Recent studies have shown that 20% of Americans fit the medical definition of obese. A random sample of 100 Americans is selected and the number of obese in the sample is determined. a. What is the sampling distribution of the sample proportion? Explain. b. What is the probability that the sample proportion is greater than 0.24?
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The sampling distribution of the sample proportion is a binomial distribution because the sample is drawn from a population and each individual in the sample can be classified as either obese or not obese. The mean of this distribution is equal to the population Show more…
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Recent studies have shown that 20% of Americans fit the medical definition of obese. A random sample of 100 Americans is selected and the number of obese in the sample is determined. What is the sampling distribution of the sample proportion? Can not be determined ^p ~ N(0.2, 0.04) ^p ~ N(10, 0.2)
Sri K.
According to the 2011 Health Interview Survey, 62% of American adults are either overweight or obese, based on their self-reported body mass index (BMI). Suppose that we take a random sample of 200 adults. Binomial Distribution X ~ Bin(n, p) n = 200 p = 0.62 x = 120 P(X ≥ x) = 0.74497 μ = E(X) = 124 σ = SD(X) = 6.864 σ² = Var(X) = 47.12 Above is shown the binomial distribution histogram that shows the probability of each value of x = 0, 1, .., 200. What is the probability of greater than or equal 120 from Normal distribution using the above information? 0.281 0.719 0.912 0.215
Apply Procedure 6.3 on page 289 to approximate the required binomial probabilities. Exercise. In the online $T I M E$ article "America's Health Checkup," A. Park reported that $40 \%$ of U.S. adults get no exercise. If 250 U.S. adults are selected at random, determine the probability that the number who get no exercise a. is exactly $40 \%$ of those sampled. b. exceeds $40 \%$ of those sampled. c. is fewer than $90 .$
The Normal Distribution
Normal Approximation to the Binomial Distribution
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