Question

Let X ~ B(6, 0.35). Find the mode of X. Find the median of X.

Answer 1

Is that B for a Binomial distribution?

Answer 2

yes.

Answer 3

@kirbykirby

Answer 4

What you can do is calculate all the probabilities for the distribution, that is find
\[P(X=0)={6 \choose 0}(0.35)^0(1-0.35)^{6-0}\\
P(X=1)={6 \choose 1}(0.35)^1(1-0.35)^{6-1}\\
\vdots\\

P(X=6)=\ldots\]

WolframAlpha gives the table of values for all these probabilties (it is just a bit annoying to do by hand) http://www.wolframalpha.com/input/?i=Binomial%286%2C+0.35%29&a=*C.Binomial-_*ProbabilityDistribution-

Then, the mode is the most frequent point that occurs. So, choose the value of X where the probability is highest. From the table of values (and the given plot), this clearly occurs at x=2.

For the median:
note that the \(100p^{\text{th}}\) percentile of a random variable \(X\) is the value \(\pi_p\) such that \[\pi_p=\inf\{x\in \mathbb{R}\mid F_X(x)\ge p\} \]
Now the median is the 50th percentile, so you will need to find\[ \pi_{0.5}=\inf\{x\in\mathbb{R}|F_X \ge 0.5\} \]
\(\inf\) is the infinum, and is just the greatest lower bound.

So, Start by finding the CDF for every X until you JUST pass 0.5:
\(P(X\le 0)=P(X=0)=0.07542\)
\(P(X\le 1)=P(X=0)+P(X=1)=0.07542+0.2437=0.31912\)
\(P(X \le 2)=P(X=0)+P(X=1)+P(X=2)=0.31912+0.328=0.647\)

So, at \(X \le 1\), you are below 0.5, and \(X \le 2\), you have just passed 0.5, so x=2 is the median (you can kind of think of the percentile definition as "the smallest value of x for when the CDF just exceeds \(p\)")