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

A data set has a mean of 50 and a standard deviation of 10. Every value in the data set is transformed using the rule y = 2x − 3. What are the mean and standard deviation of the transformed data set?

Answer: Mean = 97, standard deviation = 20

When every value is transformed by y = ax + b, the new mean equals a·(old mean) + b and the new standard deviation equals |a|·(old standard deviation) — adding a constant shifts center but not spread. Here: new mean = 2(50) − 3 = 97, and new standard deviation = 2(10) = 20. Subtracting 3 has no effect on spread.

A sorted data set of 7 values has a mean of 30 and a median of 28. The values are 12, 18, 28, 28, 35, 40, and k, where k > 40. If k is removed from the data set, which of the following correctly describes what happens to the mean and median?

Answer: The mean decreases and the median stays the same.

Since the 7 values sum to 7 × 30 = 210 and the other six values sum to 161, k = 49. Removing k = 49 lowers the sum to 161 and the new mean to 161/6 ≈ 26.8 — the mean decreases. The new 6-value data set is 12, 18, 28, 28, 35, 40; its median is the average of the 3rd and 4th values = (28 + 28)/2 = 28, unchanged. Removing an outlier from one tail always affects the mean but may leave the median intact.

Class A has 20 students with a mean exam score of 75. Class B has 30 students with a mean exam score of 85. A student claims the combined mean for all 50 students is 80. Which of the following is true?

Answer: The combined mean is 81, because the larger class pulls the average toward 85.

The combined mean must be a weighted average: (20 × 75 + 30 × 85) / 50 = (1500 + 2550) / 50 = 4050 / 50 = 81. Simply averaging 75 and 85 gives 80, which ignores the fact that Class B is larger. Because Class B contributes more students, the combined mean is pulled toward 85, yielding 81, not 80.

A data set has the following five-number summary: Min = 2, Q1 = 18, Median = 30, Q3 = 42, Max = 88. Using the 1.5 × IQR rule to identify outliers, which values are classified as outliers?

Answer: 88 only

IQR = Q3 − Q1 = 42 − 18 = 24. The lower fence is Q1 − 1.5(24) = 18 − 36 = −18, and the upper fence is Q3 + 1.5(24) = 42 + 36 = 78. Any value below −18 or above 78 is an outlier. The minimum, 2, is above −18, so it is not an outlier. The maximum, 88, exceeds 78, so it is classified as an outlier.

A researcher standardizes every score in a data set using the formula z = (x − μ) / σ, where μ is the mean and σ is the standard deviation of the original data. Which of the following correctly describes the mean and standard deviation of the resulting z-scores?

Answer: Mean = 0, standard deviation = 1

Standardizing applies a linear transformation z = (1/σ)x − (μ/σ). By the linear transformation rules, the new mean = (1/σ)·μ − μ/σ = 0, and the new standard deviation = (1/σ)·σ = 1. Every standardized distribution has mean 0 and standard deviation 1 regardless of the original values — this is the definition of the standard normal transformation.

Class X (10 students): 70, 72, 74, 74, 75, 75, 76, 76, 78, 80. Class Y (10 students): 50, 60, 65, 70, 75, 75, 80, 85, 90, 95. Which statement best compares the two distributions?

Answer: Class X and Class Y have approximately equal means, but Class X has considerably less spread.

Class X mean = 750/10 = 75; Class Y mean = 745/10 = 74.5 — virtually identical. Both distributions also share a median of 75. However, Class X has a range of 80 − 70 = 10, while Class Y has a range of 95 − 50 = 45. The two distributions have nearly equal centers but dramatically different spreads: Class X is tightly clustered, while Class Y is widely dispersed.