Data Analysis, Statistics & Basic Probability

Data Analysis, Statistics & Basic Probability

Data analysis, statistics, and probability questions test a student's ability to organize numerical information, interpret graphical data representations, and evaluate likelihoods. These topics represent approximately 15% to 20% of the HSPT Mathematics section, and they reward careful reading: most errors come from skipping a step (like ordering data) rather than from difficult arithmetic.

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Measures of Central Tendency & Spread

Statistical measures summarize the central value and distribution of a data set:

• Mean (Arithmetic Average): Sum of all data values divided by the total number of values (N). Mean = (∑ x) / N
• Missing Value Mean Problem: To find a required value to achieve a target mean, compute total required sum: Target Sum = Target Mean × N.
• Median (Middle Value): The middle number when data points are arranged in ascending numerical order. If N is even, the median is the average of the two central numbers.
• Mode (Most Frequent): The number that occurs most frequently in the data set. A set may have one mode (unimodal), multiple modes (bimodal/multimodal), or no mode.
• Range (Spread): The difference between the highest (maximum) and lowest (minimum) values: Range = Max - Min.

Central Tendency Summary

Step-by-Step Worked Example 1

Problem: A student scores 82, 86, 91, and 85 on four math tests. What score must she earn on her fifth test to achieve an overall mean test score of 88?

1. Determine total required points for 5 tests: Target Sum = 88 × 5 = 440
2. Sum current 4 test scores: Current Sum = 82 + 86 + 91 + 85 = 344
3. Calculate required score for test 5: Required Score = 440 - 344 = 96

Step-by-Step Worked Example 2

Problem: Find the mean, median, mode, and range of the data set: 7, 9, 4, 9, 6.

1. Order the data: 4, 6, 7, 9, 9.
2. Mean: (4 + 6 + 7 + 9 + 9) / 5 = 35 / 5 = 7.
3. Median: Middle (3rd) value of the ordered set = 7.
4. Mode: 9 appears twice; all others appear once → mode = 9.
5. Range: 9 - 4 = 5.

Outlier Effect: If the value 100 were added to this set, the mean would jump from 7 to 135 / 6 = 22.5, but the median would only shift to (7 + 9) / 2 = 8. The median resists outliers; the mean does not.

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Reading and Interpreting Data Graphs

• Bar Graphs: Compare discrete categorical quantities. Read column heights against the vertical axis scale.
• Line Graphs: Display trends over continuous time intervals. Slope indicates rate of change.
• Pie Charts (Circle Graphs): Represent relative proportions of a whole (100% total, corresponding to 360° total circle angle). A sector's central angle equals its percent share × 360°.

Step-by-Step Worked Example 3

Problem: A pie chart illustrates student participation in school sports across 300 students. If the Soccer sector represents 35% of the chart, how many students play soccer?

1. Convert percentage to decimal multiplier: 35% = 0.35.
2. Multiply by total student population: 300 × 0.35 = 105 students.

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Basic and Compound Probability

Probability measures the likelihood of an event (E) occurring on a scale from 0 (impossible) to 1 (certain): P(E) = Favorable Outcomes / Total Possible Outcomes

• Complement Rule: The probability that an event does not occur is P(not E) = 1 - P(E).
• Independent Compound Events: Two events A and B are independent if the occurrence of one does not affect the likelihood of the other. The joint probability is the product of individual probabilities: P(A and B) = P(A) × P(B)

With Replacement vs. Without Replacement

• With replacement: The drawn item is returned, so the total number of outcomes stays the same for every draw; the draws are independent.
• Without replacement: The drawn item is kept, so both the favorable count and the total count drop by one for the next draw; the draws are dependent.

Step-by-Step Worked Example 4

Problem: A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. If two marbles are drawn with replacement (drawn, noted, and put back), what is the probability that both marbles drawn are blue?

1. Calculate total marbles: 5 + 3 + 2 = 10 marbles.
2. Probability of blue on 1st draw: P(Blue1) = 3/10.
3. Probability of blue on 2nd draw (with replacement): P(Blue2) = 3/10.
4. Compute compound independent probability: P(Both Blue) = (3/10) × (3/10) = 9/100

Step-by-Step Worked Example 5

Problem: Using the same bag (5 red, 3 blue, 2 green), two marbles are drawn without replacement. What is the probability that both marbles drawn are red?

1. Probability of red on 1st draw: P(Red1) = 5/10 = 1/2.
2. Adjust counts for the 2nd draw: One red marble is gone, leaving 4 red out of 9 total: P(Red2) = 4/9.
3. Multiply the dependent probabilities: P(Both Red) = (5/10) × (4/9) = 20/90 = 2/9

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Fundamental Counting Principle

If a choice consists of k consecutive independent stages, where Stage 1 has m options, Stage 2 has n options, and Stage 3 has p options, the total number of distinct outcomes is: Total Outcomes = m × n × p

• Repetition allowed: Every stage keeps its full option count (a 4-digit PIN with repeatable digits: 10 × 10 × 10 × 10 = 10,000).
• Repetition not allowed: Each stage loses one option (a 4-digit PIN with no repeated digits: 10 × 9 × 8 × 7 = 5,040).

Step-by-Step Worked Example 6

Problem: A student building a lunch combination chooses 1 sandwich from 4 options, 1 side from 3 options, and 1 drink from 3 options. How many unique lunch combinations are possible?

1. Apply Fundamental Counting Principle: Total = 4 × 3 × 3 = 36 unique lunch combinations.

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Common HSPT Traps

• Median without ordering: Always sort the data set first; picking the middle of an unordered list gives a wrong answer.
• Even-count median: With an even number of values, average the two central numbers — do not just pick one of them.
• Adding instead of multiplying probabilities: P(A and B) for independent events is a product, not a sum.
• Ignoring replacement: Without replacement, the second draw's denominator shrinks by one; forgetting this is the most common probability error.