/*! 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 105 In which of the following situat... [FREE SOLUTION] | 91Ó°ÊÓ

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In which of the following situations would it be appropriate to use a Normal distribution to approximate probabilities for a binomial distribution with the given values of \(n\) and \(p ?\) (a) \(n=10, p=0.5\) (b) \(n=40, p=0.88\) (c) \(n=100, p=0.2\) (d) \(n=100, p=0.99\) (c) \(n=1000, p=0.003\)

Short Answer

Expert verified
Scenarios (a) and (c) are appropriate for normal approximation.

Step by step solution

01

Identify conditions for normal approximation

The normal distribution can approximate a binomial distribution when both \(np > 5\) and \(n(1-p) > 5\). This ensures that both the expected number of successes and failures are sufficiently large.
02

Calculate conditions for each option

Evaluate \(np\) and \(n(1-p)\) for each scenario:(a) For \(n=10, p=0.5\), \(np = 5\) and \(n(1-p) = 5\).(b) For \(n=40, p=0.88\), \(np = 35.2\) and \(n(1-p) = 4.8\).(c) For \(n=100, p=0.2\), \(np = 20\) and \(n(1-p) = 80\).(d) For \(n=100, p=0.99\), \(np = 99\) and \(n(1-p) = 1\).(e) For \(n=1000, p=0.003\), \(np = 3\) and \(n(1-p) = 997\).
03

Evaluate which conditions are met

Compare the values obtained in the previous step to the threshold of 5 for both \(np\) and \(n(1-p)\):(a) Meets both conditions, as both are equal to 5.(b) Does not meet the condition for \(n(1-p)\) (4.8 < 5).(c) Meets both conditions, as both 20 and 80 are greater than 5.(d) Does not meet the condition for \(n(1-p)\) (1 < 5).(e) Does not meet the condition for \(np\) (3 < 5).
04

Conclusion regarding normal approximation

Only scenarios (a) and (c) meet the conditions for using a normal distribution to approximate the binomial distribution because both the expected number of successes and failures are greater than 5 in these cases.

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Key Concepts

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

Binomial Distribution
A binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials. Each trial has two possible outcomes: a success or a failure. The probability of success, denoted as \( p \), remains constant across trials. This distribution is complete when the number of trials, \( n \), and \( p \) are defined ahead of time.

Key characteristics of a binomial distribution include:
  • Fixed number of trials \( n \)
  • Constant probability \( p \) of success
  • Independence of trials
The probability of obtaining exactly \( k \) successes in \( n \) trials is given by the formula: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Where \( \binom{n}{k} \) is the binomial coefficient, which calculates the number of ways to choose \( k \) successes from \( n \) trials.
Normal Distribution
A normal distribution is a continuous probability distribution that's often used in statistics due to its characteristic bell shape. This distribution is symmetric about its mean, indicating that data near the mean are more frequent in occurrence than data far from the mean.

Characteristics of a normal distribution include:
  • Bell-shaped and symmetric about the mean
  • Defined by its mean (\( \mu \)) and standard deviation (\( \sigma \))
  • 68-95-99.7 rule, meaning ~68% of data falls within one standard deviation of the mean, ~95% within two, and ~99.7% within three
The probability density function of a normal distribution is given by:\[P(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x - \mu)^2}{2\sigma^2}} \]Normal distributions are utilized in various statistical techniques, including z-tests and the central limit theorem, which states that, under certain conditions, the sum of a large number of random variables will be approximately normally distributed.
Statistics Education
Statistics education focuses on teaching students how to collect, analyze, interpret, and present data. It involves understanding and applying statistical methods to real-world scenarios. One core concept taught is the ability to differentiate between different types of distributions like the binomial and normal distributions. Understanding when and how to apply the normal approximation to a binomial distribution is a significant aspect of statistics education. In educational settings, students learn to:

  • Identify conditions that warrant using different distributions
  • Calculate probabilities using statistical formulas
  • Interpret results within the context of a given problem
Teaching often includes practical exercises to apply theoretical concepts, enhancing the student's ability to utilize statistics in problem-solving effectively. This type of education is vital for making informed decisions based on data-driven insights in various fields such as business, science, and social sciences.
Probability Approximation
The probability approximation connects how different statistical distributions can be used interchangeably under certain conditions. When working with the binomial distribution, it is sometimes approximated by a normal distribution. This is particularly beneficial when the number of trials \( n \) is large, as calculating binomial probabilities directly could be cumbersome.

The criteria for using a normal approximation for a binomial distribution include having both \( np > 5 \) and \( n(1-p) > 5 \). These conditions ensure that the distribution of outcomes is sufficiently symmetric and spread out to resemble a normal distribution. The resulting approximation allows easier calculation of probabilities using the standard normal distribution table or computational tools.

This approach is pivotal in simplifying complex probability problems and is a critical concept in applied statistics, enabling smoother transitions across different types of data analysis methods.

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Most popular questions from this chapter

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