6th Grade Math Question of the Day

1. Mean = $\frac{32}{9}\approx 3.56$; IQR = 2; Pattern: most students brought 1–4 cans clustered around 3; Striking deviation: 10 cans is much higher than the rest. (correct answer)
2. Mean = $\frac{32}{8}=4$; IQR = 1; Pattern: values cluster around 4; Striking deviation: none.
3. Mean = $\frac{32}{9}\approx 3.56$; IQR = 9; Pattern: values are evenly spread from 1 to 10; Striking deviation: none.
4. Mean = $\frac{22}{9}\approx 2.44$; IQR = 2; Pattern: most students brought around 10 cans; Striking deviation: 1 can is an outlier.

Explanation: This question tests calculating measures of center (mean) and variability (IQR), describing overall pattern in words, and noting striking deviations. Measures: mean=sum/count=(1+2+2+3+3+3+4+4+10)/9=32/9≈3.56; for IQR with 9 values, Q1 is 25th percentile (position 2.5, so between 2nd and 3rd values: Q1=2) and Q3 is 75th percentile (position 7.5, so between 7th and 8th values: Q3=4), giving IQR=4-2=2. Pattern: most students (8 of 9) brought 1-4 cans clustered around mean 3.56, forming tight group; striking deviation: 10 cans is clear outlier, approximately 6 cans above next highest value 4. Choice A correctly shows mean=32/9≈3.56, IQR=2, describes pattern as "most students brought 1-4 cans clustered around 3," and notes "10 cans is much higher than the rest." Common errors include: incorrect sum (choice C: 22/9≈2.44), confusing IQR with range (choice B: IQR=9), wrong count (choice D: 32/8=4), or incorrect IQR calculation (choice D: IQR=1). Calculating IQR: (1) find Q1=2 and Q3=4, (2) subtract: 4-2=2. Complete summary identifies 10 as outlier that pulls mean upward but doesn't affect median or IQR.