The graph, expectation and variance of abinomial distribution
In Year 12 Mathematical Methods (Queensland, QCAA), a binomial distribution \(X\sim B(n,p)\) counts the successes in \(n\) independent trials. Its column graph is centred on the mean \(np\); it is symmetric when \(p=0.5\) and skewed otherwise. The expectation is \(E(X)=np\), the variance is \(\operatorname{Var}(X)=np(1-p)\) and the standard deviation is \(\sqrt{np(1-p)}\) — and these can be reversed to find \(p\) or \(n\) from a given mean or variance.
A binomial random variable \(X\sim B(n,p)\) is the number of successes in \(n\) independent trials, each with the same probability of success \(p\). Its column graph plots the probability \(P(X=x)\) as the height of a column above each value \(x=0,1,\ldots,n\).
The expected value (mean) \(E(X)=np\) is the balance point of the graph — the columns are tallest near \(x=np\). The variance \(\operatorname{Var}(X)=np(1-p)\) measures spread, and the standard deviation \(\text{SD}=\sqrt{np(1-p)}\) is its square root, in the same units as \(X\).
The shape depends on \(p\). When \(p=0.5\) success and failure are equally likely, so the graph is symmetric about the mean. When \(p<0.5\) it is positively skewed (a tail towards larger values); when \(p>0.5\) it is negatively skewed (a tail towards smaller values).
For \(X\sim B(n,p)\), the expected value (mean):
The variance and standard deviation:
To reverse the mean and variance for \(p\) or \(n\):
Working with a binomial distribution
- Read off \(n\) and \(p\). From \(X\sim B(n,p)\), \(n\) is the number of trials and \(p\) the probability of success.
- Mean. \(E(X)=np\) — this is where the column graph is centred.
- Variance and SD. \(\operatorname{Var}(X)=np(1-p)\); the standard deviation is its square root \(\sqrt{np(1-p)}\) (do not stop at the variance).
- Shape. Compare \(p\) with \(0.5\): equal ⇒ symmetric; smaller ⇒ right-skewed; larger ⇒ left-skewed.
- Work backwards. Given the mean, \(p=\dfrac{E(X)}{n}\). Given both the mean and the variance, \(1-p=\dfrac{\operatorname{Var}(X)}{E(X)}\), then \(n=\dfrac{E(X)}{p}\).
| \(E(X)\) | \(=\) | \(80\times 0.25=20\) |
| \(\operatorname{Var}(X)\) | \(=\) | \(80\times 0.25\times 0.75=15\) |
| \(\text{SD}\) | \(=\) | \(\sqrt{15}\approx 3.87\) |
| \(48p\) | \(=\) | \(12\) |
| \(p\) | \(=\) | \(0.25\) |
| \(\operatorname{Var}(X)\) | \(=\) | \(48\times 0.25\times 0.75=9\) |
The mean is \(np=10\times 0.2=2\), so the columns are centred on \(x=2\). Since \(p=0.2<0.5\), the graph is positively skewed — a tail towards the larger values.
Divide the variance by the mean; the \(np\) cancels, leaving \(1-p\).
| \(1-p\) | \(=\) | \(\dfrac{4.8}{12}=0.4\) |
| \(p\) | \(=\) | \(0.6\) |
| \(n\) | \(=\) | \(\dfrac{12}{0.6}=20\) |
Common pitfalls
Frequently asked questions
How do you find the mean of a binomial distribution?
The mean (expected value) is \(E(X)=np\). For \(X\sim B(80,0.25)\), \(E(X)=80\times 0.25=20\).
How do you find the variance and standard deviation?
\(\operatorname{Var}(X)=np(1-p)\) and \(\text{SD}=\sqrt{np(1-p)}\). For \(B(80,0.25)\), \(\operatorname{Var}(X)=15\) and \(\text{SD}=\sqrt{15}\approx 3.87\).
When is a binomial column graph symmetric?
Only when \(p=0.5\), because then \(P(X=k)=P(X=n-k)\), so the graph is a mirror image about its mean.
How does p change the shape?
\(p<0.5\) is positively (right-) skewed, \(p>0.5\) is negatively (left-) skewed, and \(p=0.5\) is symmetric. The columns always centre near \(np\).
How do you find p from the mean?
Since \(E(X)=np\), divide by \(n\): \(p=\dfrac{E(X)}{n}\). For \(B(48,p)\) with mean \(12\), \(p=\dfrac{12}{48}=0.25\).
How do you find both n and p from the mean and variance?
Divide the variance by the mean to get \(1-p\) (the \(np\) cancels), then \(n=\dfrac{E(X)}{p}\). Mean \(12\), variance \(4.8\) gives \(p=0.6\), \(n=20\).