Question

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. use the standard normal distribution table. round the final answers to at least four decimal places and intermediate z - value calculations to two decimal places. part 1 of 2 find the probability that the mean benefit for a random sample of 29 patients is less than $3880. $p(overline{x} < 3880) = square$ part 2 of 2 find the probability that the mean benefit for a random sample of 29 patients is more than $4290. $p(overline{x} > 4290) = square$

Explanation:

Step1: Identify parameters for Part 1

Population mean $\mu = 4064$, standard deviation $\sigma = 460$, sample size $n = 29$, sample mean $\bar{x} = 3880$.

Step2: Calculate standard error for Part1

Standard error $SE = \frac{\sigma}{\sqrt{n}} = \frac{460}{\sqrt{29}} \approx 85.42$.

Step3: Compute z-score for Part1

$z = \frac{\bar{x} - \mu}{SE} = \frac{3880 - 4064}{85.42} \approx -2.15$.

Step4: Find probability for Part1

From z-table, $P(Z < -2.15) = 0.0158$.

Step5: Calculate z-score for Part2

Sample mean $\bar{x} = 4290$, $z = \frac{4290 - 4064}{85.42} \approx 2.65$.

Step6: Find probability for Part2

$P(Z > 2.65) = 1 - P(Z < 2.65) = 1 - 0.9960 = 0.0040$.

Answer:

Part1: 0.0158
Part2: 0.0040