/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q10E A marketer is interested in how ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Q10E (page 1)

A marketer is interested in how many customers arelikely to buy a particular product in a particular store. LetP be the proportion of all customers in the store who will buy the product. Let the distribution of P be uniform onthe interval [0, 1] before observing any data. The marketer

then observes 25 customers and only six buy the product. If the customers were conditionally independent given P,find the conditional distribution of P given the observed customers.

Short Answer

Expert verified

\(f\left( p \right) = {}^{25}{C_6}{p^6}{\left( {1 - p} \right)^{19}}\)

Step by step solution

01

Given Information

It is given that P is the proportionof all customers in the store who will buy the product. The distribution of P followed a uniform distribution on the interval [0,1]

\(P \sim U\left[ {0,1} \right]\)

02

Defining the conditional distribution

By definition if the joint distribution of two variables is given, or if two random variables X and Y are jointly distributed, the probability distribution of one variable, say X given the other variable value is known say Y=y is called the conditional distribution of X.

Now,\({f_{X,Y}}\left( {x.y} \right)\)is the joint distribution and\({f_X}\left( x \right)\)is the marginal distribution.

The relation with the probability distribution ofX givenY is given by:

\({f_{Y|X}}\left( {y|x} \right) = \frac{{{f_{X,Y}}\left( {x.y} \right)}}{{{f_X}\left( x \right)}}\)

03

Calculating the required distribution

The marketer observes that out of a total of 25 customers only 6 buy the product.

Note that, if the uniform distribution is conditioned on a discrete variable, then it forms a binomial distribution.

Therefore, by the condition pdf definition defined above, the conditional distribution of p given the observed customers is:

\(\)\(f\left( {p|x} \right) = {}^{25}{C_6}{p^6}{\left( {1 - p} \right)^{19}}\)

Here the binomial distribution has parameters, n and p.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that 40 percent of the students in a large population are freshmen, 30 percent are sophomores, 20 percent are juniors, and 10 percent are seniors .Suppose that10 students are selected at random from the population, and let X1, X2, X3, X4 denote, respectively, the numbers of freshmen, sophomores, juniors, and seniors that are obtained.

a. Determine 蟻(Xi, Xj ) for each pair of values i and j (i< j ).

b. For what values of i and j (i<j ) is 蟻(Xi, Xj ) most negative?

c. For what values of i and j (i<j ) is 蟻(Xi, Xj ) closest to 0?

For every collection of events\({A_i}\left( {i \in I} \right)\), show that

\({\left( {\bigcup\limits_{i \in I} {{A_i}} } \right)^c} = \bigcap\limits_{i \in I} {{A_i}^c} \)and\({\left( {\bigcap\limits_{i \in I} {{A_i}} } \right)^c} = \bigcup\limits_{i \in I} {{A_i}^c} \).

Consider a state lottery game in which each winning combination and each ticket consists of one set of k numbers chosen from the numbers 1 to n without replacement. We shall compute the probability that the winning combination contains at least one pair of consecutive numbers.

a. Prove that if\({\bf{n < 2k - 1}}\), then every winning combination has at least one pair of consecutive numbers. For the rest of the problem, assume that\({\bf{n}} \le {\bf{2k - 1}}\).

b. Let\({{\bf{i}}_{\bf{1}}}{\bf{ < }}...{\bf{ < }}{{\bf{i}}_{\bf{k}}}\)be an arbitrary possible winning combination arranged in order from smallest to largest. For\({\bf{s = 1,}}...{\bf{,k}}\), let\({{\bf{j}}_{\bf{s}}}{\bf{ = }}{{\bf{i}}_{\bf{s}}}{\bf{ - }}\left( {{\bf{s - 1}}} \right)\). That is,

\(\begin{array}{c}{{\bf{j}}_{\bf{1}}}{\bf{ = }}{{\bf{i}}_{\bf{1}}}\\{{\bf{j}}_{\bf{2}}}{\bf{ = }}{{\bf{i}}_{\bf{2}}}{\bf{ - 1}}\\{\bf{.}}\\{\bf{.}}\\{\bf{.}}\\{{\bf{j}}_{\bf{k}}}{\bf{ = }}{{\bf{i}}_{\bf{k}}}{\bf{ - }}\left( {{\bf{k - 1}}} \right)\end{array}\)

Prove that\(\left( {{{\bf{i}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{i}}_{\bf{k}}}} \right)\)contains at least one pair of consecutive numbers if and only if\(\left( {{{\bf{j}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{j}}_{\bf{k}}}} \right)\)contains repeated numbers.

c. Prove that\({\bf{1}} \le {{\bf{j}}_{\bf{1}}} \le ... \le {{\bf{j}}_{\bf{k}}} \le {\bf{n - k + 1}}\)and that the number of\(\left( {{{\bf{j}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{j}}_{\bf{k}}}} \right)\)sets with no repeats is\(\left( {\begin{array}{*{20}{c}}{{\bf{n - k + 1}}}\\{\bf{k}}\end{array}} \right)\)

d. Find the probability that there is no pair of consecutive numbers in the winning combination.

e. Find the probability of at least one pair of consecutive numbers in the winning combination

Each minute a machine produces a length of rope with mean of 4 feet and standard deviation of 5 inches. Assuming that the amounts produced in different minutes are independent and identically distributed, approximate the probability that the machine will produce at least 250 feet in one hour.

Prove that Following no. is an Integer

\(\frac{{4155 \times 4156 \times ....4250 \times 4251}}{{2 \times 3 \times .... \times 96 \times 97}}\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.