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.

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%.

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.

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.

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.

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.

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.

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.

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.

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.