This question asks about data with mean 50 and standard deviation 0. Standard deviation measures the spread of data around the mean; it equals 0 only when there is no variation at all. This means every single value must equal the mean. Since the mean is 50 and there's no deviation from it, all values must be exactly 50. Option A (all zeros) would have mean 0, not 50. Option C (evenly spread) would have positive standard deviation. Option D (half above, half below) would also have positive standard deviation. Only when all values are identical does the standard deviation equal zero.

This question compares two classes using their five-number summaries. Class 1: median = 78, IQR = Q3 - Q1 = 85 - 70 = 15. Class 2: median = 78, IQR = Q3 - Q1 = 90 - 72 = 18. Both classes have the same median (78), but Class 2 has a larger IQR (18 > 15), indicating greater spread in the middle 50% of scores. This shows that distributions can have identical centers but different spreads. The IQR is a better measure of spread than range when comparing distributions because it's not affected by extreme values. When interpreting box plots, always calculate IQR = Q3 - Q1 to measure spread.

This question compares centers and spreads of two data sets. Set A: 4, 4, 5, 5, 6, 6 has mean = (4+4+5+5+6+6)/6 = 30/6 = 5 and range = 6-4 = 2. Set B: 1, 3, 5, 5, 7, 9 has mean = (1+3+5+5+7+9)/6 = 30/6 = 5 and range = 9-1 = 8. Both sets have the same mean (5) but different ranges (2 vs 8). This illustrates that data sets can have identical centers but very different spreads. The key insight is that mean measures center while range measures spread, and these are independent characteristics. Always calculate both measures separately to fully understand a distribution.

This question asks for the mode of mile times: 7.2, 7.1, 7.0, 7.3, 7.1, 7.0, 7.1. The mode is the value that appears most frequently. Counting occurrences: 7.0 appears 2 times, 7.1 appears 3 times, 7.2 appears 1 time, 7.3 appears 1 time. Therefore, the mode is 7.1 minutes since it appears most often (3 times). A common error is confusing mode with median (middle value) or mean (average). When finding the mode, always count how many times each distinct value appears. If multiple values tie for the highest frequency, the data set can have multiple modes.

This question asks which measure of center better represents typical customer traffic when there's an outlier (50 customers). The mean is (12+15+15+16+16+18+20+20+50)/9 = 182/9 ≈ 20.2 customers. The median is the 5th value when ordered: 12, 15, 15, 16, 16, 18, 20, 20, 50, so median = 16 customers. The outlier of 50 pulls the mean up significantly (to 20.2) while the median remains at 16, which better represents the typical hour. The median is less affected by outliers because it only depends on the middle position(s), not the actual values. When data contains outliers, the median often provides a better measure of central tendency than the mean.

This question asks what happens to the median when we replace 25 with 19 in the ordered data: 6, 9, 12, 13, 15, 16, 18, 25. With 8 values, the median is the average of the 4th and 5th values: (13 + 15)/2 = 14. After replacing 25 with 19, the data becomes: 6, 9, 12, 13, 15, 16, 18, 19 (still ordered). The median is still the average of the 4th and 5th values: (13 + 15)/2 = 14. The median stays the same because we only changed a value that was not in the middle positions. This illustrates that the median is resistant to changes in extreme values - only changes to values at or near the median positions affect it.

This question asks which measure to use for typical hourly customers when there's an outlier (60). The data ordered is: 18, 20, 21, 21, 22, 22, 22, 60. For 8 values, the median is the average of the 4th and 5th values: (21+22)÷2 = 21.5. The mean would be (18+20+21+21+22+22+22+60)÷8 = 206÷8 = 25.75, which is pulled up by the outlier. The mode is 22 (appears 3 times). Since the manager wants a measure "resistant to the unusually busy hour," the median (21.5) is the best choice because it's not affected by the extreme value of 60. Always choose the median over the mean when you need a measure that resists outlier influence.

This question asks which measure of center is most affected by the outlier (20) and what the median is. To find the median of 9 values, we order them: 3, 4, 4, 5, 5, 5, 6, 6, 20, then take the 5th value (middle position), which is 5. The mean is (3+4+4+5+5+5+6+6+20)÷9 = 58÷9 ≈ 6.4, while without the outlier it would be 38÷8 = 4.75, showing the mean changes dramatically. The median remains 5 whether we include the outlier or not, demonstrating that the mean is most affected by outliers. When dealing with outliers, always check how each measure of center responds to extreme values.

The question asks for the mean of quiz scores and how it compares to the median. To find the mean: (6+7+7+8+8+8+9+10)÷8 = 63÷8 = 7.875. For the median of 8 values, we take the average of the 4th and 5th values when ordered: (8+8)÷2 = 8. Since mean (7.875) < median (8), the mean is less than the median. This makes sense because the lower scores (6, 7, 7) pull the mean down more than the single high score (10) pulls it up. When comparing mean and median, remember that in left-skewed data (with low outliers), the mean is typically less than the median.

This question compares means and standard deviations of two datasets. For set X (2,4,6,8): mean = 20÷4 = 5. For set Y (2,2,8,8): mean = 20÷4 = 5. Both have the same mean. For spread, set X has values evenly distributed (each 2 units apart), while set Y has values clustered at extremes (two 2's and two 8's). Set Y has larger deviations from the mean: values are 3 units away, while X's values are only 1 or 3 units away. Therefore Y has larger standard deviation due to values being farther from the mean. When comparing standard deviations, look at how far values are from their mean, not just the range.