Mathematics, 24.01.2021 14:00 haileyw123

Medicare Hospital Insurance The average yearly Medicare Hospital Insurance benefit per person was $4064 in a recent year. Suppose the benefits are normally distributed with a standard deviation of $460. Assume that the sample is taken from a large population and the correction factor can be ignored. Use a TI-83 Plus/TI-84 Plus calculator. Round your answer to at least four decimal places. Find the probability that the mean benefit for a random sample of 20 patients is more than $4100.
P (X > 4100) =?

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Answer from: Aleah6645

0.3632 = 36.32% probability that the mean benefit for a random sample of 20 patients is more than $4100.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

Population: $\mu = 4064, \sigma = 460$

Sample of 20: $n = 20, s = \frac{460}{\sqrt{20}} = 102.86$

Find the probability that the mean benefit for a random sample of 20 patients is more than $4100.

This is 1 subtracted by the pvalue of Z when X = 4100. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{4100 - 4064}{102.86}$

Z = 0.35

Z = 0.35 has a pvalue of 0.6368

1 - 0.6368 = 0.3632

0.3632 = 36.32% probability that the mean benefit for a random sample of 20 patients is more than $4100.

Answer from: Quest

⊂⊃⊂⊃ω $\frac{x}{y}\left[\begin{array}{ccc}1& 2& 3\\4& 5& 6\\7& 8& 9\end{array}\right] \left[\begin{array}{ccc}1& 2& 3\\4& 5& 6\\7& 8& 9\end{array}\right] \\ x^{2} \pi \pi \left[\begin{array}{ccc}1& 2& 3\\4& 5& 6\\7& 8& 9\end{array}\right] \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \int\limits^a_b {x} \, dx x_{123} \left \{ {{y=2} \atop {x=2}} \right. \sqrt[n]{x} x_{123}$

step-by-step explanation:

i hope this : )

Answer from: Quest

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.

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