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Newsweek (November 23, 1992) reported that \(40 \%\) of all U.S. employees participate in "self-insurance" health plans \((p=.40)\). a. In a random sample of 100 employees, what is the approximate probability that at least half of those in the sample participate in such a plan? b. Suppose you were told that at least 60 of the 100 employees in a sample from your state participated in such a plan. Would you think \(p=.40\) for your state? Explain.

Short Answer

Expert verified
For part a, the approximate probability that at least half of those in the sample participate in such a plan would be calculated using a binomial cumulative distribution function. For part b, if at least 60 of the 100 employees in a sample participated in such a plan, it might seem like \(p=0.60\) for your state which is much higher than the sample rate.

Step by step solution

01

Understanding binomial distribution

This is a binomial distribution problem, as we are looking at the number of 'successes' (in this case, employees participating in self-insurance health plans) in a given number of independent trials (in this case, a random sample of 100 employees). The number of trials (n) is 100, and the probability of success (p) is 0.40.
02

Calculate the probability for Part a

We need to find the probability that at least 50 employees participate in such a plan. This is equivalent to finding 1 - P(X < 50), where X is the number of employees who participate. Given that n = 100 and p = 0.40, we can calculate the probability with a binomial cumulative distribution function.
03

Apply the binomial distribution formula for Part b

Here, we are given that at least 60 out of 100 employees participated in such a plan. If 60 out of 100 people participated in the plan, the observed proportion is 0.60. This is significantly higher than the initially stated proportion of 0.40, which might suggest that the correct proportion for this state is different.

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Key Concepts

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

Binomial Cumulative Distribution Function
The binomial cumulative distribution function (CDF) is essential when dealing with binomial distribution problems, where we're interested in the probability of achieving a certain number of successes in a series of independent trials. For example, in the exercise about U.S. employees' participation in self-insurance health plans, we're asked to find the probability that at least half (50 out of 100) of the sampled employees are enrolled in such plans.

To calculate this, the binomial CDF sums the probabilities of all outcomes from 0 up to the desired threshold minus one (49 in this case). Essentially, it gives us the probability of observing up to a certain number of successes. Since we want at least 50, we subtract this cumulative probability from 1, as part of the exercise solution demonstrates. Mathematically, it's expressed as:
Probability of Success
The probability of success in a binomial distribution is denoted by 'p'. It remains constant across each trial and is crucial for computing the likelihood of various outcomes. In our exercise, 'success' is defined as an employee participating in a self-insurance health plan, with a given probability of success, here represented as 40%.

Understanding 'p' is critical since it directly influences the shape and spread of the binomial distribution. A probability of success that is high suggests a distribution skewed towards more successes, while a low 'p' indicates the opposite. It is the single-parameter that defines the expected number of successes in a fixed number of trials. When analyzing situations like the exercise example, if the observed proportion of successes deviates significantly from 'p', it might point to an underlying difference in the probability for the specific demographic or group under consideration.
Independent Trials Statistics
Independent trials are a cornerstone of binomial distribution. What it means is that the outcome of one trial does not influence or change the outcome of another. In the context of our health plan enrollment problem, this assumption would suggest that one employee's decision to participate in a self-insurance plan does not affect another's decision.

In statistics, the assumption of independence is critical when drawing inferences from data. If trials are not independent, the standard binomial model may not be applicable, and results can be misleading. This concept is beautifully illustrated in the exercise by highlighting that each employee's participation is an independent event from another’s, making the binomial model appropriate for calculating probabilities like those in the given scenario.

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Most popular questions from this chapter

Let \(x\) denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of \(x\) are \(\mu=\) 2 minutes and \(\sigma=0.8\) minute, respectively. a. If \(\bar{x}\) is the sample mean time for a random sample of \(n=9\) students, where is the \(\bar{x}\) distribution centered, and how much does it spread out about the center (as described by its standard deviation)? b. Repeat Part (a) for a sample of size of \(n=20\) and again for a sample of size \(n=100\). How do the centers and spreads of the three \(\bar{x}\) distributions compare to one another? Which sample size would be most likely to result in an \(\bar{x}\) value close to \(\mu\), and why?

A random sample is selected from a population with mean \(\mu=100\) and standard deviation \(\sigma=10\). Determine the mean and standard deviation of the \(\bar{x}\) sampling distribution for each of the following sample sizes: a. \(n=9\) b. \(n=15\) c. \(n=36\) d. \(n=50\) e. \(n=100\) f. \(\quad n=400\)

In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. In addition, there is a weight limit of 2500 pounds. Assume that the average weight of students, faculty, and staff on campus is 150 pounds, that the standard deviation is 27 pounds, and that the distribution of weights of individuals on campus is approximately normal. If a random sample of 16 persons from the campus is to be taken: a. What is the expected value of the distribution of the sample mean weight? b. What is the standard deviation of the sampling distribution of the sample mean weight? c. What average weights for a sample of 16 people will result in the total weight exceeding the weight limit of 2500 pounds? d. What is the chance that a random sample of 16 people will exceed the weight limit?

Suppose that \(20 \%\) of the subscribers of a cable television company watch the shopping channel at least once a week. The cable company is trying to decide whether to replace this channel with a new local station. A survey of 100 subscribers will be undertaken. The cable company has decided to keep the shopping channel if the sample proportion is greater than \(.25 .\) What is the approximate probability that the cable company will keep the shopping channel, even though the proportion of all subscribers who watch it is only \(.20\) ?

Explain the difference between \(\sigma\) and \(\sigma_{\bar{x}}\) and between \(\mu\) and \(\mu_{\bar{x}}^{-\text {. }}\)

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