One-Variable Data: Distributions and Measures of Center and Spread Flashcards

A frequency table shows: Score 70 (frequency 3), Score 80 (frequency 5), Score 90 (frequency 2). What is the mean score?

Answer: 79

Mean = (70×3 + 80×5 + 90×2) / (3+5+2) = (210+400+180)/10 = 790/10 = 79.

Which of the following would most increase the standard deviation of the dataset {10, 12, 14, 16, 18}?

Answer: Replace 14 with 50

Replacing the middle value with 50 introduces a very large outlier far from the mean, dramatically increasing the spread.

The range of a dataset is 30 and the minimum is 15. What is the maximum?

Answer: 45

Range = Maximum − Minimum, so Maximum = Range + Minimum = 30 + 15 = 45.

A histogram shows that 50% of values fall below the 6th bar. What does this suggest about the 6th bar's value relative to the dataset?

Answer: It is the median

The value at which 50% of the data falls below is the median by definition.

Two classes took the same test. Class A has mean 72 and standard deviation 8. Class B has mean 72 and standard deviation 3. Which statement is true?

Answer: Class B's scores are more consistent

A lower standard deviation indicates less variability, so Class B's scores are more consistent (more tightly clustered around the mean).

The dataset {x, 8, 11, 14, 17} has a mean of 12. What is the value of x?

Answer: 10

Sum = 12 × 5 = 60; x + 8 + 11 + 14 + 17 = 60, so x = 60 − 50 = 10.

A dataset has an outlier that is extremely large. Removing the outlier would most likely:

Answer: Decrease the mean and have little effect on the median

A large outlier inflates the mean; removing it decreases the mean significantly, while the median—resistant to outliers—barely changes.