/*! 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} Problem 73 Let \(x\) be a Poisson random va... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(x\) be a Poisson random variable. Using the Poisson probabilities table, write the probability distribution of \(x\) for each of the following. Find the mean, variance, and standard deviation for each of these probability distributions. Draw a graph for each of these probability distributions. a. \(\lambda=1.3\) b. \(\lambda=2.1\)

Short Answer

Expert verified
For \(\lambda = 1.3\), the probability distribution, mean, variance, and standard deviation are \(P(X=k) = \frac{e^{-1.3}1.3^k}{k!}\), \(E[X] = 1.3\), \(Var[X] = 1.3\), \(SD[X] = \sqrt{1.3}\) respectively. For \(\lambda = 2.1\), they are \(P(X=k) = \frac{e^{-2.1}2.1^k}{k!}\), \(E[X] = 2.1\), \(Var[X] = 2.1\), \(SD[X] = \sqrt{2.1}\) respectively.

Step by step solution

01

Define Poisson Distribution

Recall that the probability mass function of a Poisson random variable X is given by \(P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}\) where \(k\) is the number of occurrences of an event, and \(\lambda = E[X]\) (the expected value or 'average rate of value') is the parameter defining the distribution.
02

Definitions for lambda = 1.3

When \(\lambda = 1.3\), the probability distribution, mean, variance, and standard deviation are:Probability distribution: \(P(X=k) = \frac{e^{-1.3}1.3^k}{k!}\)Mean: \(\lambda = E[X] = 1.3\)Variance: \(\lambda = Var[X] = 1.3\)Standard Deviation: \(SD[X] = \sqrt{Var[X]} = \sqrt{1.3}\)Hence the probability distribution is dependent on \(k\), the number of occurrences of an event.
03

Definitions for lambda = 2.1

When \(\lambda = 2.1\), the probability distribution, mean, variance, and standard deviation are:Probability Distribution: \(P(X=k) = \frac{e^{-2.1}2.1^k}{k!}\)Mean: \(\lambda = E[X] = 2.1\)Variance: \(\lambda = Var[X] = 2.1\)Standard Deviation: \(SD[X] = \sqrt{Var[X]} = \sqrt{2.1}\)Hence the probability distribution is dependent on \(k\), the number of occurrences of an event.
04

Draw Graphs

To draw the probability distribution graphs, one must plot the probabilities \(P(X=k)\) against \(k\) for each \(\lambda\). The details of graph plotting are better understood through direct visual interpretation rather than through text instructions.

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

Which of the following are binomial experiments? Explain why. a. Drawing 3 balls with replacement from a box that contains 10 balls, 6 of which are red and 4 are blue, and observing the colors of the drawn balls b. Drawing 3 balls without replacement from a box that contains 10 balls, 6 of which are red and 4 are blue, and observing the colors of the drawn balls c. Selecting a few households from New York City and observing whether or not they own stocks when it is known that \(28 \%\) of all households in New York City own stocks

Let \(N=16, r=10\), and \(n=5\). Using the hypergeometric probability distribution formula, find a. \(P(x=5)\) b. \(P(x=0)\) c. \(P(x \leq 1)\)

Despite all efforts by the quality control department, the fabric made at Benton Corporation always contains a few defects. A certain type of fabric made at this corporation contains an average of \(.5\) defect per 500 yards. a. Using the Poisson formula, find the probability that a given piece of 500 yards of this fabric will contain exactly 1 defect. b. Using the Poisson probabilities table, find the probability that the number of defects in a given 500-yard piece of this fabric will be i. 2 to 4 ii. more \(\underline{\text { than }} 3\) iii. less than 2

The binomial probability distribution is symmetric for \(p=.50\), skewed to the right for \(p<.50\), and skewed to the left for \(p>.50\). Illustrate each of these three cases by writing a probability distribution table and drawing a graph. Choose any values of \(n\) (equal to 4 or higher) and \(p\) and use the table of binomial probabilities (Table I of Appendix \(\mathrm{C}\) ) to write the probability distribution tables.

An office supply company conducted a survey before marketing a new paper shredder designed for home use. In the survey, \(80 \%\) of the people who used the shredder were satisfied with it. Because of this high acceptance rate, the company decided to market the new shredder. Assume that \(80 \%\) of all people who will use it will be satisfied. On a certain day, seven customers bought this shredder. a. Let \(x\) denote the number of customers in this sample of seven who will be satisfied with this shredder. Using the binomial probabilities table (Table I, Appendix C), obtain the probability distribution of \(x\) and draw a graph of the probability distribution. Find the mean and standard deviation of \(x\). b. Using the probability distribution of part a, find the probability that exactly four of the seven customers will be satisfied.

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.