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In 2014, approximately 13% of nonelderly Americans adults had no health insurance. Suppose that a random sample of 400 such individuals was drawn. What is the probability that 15% or more had no health insurance?

User Breeden
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Final answer:

To determine the probability that in a sample of 400 nonelderly American adults 15% or more had no health insurance, a normal approximation to the binomial distribution is used. By calculating the mean and standard deviation of the distribution and then finding the z-score for 15%, we can use normal distribution tables or a calculator to find the desired probability.

Step-by-step explanation:

In 2014, approximately 13% of nonelderly American adults had no health insurance. When examining whether a sample of 400 such individuals would have 15% or more without health insurance, we are essentially looking at a probability question that can be solved using the concepts of binomial distribution or normal approximation to the binomial distribution, given the large sample size.

To find the probability that 15% or more of the sample had no health insurance, we would first find the mean (np) and standard deviation (sqrt(np(1-p))) of the distribution, where 'n' is the sample size and 'p' is the probability of success (in this case, the probability of an individual not having health insurance).

With n being 400 and p being 0.13, we would then calculate the z-score for 15% (0.15) to find out how many standard deviations away from the mean this percentage is. The probability can then be found using standard normal distribution tables or a calculator with normal distribution functions.

However, the specific calculation steps and resulting probability value are not provided here, as this answer is meant to explain generally how such a problem would be solved.

User Caster Troy
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