Statistics and Probability
This domain tests whether you can summarize and compare one-variable distributions, describe a relationship between two variables, and count and compute probabilities in situations with and without replacement.
One-variable distributions
Mean = sum of values ÷ number of values; median = middle value (the average of the two middle values when n is even); mode = most frequent value. Range = maximum − minimum; quartiles Q1 and Q3 give the interquartile range, IQR = Q3 − Q1, and standard deviation summarizes spread around the mean. A box-and-whisker plot displays median, quartiles and extremes, and flags outliers. The mean is sensitive to outliers; the median resists them. Compare distributions using a measure of centre and a measure of spread together.
Two-variable distributions
A scatter plot shows the form of a relationship. The linear correlation coefficient r lies between −1 and 1: |r| near 1 means a strong linear relation, and the sign gives the direction. A regression line supports prediction, but interpolation inside the data range is far safer than extrapolation beyond it.
Probability and counting
0 ≤ P(A) ≤ 1; P(A′) = 1 − P(A); P(A ∪ B) = P(A) + P(B) − P(A ∩ B), which reduces to a simple sum for mutually exclusive events; independent events satisfy P(A ∩ B) = P(A)·P(B). "Without replacement" changes both numerator and denominator on the second draw. Use the multiplication principle, permutations nPr = n!/(n − r)! when order matters, combinations nCr when it does not, and tree or Venn diagrams to organize.
Worked example
A bag holds 5 red, 3 blue and 2 green balls (10 in total). Two balls are drawn at random without replacement. Find P(both red) and P(at least one red). P(both red) = (5/10)(4/9) = 20/90 = 2/9 ≈ 0.222. Equivalently C(5,2)/C(10,2) = 10/45 = 2/9. P(at least one red) = 1 − P(no red) = 1 − C(5,2)/C(10,2) = 1 − 10/45 = 35/45 = 7/9 ≈ 0.778.
Traps that cost marks
- Mixing up with and without replacement.
- Adding P(A) + P(B) without subtracting the intersection.
- Assuming independence because events look unrelated.
- Using permutations when order is irrelevant; claiming causation from a strong r.