Probabilities from grouped frequencies
A grouped frequency table assigns a count to each complete class interval. Match the event boundaries to whole classes, then divide the selected frequency by the appropriate total. In the example below, the classes from 10 to 20 and 20 to 30 have frequencies 9 and 12, respectively, out of 30 observations
\[\Pr(A)\approx\frac{n(A)}{n(\text{all observations})},\quad \Pr(A\mid B)\approx\frac{n(A\cap B)}{n(B)}\]
- Match the class boundaries Use the complete classes whose boundaries make up the stated event \(\displaystyle 10\le X<30\;\Rightarrow\;[10,20)\text{ and }[20,30)\)
- Add the event frequencies The frequency of an event is the sum of the frequencies in its selected classes \(\displaystyle 9+12=21\)
- Use the total frequency For a direct grouped probability, divide the event frequency by the total frequency \(\displaystyle \Pr(10\le X<30)\approx\frac{21}{30}=\frac{7}{10}\)
- Reduce the sample space for a condition For a conditional probability, use the frequency in the conditioning event as the denominator \(\displaystyle \Pr(A\mid B)=\frac{n(A\cap B)}{n(B)}\)
Worked example
A grouped frequency table has four classes. From 0 up to but not including 5, the frequency is 4; from 5 up to but not including 10, it is 7; from 10 up to but not including 15, it is 11; and from 15 up to but not including 20, it is 8. Estimate the probability for values from 5 to 15
\[\Pr(5\le X<15)\approx\frac{7+11}{4+7+11+8}\]
- \(\displaystyle 5\le X<15\;\Rightarrow\;[5,10)\text{ and }[10,15)\)Use the complete classes between the event boundaries 5 and 15.
- \(\displaystyle n(5\le X<15)=7+11=18\)Add the frequencies in the two selected classes
- \(\displaystyle n(\text{all observations})=4+7+11+8=30\)Add every class frequency to obtain the total frequency
- \(\displaystyle \Pr(5\le X<15)\approx\frac{18}{30}=\frac{3}{5}=0.6\)Divide the event frequency by the total frequency and simplify
Common mistakes
Counting a class below the lower boundary
✕\(\displaystyle \Pr(10\le X<30)\approx\frac{6+9+12}{30}=\frac{9}{10}\)✓\(\displaystyle \Pr(10\le X<30)\approx\frac{9+12}{30}=\frac{7}{10}\)
The event starts at 10, so the class from 0 to 10 is outside the eventUsing one class frequency as the denominator
✕\(\displaystyle \Pr(10\le X<30)\approx\frac{9+12}{9}=\frac{7}{3}\)✓\(\displaystyle \Pr(10\le X<30)\approx\frac{9+12}{30}=\frac{7}{10}\)
A direct probability uses the selected event frequency over the total frequencyUsing the whole table in a conditional denominator
✕\(\displaystyle \Pr(A\mid B)\approx\frac{n(A\cap B)}{n(\text{all observations})}\)✓\(\displaystyle \Pr(A\mid B)\approx\frac{n(A\cap B)}{n(B)}\)
A conditional probability is measured within B, so the denominator is the frequency of B.Difficulty levels
introFind one interval probability from a small grouped frequency table
standardUse several classes and interpret whether an interior value is a class boundary
challengeCalculate a conditional probability from a union of whole classes and check an interior value
examUse a grouped table and histogram to solve a conditional probability item
Questions about this topic
Why is a probability from grouped data only an estimate?+
The table gives counts for whole classes, not the positions of observations inside each class. An interval ending inside a class cannot be determined exactly from the table alone
Why is the total frequency used in a direct probability?+
A direct probability compares the number of observations in the event with all observations recorded in the table
How is a conditional probability found from frequencies?+
Count observations in both events for the numerator. Count observations in the conditioning event for the denominator, then divide and simplify
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 12: Continuous probability distributions
In this chapter
12A Continuous random variables and density functions: Validate and normalise a continuous probability density function12A Continuous random variables and density functions: Estimate probabilities from grouped continuous frequency dataGroup numerical data into class intervalsConstruct a histogram from a frequency table12A Continuous random variables and density functions: Finite events, probability assignments and complements12A Continuous random variables and density functions: Conditional probability from exact representationsEvaluate and determine function rules12B Probability density validity and normalisation: Validate and normalise a continuous probability density function12B Probability density validity and normalisation: Estimate probabilities from grouped continuous frequency dataSolve a linear equation12B Probability density validity and normalisation: Evaluate a polynomial or recover a coefficient parameterWork with sets and interval notation12B Probability density validity and normalisation: Round to decimal places and significant figuresInterval and conditional probabilities from a densityCompound events, the addition rule and mutual exclusivity12C Interval and conditional probabilities: Finite events, probability assignments and complements12C Interval and conditional probabilities: Conditional probability from exact representationsMean, variance and standard deviation of a continuous random variableMedian, quartiles and percentiles of a continuous random variableExpand and factor polynomial expressions12D Mean, variance, standard deviation and percentiles: Evaluate a polynomial or recover a coefficient parameterMove between roots, factors, polynomial rules and equationsSolve a quadratic exactly with the quadratic formula12D Mean, variance, standard deviation and percentiles: Round to decimal places and significant figuresLinear transformations and totals of independent continuous random variables