/*! 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 55 Let \(x\) be a discrete random v... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(x\) be a discrete random variable that possesses a binomial distribution. Using the binomial formula, find the following probabilities. a. \(P(x=0)\) for \(n=5\) and \(p=.05\) b. \(P(x=4)\) for \(n=7\) and \(p=.90\) c. \(P(x=7)\) for \(n=10\) and \(p=.60\) Verify your answers by using Table I of Appendix \(\mathrm{C}\).

Short Answer

Expert verified
The calculated probabilities for these scenarios will be numerical values as determined by the binomial probability formula.

Step by step solution

01

Calculate Probability for Scenario a

Using the provided values \(n = 5, p = 0.05\) and using the binomial formula, calculate the probability for \(P(x=0)\) which will be \(P(x=0) = C(5,0) * (0.05)^0 * (1-0.05)^{5-0}\)
02

Calculate Probability for Scenario b

Using the provided values \(n = 7, p = 0.90\) and using the binomial formula, calculate the probability for \(P(x=4)\) which will be \(P(x=4) = C(7,4) * (0.90)^4 * (1-0.90)^{7-4}\)
03

Calculate Probability for Scenario c

Using the provided values \(n = 10, p = 0.60\) and using the binomial formula, calculate the probability for \(P(x=7)\) which will be \(P(x=7) = C(10,7) * (0.60)^7 * (1-0.60)^{10-7}\)

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.

Discrete Random Variable
A discrete random variable is a fundamental concept in probability and statistics, representing a variable that can take on a finite or countably infinite number of distinct outcomes. Unlike continuous variables, which can take any value within a range, discrete random variables have specific, countable values.
For example, when flipping a coin, the possible outcomes are heads or tails. The number of heads in ten coin flips can be a discrete random variable because it can only be certain whole numbers from 0 to 10.
Discrete random variables often appear in scenarios like counting the number of successes (like heads in coin flip trials), making them especially relevant in binomial distributions, where outcomes per trial are limited to success or failure.
Probability Calculation
Probability calculation is a crucial part of understanding how likely an event is to occur under a specific set of circumstances. For discrete random variables, probabilities are calculated by determining the likelihood of each possible outcome and summing them up to 1. This process ensures all possible events are accounted for.
In the exercises provided, calculating the probability involves using specific formulas tailored for the type of distribution being analyzed.
Understanding probability is essential in real-world applications, such as predicting the chance of rain based on weather patterns or calculating the odds in games of chance. It provides a way to quantify uncertainty and make informed decisions based on data.
Binomial Formula
The binomial formula is a key tool in probability theory used to calculate probabilities in binomial distributions. The formula is expressed as:
\[ P(x=k) = C(n, k) p^k (1-p)^{n-k} \]
Where:
  • \(C(n, k)\) refers to the combination function, calculating how many ways \(k\) successes can occur in \(n\) trials.
  • \(p\) is the probability of success on a single trial.
  • \((1-p)\) is the probability of failure.

The binomial formula helps determine the likelihood of a particular number of successes in a series of independent trials, such as flipping a coin multiple times, where each trial has exactly two possible outcomes.
Combinatorial Calculations
Combinatorial calculations involve counting the number of different ways to arrange or combine elements of a set. In probability, they help determine the number of ways certain outcomes can occur. This is central in the binomial formula, where you need the number of combinations to find the probability of achieving a particular number of successes.
The combination formula \(C(n, k)\), also known as 'n choose k,' is used here:
\[C(n, k) = \frac{n!}{k!(n-k)!}\]
Where \(!\) denotes factorial, the product of an integer and all the integers below it.
For instance, to find \(C(5, 0)\), the number of ways to choose 0 successes out of 5 trials, you'd calculate:
\[ C(5, 0) = \frac{5!}{0!(5-0)!} = 1 \]
Combinations are crucial for efficiently counting possibilities and enabling accurate probability calculations within binomial distributions.

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

According to a Harris Interactive poll, \(52 \%\) of American college graduates have Facebook accounts (http://www.harrisinteractive.com/harris_poll/pubs/Harris_Poll \(200904_{-16}\).pdf). Suppose that this result. is true for the current population of American college graduates. a. Let \(x\) be a binomial random variable that denotes the number of American college graduates in a random sample of 15 who have Facebook accounts. What are the possible values that \(x\) can assume? b. Find the probability that exactly 9 American college graduates in a sample of 15 have Facebook accounts.

The number of calls that come into a small mail-order company follows a Poisson distribution. Currently, these calls are serviced by a single operator. The manager knows from past experience that an additional operator will be needed if the rate of calls exceeds 20 per hour. The manager observes that 9 calls came into the mail-order company during a randomly selected 15 -minute period. a. If the rate of calls is actually 20 per hour, what is the probability that 9 or more calls will come in during a given 15 -minute period? b. If the rate of calls is really 30 per hour, what is the probability that 9 or more calls will come in during a given 15 -minute period? c. Based on the calculations in parts a and \(b\), do you think that the rate of incoming calls is more likely to be 20 or 30 per hour? d. Would you advise the manager to hire a second operator? Explain.

A contractor has submitted bids on three state jobs: an office building, a theater, and a parking garage. State rules do not allow a contractor to be offered more than one of these jobs. If this contractor is awarded any of these jobs, the profits earned from these contracts are $$\$ 10$$ million from the office building, $$\$ 5$$ million from the theater, and $$\$ 2$$ million from the parking garage. His profit is zero if he gets no contract. The contractor estimates that the probabilities of getting the office building contract, the theater contract, the parking garage contract, or nothing are \(.15, .30, .45\), and 10, respectively. Let \(x\) be the random variable that represents the contractor's profits in millions of dollars. Write the probability distribution of \(x\). Find the mean and standard deviation of \(x\). Give a brief interpretation of the values of the mean and standard deviation.

In how many ways can a sample (without replacement) of 9 items be selected from a population of 20 items?

A household receives an average of \(1.7\) pieces of junk mail per day. Find the probability that this household will receive exactly 3 pieces of junk mail on a certain day. Use the Poisson probability distribution formula.

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.