PSAT Math Quiz: Center Shape And Spread Of Data

What this quiz covers

This quiz focuses on Center Shape And Spread Of Data, giving you a quick way to practice the rules, question types, and explanations that matter most for PSAT Math.

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.

Question 1

1. Same median; A has larger range
2. Same median; B has larger range (correct answer)
3. A has larger median; same range
4. B has larger median; B has larger range

Explanation: 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.

Question 2

1. Mean decreases; median decreases (correct answer)
2. Mean decreases; median stays the same
3. Mean stays the same; median decreases
4. Mean increases; median decreases

Explanation: The question asks how adding a score of 20 affects the mean and median of the original 5 test scores. Original mean is (72+75+78+80+95)/5 = 400/5 = 80; new mean (400+20)/6 = 420/6 = 70, so decreases. Original median is 78 (3rd in ordered); new sorted: 20,72,75,78,80,95, median (75+78)/2 = 76.5, so decreases. Both decrease, with mean affected more by the low value. A key error is not reordering for median or miscounting positions. Remember, adding extremes pulls mean but median shifts only if position changes. As a strategy, recalculate both measures before and after to observe effects.

Question 3

1. 17
2. 18
3. 24 (correct answer)
4. 58

Explanation: The question asks for the range of cupcakes sold over 6 days: 34, 40, 41, 41, 43, 58. To find the range, subtract the minimum value (34) from the maximum (58), giving 24. This simple calculation measures the total spread of the data. Order the data if needed, but here min and max are clear. A key error is computing IQR (Q3 - Q1) instead, which for this data would be 41.5 - 40.5 = 1, not the range. Distinguish range as max minus min for overall spread, while IQR focuses on middle 50%.

Question 4

1. 0
2. 1 (correct answer)
3. 2
4. 3

Explanation: 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.

Question 5

1. IQR $=6$; farthest value is 12
2. IQR $=8$; farthest value is 20
3. IQR $=6$; farthest value is 20 (correct answer)
4. IQR $=14$; farthest value is 4

Explanation: 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.

Question 6

1. Median $=9$; mean is more affected (correct answer)
2. Median $=10$; mean is more affected
3. Median $=9$; median is more affected
4. Median $=11$; median is more affected

Explanation: 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.

Question 7

1. Mean $>$ median, and the median is less affected (correct answer)
2. Mean $<$ median, and the mean is less affected
3. Mean $=$ median, and both are equally affected
4. Mean $<$ median, and the median is less affected

Explanation: The question asks for the relationship between the mean and median of the runners' finishing times and which measure is less affected by the outlier 60 to best represent a typical time. To calculate the median, order the data: 18, 19, 19, 20, 21, 22, 60; with 7 values, the median is the 4th value, which is 20. For the mean, sum the values: 18 + 19 + 19 + 20 + 21 + 22 + 60 = 179, then divide by 7 to get approximately 25.57, showing the mean is greater than the median. The outlier 60 significantly increases the mean but has little effect on the median, making the median less affected and better for representing typical times in skewed data. A common error is miscalculating the sum or forgetting to order the data for the median. Understanding that the mean is sensitive to extreme values while the median is resistant helps in choosing the appropriate measure of center.

Question 8

1. 1
2. 2 (correct answer)
3. 3
4. 4

Explanation: This question asks for the interquartile range (IQR) of quiz scores using the median-of-halves method. First, the data is already ordered: 6, 7, 7, 8, 8, 9, 9, 10. With 8 values, split into lower half (6, 7, 7, 8) and upper half (8, 9, 9, 10). Q1 is the median of the lower half: (7+7)÷2 = 7. Q3 is the median of the upper half: (9+9)÷2 = 9. Therefore, IQR = Q3 - Q1 = 9 - 7 = 2. The IQR measures the spread of the middle 50% of data and is resistant to outliers. When using the median-of-halves method with an even number of data points, include the median values in both halves if needed.

Question 9

1. IQR 6; range 22
2. IQR 6; range 25 (correct answer)
3. IQR 9; range 22
4. IQR 9; range 25

Explanation: 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.

