50% of times. The range is maximum minus minimum, or 30 - 5 = 25, showing the full spread including outliers. These measures highlight different aspects: IQR resists extremes, while range includes them. A key error is subtracting median instead of Q1 from Q3 for IQR. When reading box plots, note that IQR focuses on central spread, making it useful for comparing distributions; verify by direct subtraction.
19. The two histograms shown display the distributions of scores on the same exam for two different groups of students, Group 1 and Group 2. Both groups have the same number of students and the same mean score. Which of the following statements is best supported by the histograms?
    Answer: Group 2 has a larger standard deviation than Group 1. Group 1 has scores concentrated near the mean (tall bars in the middle bins, small tails), giving small deviations. Group 2 has scores spread widely with tall bars in the outer bins and fewer in the center, giving larger deviations from the mean. Therefore Group 2 has a larger standard deviation.
20. Based on the box plots, by approximately how many millimeters is the median monthly rainfall in City R greater than in City S?
    Answer: About 15 mm The median line for City R is at 85 mm, and the median for City S is at 70 mm. The difference is roughly 15 mm, so choice C is correct. Other choices misread the medians.

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