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. Round the final answers to four decimal places and intermediate Z value calculations to two decimal places.

Suppose the benefits are normally distributed with a standard deviation of $460. 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 21 patients is less than $3920. P(x < 3920) -

Part 2 of 2

Find the probability that the mean benefit for a random sample of 21 patients is more than $4120. P(x > 4120)

Solution

Since the question is talking about the z-value, it implies $4064 is the population mean:

Part 1) We determine the z score for $3920, using

[tex]z=\frac{\bar X-\mu}{ \frac{ \sigma}{ \sqrt{n} } }

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We substitute to get;

[tex]z=\frac{3920-4064}{ \frac{ 406}{ \sqrt{21} } } = - 1.63

[/tex]

We read -1.63 under the z - distribution table .

The area corresponding to this -1.63 is 0.0516.

P(x < 3920)=0.0516

Part 2) Again, we need to find the z-score of $4120

[tex]z=\frac{4120-4064}{ \frac{ 406}{ \sqrt{21} } } = 0.63

[/tex]

We read 0.63 under the normal distribution table.

This corresponds to an area of 0.7357

This is the area to the left of 0.63

But the question demands the mean benefit is more than $4120

Hence P(x > 4120)=1-0.7357=0.2643