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Chapter 3: Q14E (page 117)

Find the quartiles and the median of the binomial distribution with parametersn=10 andp=0.2.

Short Answer

Expert verified

The median of the distribution is 0.268.

The quartiles of the distribution are


Step by step solution

01

Given information

Given the binomial distribution of the parameter is n=10 andp=0.2.

02

State the distribution

03

Compute the quantile function and median

Then by tabulating the values of cdf and pdf is

Then the quantiles values are given by

Hence the median is 0.268.

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Most popular questions from this chapter

Suppose that \({{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{n}}}\)form a random sample of nobservations from the uniform distribution on the interval(0, 1), and let Y denote the second largest of the observations.Determine the p.d.f. of Y.Hint: First, determine thec.d.f. G of Y by noting that

\(\begin{aligned}G\left( y \right) &= \Pr \left( {Y \le y} \right)\\ &= \Pr \left( {At\,\,least\,\,n - 1\,\,observations\,\, \le \,\,y} \right)\end{aligned}\)

Let Z be the rate at which customers are served in a queue. Assume that Z has the p.d.f.

\(\begin{aligned}f\left( z \right) &= 2e{}^{ - 2z},z > 0\\ &= 0,otherwise\end{aligned}\)

Find the p.d.f. of the average waiting time T = 1/Z.

Suppose that a Markov chain has four states 1, 2, 3, 4, and stationary transition probabilities as specified by the following transition matrix

\(p = \left[ {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{4}}&0&{\frac{1}{2}}\\0&1&0&0\\{\frac{1}{2}}&0&{\frac{1}{2}}&0\\{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\end{array}} \right]\):

a.If the chain is in state 3 at a given timen, what is the probability that it will be in state 2 at timen+2?

b.If the chain is in state 1 at a given timen, what is the probability it will be in state 3 at timen+3?

The unique stationary distribution in Exercise 9 is \({\bf{v = }}\left( {{\bf{0,1,0,0}}} \right)\). This is an instance of the following general result: Suppose that a Markov chain has exactly one absorbing state. Suppose further that, for each non-absorbing state \({\bf{k}}\), there is \({\bf{n}}\) such that the probability is positive of moving from state \({\bf{k}}\) to the absorbing state in \({\bf{n}}\) steps. Then the unique stationary distribution has probability 1 in the absorbing state. Prove this result.

Question:Suppose that a point (X,Y) is to be chosen from the squareSin thexy-plane containing all points (x,y) such that 0≤x≤1 and 0≤y≤1. Suppose that the probability that the chosen point will be the corner(0,0)is 0.1, the probability that it will be the corner(1,0)is 0.2, and the probability that it will be the corner(0,1)is 0.4, and the probability that it will be the corner(1,1)is 0.1. Suppose also that if the chosen point is not one of the four corners of the square, then it will be an interior point of the square and will be chosen according to a constant p.d.f. over the interior of the square. Determine

\(\begin{array}{l}\left( {\bf{a}} \right)\;{\bf{Pr}}\left( {{\bf{X}} \le \frac{{\bf{1}}}{{\bf{4}}}} \right)\;{\bf{and}}\\\left( {\bf{b}} \right)\;{\bf{Pr}}\left( {{\bf{X + Y}} \le {\bf{1}}} \right)\end{array}\)
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