Probability & Statistics Ch.2 Section 3 Question 7

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Let $X \sim \text{Binomial}(12, \theta)$ . For what value of $\theta$ is $P(X = 11)$ maximized?

Solution

From Example 2.3.3 in the textbook, we know:

p_X(x) = P(X=x) = \binom{n}{x}\theta^x(1-\theta)^{n-x}

We plug in X = 11 and n = 12:

P(X=11) = \binom{12}{11}\theta^{11}(1-\theta)^{12-11} = 12\theta^{11}(1-\theta)

We differentiate to locate the peak. First, expand:

P(X=11) = 12(\theta^{11} - \theta^{12})

Apply the power rule:

\frac{d}{d\theta}P(X=11) = 12(11\theta^{10} - 12\theta^{11})

Factor and set equal to zero:

0 = 12\theta^{10}(11 - 12\theta)

This gives us:

$\theta^{10} = 0 \Rightarrow \theta = 0$
$11 - 12\theta = 0 \Rightarrow \theta = \frac{11}{12}$

Since $\theta = 0$ is not a valid solution (it would make it impossible to have any of the 11 desired successes), the answer is:

\theta = \frac{11}{12}