Chapter 2 Section 3 Question 7
Let $X$ ~ $Binomial(12, \theta )$. For what value of $\theta$ is $P(X = 11)$ maximized?
Author: Mohammad-Ali Bandzar| Date:Oct 15 2020
From Example 2.3.3 in the book we know: $$p\scriptsize{X}\normalsize{(x)=P(X=x)={n\choose x}\theta ^x(1-\theta)^{n-x}}$$
we shall now plug in X=11 and n=12 $$P(X=11)={12\choose 11}\theta ^11(1-\theta)^{12-11}$$ $$P(X=11)=12\theta ^11(1-\theta)$$ we can now differentiate the function to try and locate our peak $$P'(X=11)=12*\theta ^11(1-\theta)$$ expand out: $$P'(X=11)=12*(\theta ^11-\theta ^12)$$ power rule: $$P'(X=11)=12*(11\theta ^10-12\theta ^11)$$ we will now factor and set this equal to zero $$0=12\theta^10*(11-12\theta)$$ Therefore:
$$12\theta^10=0$$ and $$11-12\theta=0$$ $$\theta=\frac{11}{12}$$ since $\theta=0$ is not a valid solution as it would make it impossible to have any of the 11 desired successes
therefore $\theta=\frac{11}{12}$ is the correct solution