Question

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 at least four decimal places and intermediate z-value calculations to two decimal places.Find the probability that the mean benefit for a random sample of 27 patients is less than $3810.Find the probability that the mean benefit for a random sample of 27 patients is more than $4290.

Answer 1

Step 1:

In this question, we are given the following:

When σ Is Known

If the standard deviation, σ, is known, we can transform to an approximately standard normal variable, Z:

`\text{Z = }\frac{X_i-\mu}{(\frac{\sigma}{\sqrt[]{n}})}`

Step 2:

Now, we have that:

Find the probability that the mean benefit for a random sample of 27 patients is less than $3810.

`\begin{gathered} X_i=\text{ 3810} \\ \mu\text{ = 4064} \\ \sigma\text{ = 460} \\ n\text{ = 27} \end{gathered}`
`Z\text{ = }\frac{3810\text{ - 4064}}{\frac{460}{\sqrt[]{27}}}`
`\text{ Z= }(-254)/(88.52704128)`
`\begin{gathered} Z=-2.\text{ 869179816} \\ Z\text{ }\approx-\text{ 2. 87 ( 2 decimal places)} \end{gathered}`

Hence,

`P\text{ ( x < Z) = P ( x < -2.87 ) = ? }`

From the Probability table, P(x

`Z=\frac{4290\text{ - 4064}}{\frac{460}{\sqrt[]{27}}}`
`\begin{gathered} Z\text{ = }(226)/(88.52704128) \\ Z\text{ = }2.552892277 \\ Z\text{ }\approx\text{ 2.55 ( 2 decimal places)} \end{gathered}`
`\text{ P( x > Z ) = P ( x > 2. 55 ) = ?}`

From the Probability table, we have that:

Given Z = 2.55,

P(x>Z) = 0.0053861

`P\text{ ( x > Z ) = 0.0054 ( 4 decimal places)}`