/*! 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 2 Consider two binomial distributi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider two binomial distributions, with \(n\) trials each. The first distribution has a higher probability of success on each trial than the second. How does the expected value of the first distribution compare to that of the second?

Short Answer

Expert verified
The expected value of the first distribution is greater than that of the second.

Step by step solution

01

Understand the parameters of a binomial distribution

A binomial distribution is defined by two parameters: the number of trials \(n\) and the probability of success \(p\). It is denoted as \(B(n, p)\). The expected value (or mean) of a binomial distribution is calculated using the formula \(E(X) = n \cdot p\), where \(X\) is a random variable representing the number of successes.
02

Define the parameters for the two distributions

Consider the first binomial distribution \(B(n, p_1)\) and the second binomial distribution \(B(n, p_2)\). The problem states that the probability of success for the first distribution \(p_1\) is greater than that of the second distribution \(p_2\). Thus, \(p_1 > p_2\).
03

Calculate the expected values of both distributions

For the first distribution \(B(n, p_1)\), the expected value \(E(X_1)\) is given by \(E(X_1) = n \cdot p_1\). For the second distribution \(B(n, p_2)\), the expected value \(E(X_2)\) is \(E(X_2) = n \cdot p_2\).
04

Compare the expected values

Since the probability of success \(p_1\) for the first distribution is greater than \(p_2\) for the second, this implies \(n \cdot p_1 > n \cdot p_2\). Therefore, the expected value of the first distribution \(E(X_1)\) is greater than the expected value of the second distribution \(E(X_2)\).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
The concept of the expected value is crucial when dealing with binomial distributions. In simpler terms, the expected value represents the average number of successes over many trials in a given distribution. For a binomial distribution, which is notated as \( B(n, p) \), the expected value is calculated by multiplying the total number of trials \( n \) by the probability of success \( p \). Thus, the formula is \( E(X) = n \cdot p \), where \( X \) denotes the random variable for the number of successes.

Understanding this helps us determine what to expect, on average, from our random process. So, if one distribution has a higher expected number due to a higher probability of success, it effectively means that on average, it will have more successes compared to a distribution with a lower probability.
Probability of Success
The probability of success \( p \) is a fundamental part of the binomial distribution. It quantifies the likelihood of a success in a single trial. For example, if you are flipping a coin, the probability of it landing on heads (if heads are considered success) is 0.5.

In comparing two distributions within the binomial framework, this probability plays a key role. If \( p_1 \) is the probability of success for the first distribution and \( p_2 \) for the second, and if \( p_1 > p_2 \), it automatically leads to the expected value of the first distribution being higher than the second. This directly stems from the relationship between probability and expected outcomes.

The higher the probability \( p \), the more likely it is to achieve a greater number of successes over a series of trials.
Comparing Distributions
When comparing two binomial distributions, it's helpful to pay attention to both the expected value and the probability of success. Using the exercise, compare the distributions based on the given information.

1. **Assess the probabilities**: Knowing that \( p_1 > p_2 \) informs us that each trial in the first distribution has a greater chance of success.

2. **Calculate the expected values**: With \( E(X_1) = n \cdot p_1 \) and \( E(X_2) = n \cdot p_2 \), if \( p_1 \) is greater, then naturally, \( E(X_1) > E(X_2) \).

So, in practical learning or when conducting experiments, if you want to maximize the number of expected successes, always aim for configurations with higher probability values. This clear understanding of the inequalities and how expected values respond to different probabilities enhances decision-making based on statistical data.

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

USA Today reported that about 20\% of all people in the United States are illiterate. Suppose you interview seven people at random off a city street. (a) Make a histogram showing the probability distribution of the number of illiterate people out of the seven people in the sample. (b) Find the mean and standard deviation of this probability distribution. Find the expected number of people in this sample who are illiterate. (c) How many people would you need to interview to be 98\% sure that at least seven of these people can read and write (are not illiterate)?

The Denver Post reported that, on average, a large shopping center has had an incident of shoplifting caught by security once every 3 hours. The shopping center is open from 10 A.M. to 9 P.M. (11 hours). Let \(r\) be the number of shoplifting incidents caught by security in the 11 -hour period during which the center is open. (a) Explain why the Poisson probability distribution would be a good choice for the random variable \(r\). What is \(\lambda ?\) (b) What is the probability that from 10 A.M. to 9 P.M. there will be at least one shoplifting incident caught by security? (c) What is the probability that from 10 A.M. to 9 P.M. there will be at least three shoplifting incidents caught by security? (d) What is the probability that from 10 A.M. to 9 P.M. there will be no shoplifting incidents caught by security?

USA Today reported that for all airlines, the number of lost bags was May: 6.02 per 1000 passengers December: 12.78 per 1000 passengers Note: A passenger could lose more than one bag. (a) Let \(r=\) number of bags lost per 1000 passengers in May. Explain why the Poisson distribution would be a good choice for the random variable \(r\) What is \(\lambda\) to the nearest tenth? (b) In the month of May, what is the probability that out of 1000 passengers, no bags are lost? that 3 or more bags are lost? that 6 or more bags are lost? (c) In the month of December, what is the probability that out of 1000 passengers, no bags are lost? that 6 or more bags are lost? that 12 or more bags are lost? (Round \(\lambda\) to the nearest whole number.)

In the western United States, there are many dry-land wheat farms that depend on winter snow and spring rain to produce good crops. About \(65 \%\) of the years, there is enough moisture to produce a good wheat crop, depending on the region (Reference: Agricultural Statistics, U.S. Department of Agriculture). (a) Let \(r\) be a random variable that represents the number of good wheat crops in \(n=8\) years. Suppose the Zimmer farm has reason to believe that at least 4 out of 8 years will be good. However, they need at least 6 good years out of 8 to survive financially. Compute the probability that the Zimmers will get at least 6 good years out of \(8,\) given what they believe is true; that is, compute \(P(6 \leq r | 4 \leq r) .\) See part (d) for a hint.

Jim is a real estate agent who sells large commercial buildings. Because his commission is so large on a single sale, he does not need to sell many buildings to make a good living. History shows that Jim has a record of selling an average of eight large commercial buildings every 275 days. (a) Explain why a Poisson probability distribution would be a good choice for \(r=\) number of buildings sold in a given time interval. (b) In a 60 -day period, what is the probability that Jim will make no sales? one sale? two or more sales? (c) In a 90 -day period, what is the probability that Jim will make no sales? two sales? three or more sales?

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.