Question 10

1. Mean increases; median stays the same (correct answer)
2. Mean stays the same; median increases
3. Both mean and median increase by the same amount
4. Mean decreases; median stays the same

Explanation: The question asks how adding a score of 40 affects the mean and median of points scored in 7 games: 8, 9, 9, 10, 10, 11, 13. The original mean is the sum (70) divided by 7, equaling 10; after adding 40, the new sum is 110 divided by 8, or 13.75, so the mean increases. The original median is the 4th value in the ordered list, which is 10; with 8 values now (8, 9, 9, 10, 10, 11, 13, 40), the median is the average of the 4th and 5th (both 10), staying the same at 10. This shows the mean is sensitive to high outliers, while the median resists them. A key error is assuming the median shifts with the added value without reordering and recounting positions. When adding data points, recalculate both measures carefully to observe differential impacts on center.

Question 11

1. About 5 mm
2. About 10 mm
3. About 15 mm (correct answer)
4. About 25 mm

Explanation: 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.

Question 12

1. 38 pounds
2. 42 pounds
3. 45 pounds
4. 52 pounds (correct answer)

Explanation: Shelter A median = 40 (20 dogs), Shelter B median = 48 (30 dogs). When combined (50 dogs), the median is the average of the 25th and 26th values. The combined median must lie between the two individual medians, i.e., between 40 and 48 (inclusive) under typical conditions, but can be pulled toward the shelter with more data. The combined median cannot exceed 48 (Shelter B's median), because at least half of Shelter B (15 dogs) are ≤48 and all 20 of Shelter A could be up to 40-ish, so at least 15+? Precisely: at least 10 (half of A) are ≤40 and at least 15 (half of B) are ≤48, giving at least 25 values ≤48. With 50 total, the 25th and 26th values are ≤48. So 52 is impossible.

Question 13

1. 14
2. 17 (correct answer)
3. 20
4. 24

Explanation: 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.

Question 14

1. The mean
2. The median (correct answer)
3. The range
4. The standard deviation

Explanation: 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.

Question 15

1. Range $=26$; mean best represents typical
2. Range $=25$; median best represents typical
3. Range $=26$; median best represents typical (correct answer)
4. Range $=40$; mean best represents typical

Explanation: 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.

Question 16

1. Group X has a larger standard deviation
2. Group X has a smaller standard deviation (correct answer)
3. Both groups must have the same standard deviation
4. Standard deviation cannot be compared from histograms

Explanation: The question asks to compare the standard deviations of reading times for Group X and Group Y based on their histograms. Group X's values are clustered narrowly between 10–20 minutes, indicating low variability. Group Y's values spread widely across 0–40 minutes, showing higher variability. Standard deviation measures average deviation from the mean, so a narrower spread means smaller standard deviation for Group X. A key error is assuming equal spreads without considering the clustering. When comparing histograms, assess the width and clustering to infer variability measures like standard deviation.

Question 17

1. 0–10 minutes
2. 10–20 minutes
3. 20–30 minutes (correct answer)
4. 30–40 minutes

Explanation: The question asks for the median bin in a histogram of homework times for 30 students, with given bin frequencies: 0–10: 4, 10–20: 8, 20–30: 10, 30–40: 6, 40–50: 2. For 30 values, the median is the average of the 15th and 16th values in the ordered distribution. Use cumulative frequencies: up to 10: 4, up to 20: 12, up to 30: 22, so the 15th and 16th fall in the 20–30 bin. Thus, the median bin is 20–30 minutes. A key error is miscounting cumulatives or forgetting to average positions for even counts. When working with histograms, build cumulatives carefully to locate the median accurately.

Question 18

1. Mean (correct answer)
2. Median
3. They change by the same amount
4. Neither changes

Explanation: 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.

Question 19

1. 7
2. 8 (correct answer)
3. 8.5
4. 9

Explanation: 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.

Question 20

1. The mean is better because it is not affected by outliers.
2. The median is better because the outlier increases the mean more than the median. (correct answer)
3. The mean and median are equal, so either works equally well.
4. The mode is better because it always equals the middle value.

Explanation: 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.