When a constant is added to every value in a dataset, the mean increases by that same constant. If the original mean is 20 and we add 5 to each value, the new mean = 20 + 5 = 25. This is because mean = sum/count, and adding 5 to each of n values increases the sum by 5n, so the new mean = (sum + 5n)/n = sum/n + 5 = 20 + 5 = 25. This property makes it easy to adjust means without recalculating from scratch. The same rule applies to subtraction: subtracting a constant from each value decreases the mean by that constant.

To find the mean points scored per game, we sum all values and divide by the count. Sum = 18 + 22 + 25 + 19 + 21 + 24 + 20 + 23 = 172 points. Mean = 172 ÷ 8 = 21.5 points per game. The calculation requires careful addition of all values - a common error is miscounting or missing a value. When computing means, double-check your sum before dividing. For test-taking efficiency, you can sometimes eliminate obviously wrong answers before completing the full calculation.

This question compares the standard deviations (spread around the mean) of two classes. Class A: 6,7,7,8,8,8,9,9 has values clustered near the center. Class B: 2,6,7,8,8,8,9,10 has more extreme values (2 and 10) farther from the center. The standard deviation measures how far values typically are from the mean - Class B's extreme values (2 and 10) create larger deviations from the mean than Class A's values. Even with the same median, the presence of extreme scores increases standard deviation. When comparing spreads visually, look for values far from the center of the distribution.

Data set X has all values equal to 10, so mean = 10 and standard deviation = 0 (no variation). Data set Y: 6,8,10,12,14 has mean = (6+8+10+12+14)/5 = 50/5 = 10 (same as X). However, Y's values vary from the mean: deviations are -4,-2,0,+2,+4, giving a positive standard deviation. Therefore, both sets have the same mean (10), but Y has the larger standard deviation because X has zero spread. This illustrates that datasets can have identical means but very different spreads. Standard deviation measures variability, not the center.

Given: 6 numbers with median 10, range 8, and minimum 6. Using range = max - min, we have 8 = max - 6, so max = 14. This uses the fundamental relationship between range, maximum, and minimum values. The median information confirms this is reasonable - with 6 values, the median of 10 falls between the 3rd and 4th values when ordered, which is consistent with a minimum of 6 and maximum of 14. When given range and one extreme value, you can always find the other extreme value by adding or subtracting appropriately.

With the 60-year-old: mean = (18+19+19+20+22+24+24+60)/8 = 206/8 = 25.75 years. Without the 60-year-old: mean = (18+19+19+20+22+24+24)/7 = 146/7 ≈ 20.86 years, a change of about 4.89 years. The median with all 8 values is (20+22)/2 = 21; without the outlier it's 20, a change of only 1 year. The mode (19) and minimum (18) don't change at all. The mean changes most because it's sensitive to extreme values, while median is resistant to outliers. This demonstrates why median is preferred when outliers are present.

The mode is the value that appears most frequently in a dataset. Looking at the scores: 6 (once), 7 (twice), 8 (three times), 9 (once), 10 (once). The value 8 appears three times, more than any other score, making it the mode. A dataset can have one mode, multiple modes, or no mode at all. Common errors include confusing mode with median (middle value) or mean (average). When finding the mode, systematically count the frequency of each unique value.

This dataset has an outlier salary of 120 thousand. For the median: ordering gives 38,40,41,42,43,44,45,46,120, so median = 43 (5th value). For the mean: sum = 38+40+41+42+43+44+45+46+120 = 459, so mean = 459/9 = 51. The mean (51) is greater than the median (43) because the outlier pulls the mean upward. This illustrates that outliers affect the mean more than the median. When a distribution has a high outlier, expect mean > median; with a low outlier, mean < median.

This question asks which measure of center is most affected by the outlier 30 in the data: 4, 5, 5, 6, 6, 6, 7, 30. The mean without the outlier would be about 5.7, but with it: (4+5+5+6+6+6+7+30)÷8 = 69÷8 = 8.625, showing a large increase. The median would be 6 without the outlier and remains 6 with it (middle values of ordered data). The mode stays at 6 (most frequent value) regardless. The mean is most sensitive to outliers because it uses all values in its calculation. When analyzing data with outliers, consider reporting both mean and median to show the full picture.

This question compares IQRs and medians from two five-number summaries. Class M: IQR = Q3 - Q1 = 75 - 55 = 20, median = 65. Class N: IQR = Q3 - Q1 = 70 - 60 = 10, median = 65. Therefore, Class M has the larger IQR (20 > 10) and the medians are equal (both 65). The IQR tells us about the spread of the middle 50% of scores - Class M's scores are more spread out in the middle. When comparing box plots, always calculate IQRs separately and compare medians directly from the five-number summaries.