4MA1: Statistics and probability. Version: Issue 2, November 2017; first assessment June 2018; linear Mathematics A.
Foundation teaching and Higher additions are labelled below. This reference packages the existing native-lesson crosswalk. It does not certify unreviewed specification rows or a whole qualification. Original diagnostics are separate and are not reproduced.
Centre, spread and data displays · Foundation
The mean is total divided by count. The median is the central value after sorting. The range is maximum minus minimum. Use frequency tables, bar charts and suitable comparisons.
For 2,4,4,6,9, total=25 and count=5, so mean=5. The central value is 4, so median=4. The range is 9-2=7. Explain both a typical value and the spread.
The tallest histogram bar need not contain the most observations. A grouped mean is an estimate. Correlation does not prove causation, and extrapolation extends beyond the observed range.
A bar chart uses separate bars for categories. Unequal-class-width histograms and formal density calculations are outside this Foundation/Core lesson.
Data summaries, histograms and interpretation · Higher
Compare an appropriate average and spread in context. A histogram uses area for frequency, so height=frequency/class width. Grouped estimates assume representative values within intervals.
A class from 10 to 20 with frequency 30 has density 30/10=3. A class from 20 to 40 with frequency 20 has density 20/20=1. Its wider bar must not be mistaken for a larger density. For values 2,4,4,6,9, the median is 4 and mean is 5.
The tallest histogram bar need not contain the most observations. A grouped mean is an estimate. Correlation does not prove causation, and extrapolation extends beyond the observed range.
Choose a display that fits the data type. Give both a numerical comparison and what it means for the population; do not infer more precision than the sample supports.
Probability trees and outcomes · Foundation
A probability lies between 0 and 1. Exhaustive, mutually exclusive outcomes have probabilities summing to 1. Multiply successive branch probabilities and add separate routes to an outcome.
A bag contains 3 red and 2 blue counters. With replacement, P(two red)=3/5×3/5=9/25=0.36. Without replacement, the red-red branch is 3/5×2/4=0.3. Label each branch before multiplying.
Mutually exclusive means no overlap; independent means that knowing one event does not change the other's probability. Two disjoint events with positive probability are not independent.
Use a frequency table or a simple tree before calculating. Formal conditional probability formulae are outside this Foundation/Core support lesson.
Probability, trees and conditional reasoning · Higher
Multiply along a tree branch and add disjoint branches. With replacement, the composition stays fixed. Conditional probability is P(A given B)=P(A∩B)/P(B), for P(B)>0.
Without replacement, P(two red)=3/5×2/4=3/10. P(one of each)=3/5×2/4+2/5×3/4=3/5. If P(A∩B)=0.12 and P(B)=0.3, P(A given B)=0.4.
Mutually exclusive means no overlap; independent means that knowing one event does not change the other's probability. Two disjoint events with positive probability are not independent.
A two-way table makes the restricted denominator visible. Before using a product P(A)P(B), justify independence from the context or the supplied information.
Cumulative frequency and box plots · Higher
A cumulative frequency counts observations below successive class boundaries. Read quartiles at one quarter, one half and three quarters of the total frequency. A box plot represents minimum, lower quartile, median, upper quartile and maximum.
For 80 observations, read Q1 at cumulative frequency 20, median at 40 and Q3 at 60. If Q1=12,Q3=21, then IQR=9. Compare the medians for typical journey time and the IQRs for consistency.
Plot against class boundaries rather than midpoints. Grouped quartiles are estimates. The range is sensitive to extremes; the IQR describes only the middle half.
Explain a comparison in the context of the measured quantity. An outlier rule may use Q1-1.5IQR and Q3+1.5IQR; use the rule specified in the task rather than assuming every graph follows it.
