A normal probability followed by a binomial model
First find the probability that one normal trial is a success. That probability becomes p in a binomial model for the number of successes in the repeated independent trials. In the five-trial example, the normal mean is 30, the standard deviation is 5, and readings below 35 count as successes. These examples use empirical-rule estimates 0.84 and 0.16 as the binomial probabilities
- Standardise the success boundary Subtract the normal mean and divide by the normal standard deviation \(\displaystyle z=\frac{x-\mu}{\sigma}\)
- Find the one-trial probability Use the normal tail that matches the success condition. Here, the 68% empirical rule gives the estimate 0.84 \(\displaystyle X<35,\quad z=1\;\Rightarrow\;p=\Pr(Z<1)\approx0.84\)
- Model the repeated count Independent trials with two outcomes and common probability use a binomial distribution \(\displaystyle Y\sim\operatorname{Bin}(5,0.84)\)
- Choose the binomial calculation Use one term for an exact count, a cumulative sum or complement for a range, and np for the expected count \(\displaystyle E(Y)=np=5\times0.84=4.2\)
Worked example
A device reading has a normal distribution with mean 52 and variance 36. A success is a reading below 46. Six independent trials use this same model. Use the empirical-rule estimate for the one-trial probability to estimate the probability of exactly two successes
- \(\displaystyle z=\frac{46-52}{6}=-1\)Standardise the success boundary using μ = 52 and σ = 6
- \(\displaystyle p=\Pr(X<46)=\Pr(Z<-1)\approx0.16\)The lower-tail empirical estimate is 0.16. Use this supplied approximation as p for the binomial calculation
- \(\displaystyle Y\sim\operatorname{Bin}(6,0.16),\quad \Pr(Y=2)=\binom{6}{2}(0.16)^2(0.84)^4\)Use the binomial exact-count formula with two successes and four failures
- \(\displaystyle \Pr(Y=2)=15(0.16)^2(0.84)^4\approx0.1912\)Using the empirical-rule estimate, the probability of exactly two successes is approximately 0.1912. A calculation retaining the full normal probability gives a different approximation
Common mistakes
Difficulty levels
Questions about this topic
Why is p a normal probability rather than a standard score?+
A standard score gives the location of the boundary in standard-deviation units. The binomial model needs the probability of the success event, such as the empirical-rule estimate of 0.84 for a boundary one standard deviation above the mean
How is the probability of exactly k successes calculated?+
Use the binomial exact-count formula, where n is the number of trials and p is the one-trial success probability
What is the expected number of successes?+
For a binomial count, the expected number is the number of trials multiplied by the one-trial success probability. This is the long-run average count over repeated groups of n trials, so it need not be a whole number. An observed count in one group is always an integer
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