/*! 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} Q3E A certain market has both an exp... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A certain market has both an express checkout line and a superexpress checkout line. Let \({{\rm{X}}_{\rm{1}}}\) denote the number of customers in line at the express checkout at a particular time of day, and let \({{\rm{X}}_{\rm{2}}}\) denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of \({{\rm{X}}_{\rm{1}}}\)and\({{\rm{X}}_{\rm{2}}}\) is as given in the accompanying table

a. What is \({\rm{P(}}{{\rm{X}}_{\rm{1}}}{\rm{ = 1}}{{\rm{X}}_{\rm{2}}} = 1)\), that is, the probability that there is exactly one customer in each line?

b. What is \({\rm{P(}}{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{X}}_{\rm{2}}})\),that is, the probability that the numbers of customers in the two lines are identical? c. Let A denote the event that there are at least two more customers in one line than in the other line. Express A in terms of \({{\rm{X}}_{\rm{1}}}\) and \({{\rm{X}}_{\rm{2}}}\), and calculate the probability of this event.

d. What is the probability that the total number of customers in the two lines is exactly four? At least four?

Short Answer

Expert verified
  1. The probability is \({\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 1,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 1}}} \right){\rm{ = 0}}{\rm{.15;}}\)
  2. The probability is \({\rm{ P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.4;}}\)
  3. The probability is \({\rm{ P(A) = 0}}{\rm{.22;}}\)
  4. The probability is \({\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}} \ge {\rm{4}}} \right){\rm{ = 0}}{\rm{.46}}{\rm{.}}\)

Step by step solution

01

Definition of Probability

Probability is a metric for determining the possibility of an event occurring. Many things are impossible to forecast with\({\rm{100\% }}\)accuracy. Using it, we can only anticipate the probability of an event occurring, or how probable it is to occur. Probability can range from\({\rm{0}}\)to\({\rm{1}}\), with\({\rm{0}}\)indicating an improbable event and 1 indicating a certain event. Possibility of...

02

Find the probability that there is exactly one customer in each line?

(a):

As an entry, the following probability can be read from the given table.\((1,1)\)-first column/first row

\(\begin{aligned}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 1,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 1}}} \right)\\{\rm{ = p(1,1)}}\\{\rm{ = 0}}{\rm{.15}}{\rm{.}}\end{aligned}\)

Therefore , the probability is \( = 0.15\).

03

 Find the probability that the numbers of customers?

(b):

The following holds

\(\begin{aligned}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = p(0,0) + p(1,1) + p(2,2) + p(3,3)}}\\{\rm{ = 0}}{\rm{.08 + 0}}{\rm{.15 + 0}}{\rm{.1 + 0}}{\rm{.7}}\\{\rm{ = 0}}{\rm{.4}}\end{aligned}\)

Therefore , the probability is \( = 0.4\).

04

Calculate the probability of this event?

(c):

The following two events can be combined to form the provided event.

\(\left\{ {{{\rm{X}}_{\rm{1}}} \ge {\rm{2 + }}{{\rm{X}}_{\rm{2}}}} \right\}{\rm{ and }}\left\{ {{{\rm{X}}_{\rm{2}}} \ge 2 + {{\rm{X}}_{\rm{1}}}} \right\}.\)

Therefore, we have

\({\rm{A = }}\left\{ {\left\{ {{{\rm{X}}_{\rm{1}}} \ge 2 + {{\rm{X}}_{\rm{2}}}} \right\} \cup \left\{ {{{\rm{X}}_{\rm{2}}} \ge 2 + {{\rm{X}}_{\rm{1}}}} \right\}} \right\}\)

The event \({\rm{A}}\)can be represented as the sum of nine discrete occurrences in form.

\(\left\{ {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}} \right\}{\rm{,}}\left\{ {{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}} \right\}\)

where \({{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}\) satisfy the event \({\rm{A}}\)'s specified property. As a result, the chance can be computed as follows:

\(\begin{aligned}{\rm{P(A) = P}}\left( {\left\{ {{X_1} \ge 2 + {X_2}} \right\} \cup \left\{ {{X_2} \ge 2 + {X_1}} \right\}} \right)\\{\rm{ = p(2,0) + p(3,0) + p(4,0) + p(3,1) + p(4,1) + p(4,2) + p(0,2) + p(0,3) + p(1,3)}}\\{\rm{ = 0}}{\rm{.05 + 0}}{\rm{.00 + 0}}{\rm{.00 + 0}}{\rm{.03 + 0}}{\rm{.01 + 0}}{\rm{.05 + 0}}{\rm{.04 + 0}}{\rm{.00 + 0}}{\rm{.04}}\\{\rm{ = 0}}{\rm{.22}}{\rm{.}}\end{aligned}\)

Therefore , the probability is \({\rm{ = }}{\rm{.22}}\).

