Concept understanding — Mean and Variance of Binomial Distribution
For X∼B(n,p), three summary numbers describe the whole distribution without listing every term of the p.m.f.: the mean μ=E(X)=np (the average number of successes over many repetitions of the n-trial experiment), the variance Var(X)=npq (how spread out the successes are around that mean), and the standard deviation SD(X)=σX=npq. These are stated as formulae without proof. They are used two ways: forward, when n and p are known, to compute E(X) and Var(X) directly; and backward, when two of E(X), Var(X), n, p are given and the other two must be recovered - typically by dividing Var(X)=npq by E(X)=np to isolate q=Var(X)/E(X) first (since the n cancels), then p=1−q, and finally n=E(X)/p.
n=10, E(X)=5; find p and Var(X).
✓Final answer
p=0.5, Var(X)=2.5.
E(X)=np=5 with n=10 gives p=105=0.5, so q=1−0.5=0.5.
Var(X)=npq=10×0.5×0.5=2.5.
✓Final answer
p=0.5, Var(X)=2.5.
Solve p = E(X)/n from the mean formula, then substitute into npq for the variance.
Assuming p is automatically 0.5 without deriving it from n and E(X) (it only comes out to 0.5 here because of these particular numbers).