Digital SAT Math Prep: Real-World Linear Equations Guide

A real estate agent tracks the listing prices of 20 houses in a specific zip code. The table below summarizes the distribution of the listing prices.

The agent discovers that the house listed at $1,200,000 was entered in error and removes it from the dataset. Which of the following statements best describes the effect of removing this data point?

• A) The mean decreases, and the median decreases.
• B) The mean decreases, and the median remains the same.
• C) The mean remains the same, and the median decreases.
• D) The mean increases, and the median remains the same.

The SAT “Shortcut” Method (Conceptual Logic)

Because the SAT is a timed exam, high-scoring students learn to solve outlier questions conceptually without performing heavy arithmetic:

• The Mean: The $1,200,000 home is a massive outlier on the high end of the dataset. It acts like a heavy anchor pulling the average (mean) upward. Removing this extreme high value will cause the mean to decrease.
• The Median: The median is the middle data point. In a group of 20 homes, the middle falls between the 10th and 11th value. Looking at the frequency column, the first 13 homes (5 + 8) are all $275,000 or less. This means both the 10th and 11th homes are exactly $275,000, making the original median $275,000. If you remove the top house, you now have 19 homes. The middle position shifts to the 10th home, which is still $275,000. The median remains the same.

📐 The Long Math Proof (For Reviewing Details)

• Original Dataset (n=20):

• • Sum = (5 ∙ 250,000) + (8 ∙ 275,000) + (6 ∙ 300,000) + 1,200,000 = $5,450,000
  • Original Mean = 5,450,000/20 = $272,500
  • Original Median = $275,000

• New Dataset (n=19):

• • New Sum = $5,450,000 – $1,200,000  = $4,250,000
  • New Mean = 4,250,000/19 ≈ $223,684} (The mean decreased)
  • New Median = $275,000 (The median stayed the same)

• Correct Answer: B. The mean decreases, and the median remains the same.

The dot plots show the distribution of daily high temperatures, in degrees Fahrenheit, recorded in City A and City B over a 14-day period in July.

Which statement truly describes the relationship between their means (μ) and standard deviations (σ)?

• A) μ A = μ B and σ A > σ B
• B) μ A = μ B and σ A < σ B
• C) μ A > μ B and σ A = σ B
• D) μ A < μ B and σ A < σ B

🧭 The SAT Concept: Skewness vs. Symmetry

On the digital SAT, you don’t need to waste time calculating standard deviation or means by hand. You can solve this purely by looking at the balance of the data points.

Step 1: Compare the Means (μ). 

• • City B is completely symmetrical. The data points on the left perfectly mirror the data points on the right, which places its balancing center (mean) exactly at 88 degrees.
  • City A is skewed to the left. It has a heavy cluster of low temperatures (three days at 85) pulling the average down, with fewer high-temperature days to balance it out. 
  • Because the bulk of the data is on the lower end, City A’s mean is smaller than City B’s mean (). This immediately narrows our choices down to Option D!
  •  

• Step 2: Verify the Standard Deviation (σ). Standard deviation measures the variation or dispersion of the data.
  • In City A, the vast majority of the data is packed tightly together at the low end (7 out of the 11 days are clustered between 85 and 87).
  • In City B, the data is pushed out heavily to the extreme outer edges (3 days at the absolute minimum of 85, and 3 days at the absolute maximum of 91), creating a much wider total spread.
  • Because City B’s data points are more spread out from its center, City B has a larger standard deviation. Therefore

σ A < σ B.

• Correct Answer: D. μ A < μ B and σ A < σ B.