05

Find the probability that the total number of customers?

(d):

The following holds

\(\begin{aligned}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}{\rm{ = 4}}} \right){\rm{ = p(1,3) + p(2,2) + p(3,1) + p(4,0)}}\\{\rm{ = 0}}{\rm{.04 + 0}}{\rm{.1 + 0}}{\rm{.03 + 0}}{\rm{.00}}\\{\rm{ = 0}}{\rm{.17}}{\rm{.}}\end{aligned}\)

And for the second part the following is true

\(\begin{aligned}{l}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}} \ge {\rm{4}}} \right){\rm{ = p(1,3) + p(2,2) + p(3,1) + p(4,0) + p(4,1) + p(4,2) + p(4,3) + p(3,2) + p(3,3) + p(2,3)}}\\{\rm{ = 0}}{\rm{.46}}\end{aligned}\)

All values can be found at the given table (see \({\rm{(a)}}\)).

Therefore , the probability is \({\rm{ = 0}}{\rm{.46}}\).

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 91Ó°ÊÓ!

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

Carry out a simulation experiment using a statistical computer package or other software to study the sampling distribution of \({\rm{\bar X}}\) when the population distribution is Weibull with \({\rm{\alpha = 2}}\) and\({\rm{\beta = 5}}\), as in Example\({\rm{5}}{\rm{.20}}\).[A1] Consider the four sample sizes, and\({\rm{30}}\), and in each case use \({\rm{1000}}\) replications. For which of these sample sizes does the \({\rm{\bar X}}\) sampling distribution appear to be approximately normal?

Suppose that when the pH of a certain chemical compound is\(5.00\), the pH measured by a randomly selected beginning chemistry student is a random variable with a mean of\(5.00\)and a standard deviation .2. A large batch of the compound is subdivided and a sample is given to each student in a morning lab and each student in an afternoon lab. Let\(X = \)the average pH as determined by the morning students and\(Y = \)the average pH as determined by the afternoon students.

a. If pH is a normal variable and there are\(25\)students in each lab, compute\(P\left( { - .1 \le X - Y \le - .1} \right)\)

b. If there are\(36\)students in each lab, but pH determinations are not assumed normal, calculate (approximately)\(P\left( { - .1 \le X - Y \le - .1} \right)\).

a. Compute the covariance between \({\rm{X}}\) and\({\rm{Y}}\).

b. Compute the correlation coefficient \({\rm{\rho }}\) for this \({\rm{X}}\) and \({\rm{Y}}\).

A health-food store stocks two different brands of a certain type of grain. Let \(X = \)the amount (lb) of brand A on hand and \(Y = \)the amount of brand B on hand. Suppose the joint pdf of X and Y is

\(f(x,y) = \left\{ {\begin{array}{*{20}{c}}{kxy\;\;\;\;\;\;\;x \ge 0,\;y \ge 0,\;20 \le x + y \le 30}\\{0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise}\end{array}} \right\}\)

a. Draw the region of positive density and determine the value of k.

b. Are X and Y independent? Answer by first deriving the marginal pdf of each variable.

c. Compute \(P\left( {X + Y \le 25} \right)\).

d. What is the expected total amount of this grain on hand?

e. Compute \(Cov\left( {X, Y} \right)\)and\(Corr\left( {X, Y} \right)\).

f. What is the variance of the total amount of grain on hand?

A company maintains three offices in a certain region, each staffed by two employees. Information concerning yearly salaries (\({\rm{1000}}\)s of dollars) is as follows:

\(\begin{array}{*{20}{c}}{{\rm{ Office }}}&{\rm{1}}&{\rm{1}}&{\rm{2}}&{\rm{2}}&{\rm{3}}&{\rm{3}}\\{{\rm{ Employee }}}&{\rm{1}}&{\rm{2}}&{\rm{3}}&{\rm{4}}&{\rm{5}}&{\rm{6}}\\{{\rm{ Salary }}}&{{\rm{29}}{\rm{.7}}}&{{\rm{33}}{\rm{.6}}}&{{\rm{30}}{\rm{.2}}}&{{\rm{33}}{\rm{.6}}}&{{\rm{25}}{\rm{.8}}}&{{\rm{29}}{\rm{.7}}}\end{array}\)

a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary\({\rm{\bar X}}\).

b. Suppose one of the three offices is randomly selected. Let\({{\rm{X}}_{\rm{1}}}\)and\({{\rm{X}}_{\rm{2}}}\)denote the salaries of the two employees. Determine the sampling distribution of\({\rm{\bar X}}\).

c. How does \({\rm{E(\bar X)}}\) from parts (a) and (b) compare to the population mean salary\({\rm{\mu }}\)?

See all solutions

Recommended explanations on Math Textbooks

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.