Bluebook SAT Test One-Variable Data: Distributions and Measures of Center and Spread 5

Question 1

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?

Accepted Answer

Mean = 97, standard deviation = 20

Answer

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.

Question 2

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?

Accepted Answer

The mean decreases and the median stays the same.

Answer

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.

Question 3

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?

Accepted Answer

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

Answer

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.

Question 4

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?

Accepted Answer

88 only

Answer

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.

Question 5

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?

Accepted Answer

Mean = 0, standard deviation = 1

Answer

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.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test One-Variable Data: Distributions and Measures of Center and Spread 5