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Chapter 6: Problem 20

Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us. (a) Suppose \(n=33\) and \(p=0.21\). Can we approximate the \(\hat{p}\) distribution by a normal distribution? Why? What are the values of \(\mu_{\hat{p}}\) and \(\sigma_{\hat{p}} ?\) (b) Suppose \(n=25\) and \(p=0.15 .\) Can we safely approximate the \(\hat{p}\) distribution by a normal distribution? Why or why not? (c) Suppose \(n=48\) and \(p=0.15 .\) Can we approximate the \(\hat{p}\) distribution by a normal distribution? Why? What are the values of \(\mu_{\hat{p}}\) and \(\sigma_{\hat{p}} ?\)

Short Answer

Expert verified
(a) Yes, values: \(\mu_{\hat{p}} = 0.21\), \(\sigma_{\hat{p}} \approx 0.0735\); (b) No; (c) Yes, values: \(\mu_{\hat{p}} = 0.15\), \(\sigma_{\hat{p}} \approx 0.0523\).

Step by step solution

01

Understanding the Binomial Approximation Requirement

A binomial distribution can be approximated by a normal distribution if both the expected number of successes \( np \) and the expected number of failures \( n(1-p) \) are greater than 5. This rule of thumb ensures that the distribution is sufficiently symmetric for the normal approximation to be reasonable.
02

Calculating for Part (a)

For part (a), we have \( n = 33 \) and \( p = 0.21 \). We calculate \( np = 33 \times 0.21 = 6.93 \) and \( n(1 - p) = 33 \times 0.79 = 26.07 \). Both values are greater than 5, so we can use the normal distribution to approximate \( \hat{p} \). The mean \( \mu_{\hat{p}} = p = 0.21 \) and the standard deviation \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.21 \times 0.79}{33}} \approx 0.0735 \).
03

Evaluating Part (b) Conditions

For part (b), we have \( n = 25 \) and \( p = 0.15 \). We calculate \( np = 25 \times 0.15 = 3.75 \) and \( n(1 - p) = 25 \times 0.85 = 21.25 \). Here, \( np \) is less than 5, which implies that the normal approximation is not valid because the expected number of successes is too low.
04

Assessing Feasibility for Part (c)

For part (c), we use \( n = 48 \) and \( p = 0.15 \). We find \( np = 48 \times 0.15 = 7.2 \) and \( n(1 - p) = 48 \times 0.85 = 40.8 \). Since both values exceed 5, a normal approximation is valid. The mean \( \mu_{\hat{p}} = p = 0.15 \) and the standard deviation \( \sigma_{\hat{p}} = \sqrt{\frac{0.15 \times 0.85}{48}} \approx 0.0523 \).

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Key Concepts

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

Normal Approximation
In statistics, the normal approximation is a method used to simplify the interpretation of a binomial distribution. The binomial distribution shows the probability of achieving a certain number of successes in a sequence of independent experiments. However, calculating probabilities using the binomial formula can be quite complex, particularly with a large number of trials.
Normal approximation comes to the rescue by allowing the binomial distribution to be approximated using a normal distribution, which is easier to work with. This approximation can be applied when both the expected number of successes \(np\) and the expected number of failures \(n(1-p)\) are greater than 5.
This rule of thumb ensures that the distribution is nearly symmetric, making the normal curve a reasonable approximation. In practical scenarios, this means fewer computations and quicker analysis, often critical in experimental findings.
Expected Number of Successes
The expected number of successes, in a binomial distribution, is a straightforward concept that refers to the average number of successful outcomes you'd anticipate over a large number of trials. It is calculated by multiplying the number of trials \(n\) by the probability of success in each trial \(p\), represented by the formula \( np \).
For instance, if you're flipping a coin 10 times, and the probability of getting heads (success) is 0.5, you would expect to get heads about 5 times, since \( np = 10 \times 0.5 = 5 \). This formula is pivotal because it determines if a normal approximation can be used.
If the expected number of successes is greater than 5 (and the same is true for failures), then the distribution can likely be simplified using the normal approximation, facilitating easier computation of probabilities.
Standard Deviation
The standard deviation in the context of a binomial distribution describes how much variation or "spread" there is from the expected number of successes. A higher standard deviation means more variation in outcomes.
It is calculated using the formula \( \sigma = \sqrt{np(1-p)} \), where \(np\) is the expected number of successes and \(p\) is the probability of success. This formula takes into account both the proportion of successes and failures, providing a measure of variability around the predicted mean.
Understanding standard deviation helps in interpreting the distribution's spread, predicting the variability in repeated experiments, and assessing whether the normal approximation is appropriate.
Mean of Distribution
The mean of a binomial distribution, also known as the expected value, is crucial as it represents the average result expected from a series of trials. In a binomial setting, this mean is denoted as \( \mu = np \), where \(n\) is the number of trials and \(p\) is the probability of success on each trial.
For example, in an experiment with 20 trials and a success probability of 0.3, the mean \( \mu \) would be \( 20 \times 0.3 = 6 \). This indicates that, on average, you'd expect 6 successful outcomes.
The mean is vitally important because it anchors the distribution, providing a central value that tells us where most of the probabilities are clustered, especially when considering the binomial distribution鈥檚 potential normal approximation.

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Most popular questions from this chapter

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