/*! 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 situa... [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) \(\quad n=100, p=0.2\) (d) \(n=100, p=0.99\) (e) \(n=1000, p=0.003\)

Short Answer

Expert verified
Use Normal approximation for (a) and (c).

Step by step solution

01

Check Conditions for Normal Approximation

To use a Normal distribution to approximate a Binomial distribution, the sample size \(n\) and probability \(p\) must satisfy the conditions: \(np \geq 5\) and \(n(1-p) \geq 5\). These conditions ensure that both the number of expected successes and failures are sufficiently large for a Normal approximation to be valid.
02

Evaluate Each Option

For each option, check whether both conditions \(np \geq 5\) and \(n(1-p) \geq 5\) are satisfied:- (a) \(n=10, p=0.5\): \(np = 5\) and \(n(1-p) = 5\) - (b) \(n=40, p=0.88\): \(np = 35.2\) and \(n(1-p) = 4.8\)- (c) \(n=100, p=0.2\): \(np = 20\) and \(n(1-p) = 80\)- (d) \(n=100, p=0.99\): \(np = 99\) and \(n(1-p) = 1\)- (e) \(n=1000, p=0.003\): \(np = 3\) and \(n(1-p) = 997\)Check which options satisfy both conditions.
03

Determine Valid Situations

From the evaluations, identify which situations allow for normal approximation:- Option (a) is valid because both conditions are satisfied.- Option (b) is not valid because \(n(1-p) = 4.8 < 5\).- Option (c) is valid as both conditions are satisfied.- Option (d) is not valid because \(n(1-p) = 1 < 5\).- Option (e) is not valid because \(np = 3 < 5\).Thus, options (a) and (c) are situations where the Normal approximation is appropriate.

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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 common probability distribution. It's used to model scenarios where there are two possible outcomes for each trial. These outcomes are often termed as "success" and "failure". The distribution depends on two parameters: the number of trials, denoted as \( n \), and the probability of success for each trial, represented by \( p \). To imagine this in real life, consider flipping a coin \( n \) times. Each flip has a success probability \( p \) (e.g., heads), and the result is randomized and follows Binomial distribution characteristics.Binomial distribution graphs often show an initial steep peak that tapers off. The shape can vary significantly based on \( n \) and \( p \), and this variability makes it essential to have methods like Normal approximation for certain calculations.
Probability
Probability is a way of measuring how likely an event is to occur. In Binomial distribution problems, the probability of success \( p \) plays a crucial role. Each trial in a Binomial experiment is independent, and the outcome does not affect the probability of the next trial. This makes Binomial distribution quite predictable and suitable for probability modeling.For example, if you are rolling a die, the probability of rolling a four is the same each time: \( rac{1}{6} \). When dealing with large datasets or many trials, calculating exact probabilities can be complex. Here, tools like Normal approximation can simplify the process significantly by translating the problem into one solvable with Normal distribution techniques.
Sample Size Conditions
To apply a Normal approximation on a Binomial distribution, the sample size conditions need to be met. These conditions ensure a good similarity between Binomial and Normal distributions. The rules are:
  • \( np \geq 5 \)
  • \( n(1-p) \geq 5 \)
If both conditions hold, it implies enough trials contribute to both successes and failures, providing a more 'symmetrical' distribution that is closer to a Normal shape. If these conditions are not met, the probabilities calculated using Normal approximation may lead to significant errors. Consider a situation where \( n = 40 \) and \( p = 0.88 \). Not meeting \( n(1-p) \geq 5 \) makes it challenging to apply a Normal approximation reliably.
Expected Successes and Failures
When using a Binomial distribution, expected successes and failures play a significant role in understanding the distribution. Expected successes are calculated with \( np \), while expected failures are calculated with \( n(1-p) \). These calculations help us verify whether the sample size conditions for Normal approximation are satisfied. Knowing these numbers in advance can guide whether a Binomial distribution skews heavily towards one outcome or another, affecting the potential accuracy of any Normal approximation.For instance, if you are conducting 100 trials with a success probability of 0.2, your expected successes are \( 100 \times 0.2 = 20 \). Similarly, the expected number of failures is \( 100 \times 0.8 = 80 \). Both numbers satisfy the sample size condition, making the Normal approximation a valid method in such a case.

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