This question tests descriptive statistics by evaluating changes in mean and median after removing an outlier. Mean is the average, sensitive to extremes, while median is the middle value, resistant to outliers. Original set {3,3,4,4,4,5,20} sorted, median is 4 (fourth value), sum=3+3+4+4+4+5+20=43, mean=43/7≈6.143. Removing 20 gives {3,3,4,4,4,5}, sum=23, mean=23/6≈3.833 (decreases), median now average of 3rd and 4th: 4 and 4, so 4 (same). Thus, mean decreases, median stays the same, as removal of high outlier lowers average but middle unchanged. Tempting to think median changes, like in A, but with even count now, it's still 4. Mean vs median confusion might lead to thinking both decrease, but median is robust.

This question tests descriptive statistics by comparing mean and median in skewed datasets. Mean is pulled by outliers, median is middle value resistant to them. P {1,2,3,4,100} median 3 (third), mean (1+2+3+4+100)/5=110/5=22. Q {1,2,3,4,5} median 3, mean 15/5=3. So medians equal at 3, P has greater mean due to 100. The outlier inflates P's mean but not median, keeping medians same. Tempting to ignore outlier for mean, like E, but mean includes all values. Another error is thinking P has greater median due to largest value, like A.

This question tests descriptive statistics by finding mean of a subset after knowing overall mean. Mean is sum divided by count; if overall mean 10 for 4 numbers, sum=40. If one is 2, sum of remaining 3 is 40-2=38, mean=38/3≈12.667. Thus, the mean of the remaining 3 is 38/3, as subtracting the known value adjusts the sum accordingly. A common mistake is thinking it's still 10, like C, ignoring the removal. Confusion might lead to 40/3 like E, but that's including the 2 incorrectly.

This question tests descriptive statistics by comparing standard deviation based on spread, despite equal means. Standard deviation quantifies dispersion from mean; greater spread means higher SD. Both have mean 50, but R clustered near 50 (low SD), S from 10 to 90 (high spread, high SD). Thus, S has greater standard deviation due to wider range of values. A common error is thinking equal means imply equal SD, like C, but SD measures variability, not location. Another mistake is focusing on median instead of SD, like D or E.

This question tests descriptive statistics by comparing variability in datasets with same mean. Standard deviation measures spread from the mean, higher when values are more dispersed. Dataset A {10,10,10,10,10} has mean 10, all values at mean, so SD=0. Dataset B {8,9,10,11,12} has mean (8+9+10+11+12)/5=50/5=10, but values spread out, so higher SD. B has greater standard deviation because its values deviate more from the mean, while A has none. A common mistake is thinking equal means imply equal SD, like in E, but SD is independent of mean value. Another error is focusing on median or equality instead of spread.

This question tests descriptive statistics by calculating mean after removing a value. Mean is sum over count; original 8 values mean 14, sum=112. Remove 30, sum=112-30=82, remaining 7 mean=82/7≈11.714. Thus, the mean is 82/7, as adjusting sum and count gives the new average. Tempting to think it's 112/7=16, like E, but that's without removing. Mean vs total confusion might lead to 11 or 12, like A or B.

This question tests descriptive statistics by assessing mean change after score correction. Mean is total sum over count; original mean 80 for 10 students, sum=800. Correcting from 60 to 90 increases sum by 30, new sum=830, new mean=830/10=83, increase by 3. The mean increases by 3, as the correction adds 30 to total, divided by 10. A common error is thinking increase by 30, like B, forgetting to divide by n. Another mistake is assuming no change since only one score, like E.

This question tests descriptive statistics by examining mean and median after replacing a value. Mean sensitive to changes, median to order; original {1,2,2,2,9} sum=16, mean=3.2, median 2. Replace 9 with 3: {1,2,2,2,3} sum=10, mean=2 (decreases), median 2 (same). Mean decreases, median stays same, as replacement lowers sum but middle unchanged. A tempting option is thinking median increases to 2.5 or something, but sorted it's 1,2,2,2,3, middle 2. Confusion with mode might suggest mean same, like D.

This question tests descriptive statistics by examining changes in central tendency measures after data modification. The key concepts are mean, which is the average of the values, and median, which is the middle value in an ordered list. Originally, the set {6, 7, 7, 8, 8, 9, 20} has a sum of 65 and mean of 65/7 ≈ 9.29, with median 8 as the fourth value. Replacing 20 with 10 gives a new sum of 55 and mean of 55/7 ≈ 7.86, while the ordered set {6, 7, 7, 8, 8, 9, 10} still has median 8. Thus, the mean decreases, but the median stays the same, as the replacement affects the average but not the middle position. A tempting mistake is thinking the median changes, like in option A, but median depends on order, not extreme values. Confusion between mean and median often leads to incorrect choices, such as assuming both decrease when only the mean is pulled down by the lower value.

This question tests descriptive statistics by exploring properties of median in even and odd datasets. Median for odd n is middle value; for even, average of two middle. Original 9 values, median 15 (5th value). Adding one makes 10 values, median average of 5th and 6th =15. Thus, the 5th and 6th in ordered new set average to 15, which must be true regardless of added value's position. This holds because the median definition requires it for the new even count. A tempting incorrect option is A, assuming added value is 15, but it could be anything as long as middle two average 15. Another error is thinking original mean is 15, like D, confusing mean and median.