One-Variable Data: Distributions and Measures of Center and Spread Flashcards
7 cards from real Bluebook SAT Test practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 7 One-Variable Data: Distributions and Measures of Center and Spread flashcards as text
The mode of the dataset {2, 3, 3, 5, 7, 7, 7, 9} is:
Answer: 7
The mode is the value that appears most often; 7 appears three times, more than any other value.
Two datasets each have 10 values and the same mean of 20. Dataset X has values all between 18 and 22, while Dataset Y has values ranging from 5 to 35. Which has the greater standard deviation?
Answer: Dataset Y
Dataset Y's values are much more spread out from the mean, resulting in a greater standard deviation.
A box plot has whiskers extending from 10 to 90, with Q1=30 and Q3=70. A data point at 95 would be classified as:
Answer: A mild outlier
The IQR is 40, and 1.5 × 40 = 60, so the upper fence is Q3 + 60 = 130; 95 is outside the whisker (90) but below the fence, making it a mild outlier.
A dataset of 9 values has a median of 15. A 10th value of 15 is added. What happens to the median?
Answer: The median remains 15
Adding 15 places it in the middle of the sorted dataset; the median of the 10-value dataset averages the 5th and 6th values, both near 15, keeping the median at 15.
The histogram below shows the distribution of test scores in a class. The distribution is left-skewed (skewed toward lower values). Which is most likely true?
Answer: Mode > Median > Mean
In a left-skewed distribution, the long tail pulls the mean left, so Mode > Median > Mean.
Students' weekly study hours: 2, 4, 4, 5, 6, 6, 6, 8, 10. What is the mean number of study hours?
Answer: 5.67
The sum is 2+4+4+5+6+6+6+8+10 = 51; dividing by 9 gives 51/9 ≈ 5.67.
If a constant of 10 is added to every value in a dataset, what happens to the mean and standard deviation?
Answer: Mean increases by 10; standard deviation stays the same
Adding a constant shifts every value equally, increasing the mean by that constant while leaving the spread (standard deviation) unchanged.