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Conditional probabilities with normal intervals

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Which interval is the intersection of the two events?

\[A:10<X<20,\quad B:14<X<25\]
Year 12, VCE Units 3 & 4, Chapter 13, 4 difficulty levels, Calculator allowedBuild a worksheet

Conditional normal probabilities

A conditional probability restricts attention to the event after the vertical bar. Find the overlap of the two normal intervals, calculate its probability, and divide by the probability of the conditioning event. Use the empirical rule only when requested; otherwise calculate normal areas using technology and retain precision until the final answer

\[\Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}\]
  1. Read the conditioning event The event after the vertical bar supplies the denominator, which must be positive \(\displaystyle \Pr(A\mid B)\;\Rightarrow\;B\text{ is the conditioning event}\)
  2. Find the intersection Keep the larger lower endpoint and the smaller upper endpoint. For these open intervals, if the lower endpoint is not below the upper endpoint, the intersection is empty \(\displaystyle 10<X<20\text{ and }14<X<25\;\Rightarrow\;14<X<20\)
  3. Standardise both endpoints Use the mean and standard deviation to convert each endpoint to a standard score \(\displaystyle z=\frac{x-\mu}{\sigma}\)
  4. Divide the two probabilities The numerator is the intersection probability and the denominator is the conditioning probability. The displayed estimate uses empirical-rule areas \(\displaystyle \Pr(A\mid B)\approx\frac{0.34}{0.68}=0.5\)

Worked example

Estimate the conditional probability using the 68% empirical rule, then check the result by symmetry

\[\Pr(A\mid B),\quad A:80<X<84,\quad B:76<X<84,\quad X\sim N(80,16)\]
  1. \(\displaystyle A\cap B:80<X<84\)The interval for A is contained in B, so the intersection is A.
  2. \(\displaystyle \Pr(A\cap B)=\Pr(0<Z<1)\approx0.84-0.50=0.34\)The second parameter 16 is the variance, so σ = 4. Standardise 80 and 84, then use the empirical-rule estimate for the area from the mean to one standard deviation above it
  3. \(\displaystyle \Pr(B)=\Pr(-1<Z<1)\approx0.84-0.16=0.68\)Standardise 76 and 84 and use the empirical estimate 0.68 for the central one-standard-deviation area
  4. \(\displaystyle \Pr(A\mid B)\approx\frac{0.34}{0.68}=0.5\)Divide the two empirical area estimates. Independently, A is the upper half of the symmetric interval B, so the exact conditional probability is also 0.5

Common mistakes

Union used as the intersection
✕\(\displaystyle A:10<X<20,\ B:14<X<25\;\Rightarrow\;A\cap B:10<X<25\)✓\(\displaystyle A:10<X<20,\ B:14<X<25\;\Rightarrow\;A\cap B:14<X<20\)
An intersection keeps only values that satisfy both events, so it uses the larger lower boundary and smaller upper boundary
Wrong conditioning denominator
✕\(\displaystyle \Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(A)}\)✓\(\displaystyle \Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}\)
The event after the vertical bar is the condition and therefore supplies the denominator
Raw endpoints not standardised
✕\(\displaystyle X\sim N(80,16),\quad 80<X<84\;\Rightarrow\;0<Z<4\)✓\(\displaystyle z=\frac{x-80}{4},\quad 80<X<84\;\Rightarrow\;0<Z<1\)
Each boundary must be centred by subtracting the mean and scaled by dividing by the standard deviation

Difficulty levels

introFind a conditional probability when one interval with whole-number standard-score boundaries is contained in the other
standardUse overlapping intervals and either conditional direction
challengeHandle disjoint intervals or non-integer normal boundaries
examChoose the correct intersection or denominator before calculating the conditional value

Questions about this topic

Why does the conditioning event supply the denominator?+

The condition B defines the restricted set of outcomes being considered. The numerator is the part of B that also belongs to A, so it is divided by the probability of B.

How is the intersection of two open intervals found?+

Take the larger of the two lower endpoints and the smaller of the two upper endpoints. For these open intervals, if the resulting lower endpoint is not below the upper endpoint, the intersection is empty

Do strict and non-strict endpoints change a normal probability?+

No. A continuous normal variable has probability zero at any single endpoint, so < and ≤ give the same interval probability

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 13: Normal distribution

In this chapter

Read normal parameters and describe their graph effectCalculate probabilities for normal distributionsCalculate a standard score or actual valueUse z-score boundaries to estimate normal-model percentages and countsCompare observations from different distributions using z-scoresInverse normal calculationsInfer a normal-model mean or standard deviationConditional probabilities with normal intervalsMixed probability applications with the normal distribution