SOCS Check 🧦
A distribution of daily temps has a mean of 75°F, but most values are between 60–70°F with a few very hot days around 100°F. Which SOCS features should you mention?
Shape = skewed right, Outliers = 100°F values, Center ≈ mean 75 or median ≈ 65–70, Spread = wide.
Dotplot has 20 dots. 5 are above “10,” rest are below. Is mean larger, smaller, or about equal to median?
Median is below 10. Since only 5 high values, mean is pulled upward, so mean > median.
Histogram: cluster left, long tail right. Shape & mean vs median?
Shape = skewed right. Mean > median.
Data = 2, 4, 6, 8, 100. Find mean.
(2+4+6+8+100) ÷ 5 = 120 ÷ 5 = 24. Outlier 100 pulls mean up.
Median of 7, 9, 3, 12, 8?
Order: 3, 7, 8, 9, 12. Median = 8.
Two datasets have the same mean but one has much larger spread. How would this show up in dotplots or histograms?
The shapes might look similar, but the wider spread set will have values more spread out along the axis (flatter histogram, dots farther apart).
Dotplot clustered at 40–50, one dot at 90. How does outlier affect mean vs median?
Mean increases noticeably, median stays nearly the same.
Draw symmetric histogram with mean ≈ 50.
Example: bars evenly spread around 50. (Any symmetric shape centered on 50 works.)
20 students avg = 70. New student 100. New mean?
Total = 20×70 = 1400. New total = 1500. ÷ 21 = ≈ 71.4.
Median of 4, 8, 10, 12, 20, 22.
Middle two = 10, 12 → median = 11.
A dataset has an outlier far above the rest. What happens to mean vs median? Which is more reliable?
Mean increases a lot (pulled toward outlier), median stays stable. Median is more reliable for center.
Create 6 numbers for a left-skewed dotplot.
Example: 1, 1, 2, 5, 6, 7 (tail on the left).
Class A: tightly around 80. Class B: spread 50–100, both same mean. Which class for guaranteed B?
Class A — less spread = more consistent scores near 80.
5 numbers, mean = 12. Four are 10, 11, 12, 14. Missing?
Total = 60. Known = 47. Missing = 13.
Dataset of 7 values: median = 10, mean = 12. What does this say about shape?
Mean > median → skewed right.
Compare these two sets with SOCKS:
A: 2, 2, 2, 2, 10
B: 4, 4, 4, 4, 4
Set A: Shape = skewed right, Outlier = 10, Center = mean > median, Spread = large.
Set B: Shape = uniform/symmetric, no outliers, Center = mean = median = 4, Spread = 0.
Dotplot: peak at 8, scores 4–10. Describe with SOCS.
Shape = roughly symmetric/unimodal, Outliers = none, Center ≈ 8, Spread = 4–10.
Change bin width from 10 → 5. How could shape change?
More bins = histogram looks bumpier/detailed. Same data, different appearance.
Goals = 0, 1, 2, 4, ?. Mean = 3. Missing?
Total needed = 15. Current = 7. Missing = 8.
11 values, median = 6th value = 10. If skewed right, mean bigger or smaller?
Mean > 10. Outliers pull it higher.
A distribution is symmetric and unimodal. Why are mean & median the same, and what if an outlier is added?
In symmetric unimodal data, balance point = middle = same value. An outlier would pull the mean away, but not the median.
Two classes: same shape & center, but one has wider spread. What does that mean?
Wider spread class has less consistent scores (more variability).
Create 2 histograms with mean 60 but diff spreads.
One tight cluster around 60 (low spread). One spread wide 20–100 (high spread). Both balance at 60 but look different.
10 students, mean = 80. Total = 800. Remove one score, new mean = 77 for 9 students. Find removed score.
New total = 693. Removed score = 107.
9 values, first 8 are 2, 4, 5, 6, 8, 10, 12, 14. What must 9th be for median = 9?
Median = 5th value when ordered. Current 5th = 8. To shift median to 9, the 9th number must be ≥ 9 and inserted between 8 and 10. Example: 9.