Topic 3

Data and Probability in Context

SAT questions about data analysis and probability rarely test a single skill in isolation. Instead, they embed statistical reasoning inside a real-world scenario and require you to combine reading data (from tables, descriptions, or summary statistics) with algebraic or proportional reasoning to reach an answer. This topic focuses on those multi-step combinations.

Mean, median, and weighted averages are the most common statistical tools on the SAT. The mean (average) of a data set is the sum of all values divided by the number of values. A weighted average applies when groups of different sizes each have their own average, and you need the overall average:

\text{Overall mean} = \frac{n_1 \cdot \bar{x}_1 + n_2 \cdot \bar{x}_2}{n_1 + n_2}

A common SAT pattern gives you the mean of a group and asks what value must be added (or removed) to produce a new target mean. This requires setting up an equation with the unknown value.

The median is the middle value when data is sorted. For an even count of values, the median is the average of the two middle values. SAT problems often provide data in a frequency table or grouped format, requiring you to count through frequencies to locate the median position.

Probability on the SAT is almost always classical probability: favorable outcomes divided by total outcomes. Multi-step probability problems ask you to combine information from two-way tables, conditional statements, or sequential events. Conditional probability is the probability of an event given that another event has occurred:

P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)} = \frac{\text{number in both } A \text{ and } B}{\text{number in } B}

Expected value and interpreting statistical claims (margin of error, sampling methods, correlation vs.\ causation) also appear. The SAT tests whether you can distinguish between a result that applies to the sample versus the population, and whether an observed association implies a causal relationship. Correlation does not imply causation; only a randomized controlled experiment can establish causation.

Core Skills

Solve for unknown values using mean and weighted-average equations.
Find the median from a frequency table or a described data set.
Calculate probabilities from two-way tables, including conditional probabilities.
Interpret margin of error, sampling bias, and the limits of statistical inference.
Combine data-reading with algebraic reasoning in multi-step problems.

Example 1: Finding a Missing Value from the Mean

Five test scores have a mean of 82. Four of the scores are 78, 85, 90, and 76. What is the fifth score? Step 1: The sum of all five scores is

5 \times 82 = 410

. Step 2: The sum of the four known scores is

78 + 85 + 90 + 76 = 329

. Step 3: The fifth score is

410 - 329 = 81

\boxed{81}

Example 2: Weighted Average

A class of 20 students has a mean exam score of 74. Another class of 30 students has a mean exam score of 82. What is the combined mean for all 50 students? Step 1: Total points from class 1:

20 \times 74 = 1480

. Total from class 2:

30 \times 82 = 2460

. Step 2: Combined mean:

\frac{1480 + 2460}{50} = \frac{3940}{50} = 78.8

\boxed{78.8}

Example 3: Median from a Frequency Table

A survey recorded the number of books read last month by 25 people. The results are shown below.

\begin{array}{|c|c|} \hline \textbf{Books read} & \textbf{Frequency} \\ \hline 0 & 3 \\ 1 & 7 \\ 2 & 6 \\ 3 & 5 \\ 4 & 4 \\ \hline \end{array}

What is the median number of books read? Step 1: There are 25 data points, so the median is the 13th value when sorted. Step 2: Count through the frequencies: positions 1--3 are 0 books, positions 4--10 are 1 book, positions 11--16 are 2 books. The 13th value falls in the ``2 books'' group.

\boxed{2}

Example 4: Conditional Probability from a Two-Way Table

A survey of 200 students asked whether they preferred dogs or cats and whether they lived in a house or an apartment.

\begin{array}{|c|c|c|c|} \hline & \textbf{Dog} & \textbf{Cat} & \textbf{Total} \\ \hline \textbf{House} & 70 & 30 & 100 \\ \hline \textbf{Apartment} & 50 & 50 & 100 \\ \hline \textbf{Total} & 120 & 80 & 200 \\ \hline \end{array}

What is the probability that a randomly selected student prefers dogs, given that the student lives in a house? Step 1: We want

P(\text{Dog} \mid \text{House})

. Restrict to the ``House'' row: 100 students total, 70 prefer dogs. Step 2:

P(\text{Dog} \mid \text{House}) = \frac{70}{100} = 0.70

\boxed{0.70}

Example 5: Interpreting a Statistical Study

A researcher surveys 500 randomly selected adults in a city and finds that

62\%

support a new park, with a margin of error of

\pm 4\%

. Which of the following conclusions is supported? (a) Between

58\%

and

66\%

of all adults in the city likely support the new park. (b) Exactly

62\%

of all adults in the city support the new park. (c) The new park will be built. Step 1: The margin of error means we are confident the true proportion is between

62 - 4 = 58\%

and

62 + 4 = 66\%

. This supports conclusion (a). Conclusion (b) is too precise, and (c) is a policy prediction, not a statistical conclusion. The supported conclusion is

\boxed{(a)}

Key Takeaways

When a problem gives you a mean and asks for a missing value, use the equation $\text{sum} = \text{mean} \times \text{count}$ and solve for the unknown. Do not average the known values and guess.
For weighted averages, never simply average the two group means unless the groups are the same size. Weight by group size.
In a two-way table, conditional probability restricts the denominator to the given condition's row or column total, not the grand total.
Margin of error defines a confidence interval. Results from a random sample can be generalized to the population, but results from a voluntary or convenience sample cannot. Correlation from an observational study does not establish causation.