Answer :
SOLUTION:
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:
[tex]\text{Z = }\frac{X_i-\mu}{(\frac{\sigma}{\sqrt[]{n}})}[/tex]Step 2:
Now, we have that:
Find the probability that the mean benefit for a random sample of 27 patients is less than $3810.
[tex]\begin{gathered} X_i=\text{ 3810} \\ \mu\text{ = 4064} \\ \sigma\text{ = 460} \\ n\text{ = 27} \end{gathered}[/tex][tex]Z\text{ = }\frac{3810\text{ - 4064}}{\frac{460}{\sqrt[]{27}}}[/tex][tex]\text{ Z= }\frac{-254}{88.52704128}[/tex][tex]\begin{gathered} Z=-2.\text{ 869179816} \\ Z\text{ }\approx-\text{ 2. 87 ( 2 decimal places)} \end{gathered}[/tex]Hence,
[tex]P\text{ ( x < Z) = P ( x < -2.87 ) = ? }[/tex]From the Probability table, P(x[tex]P(x
Step 3:
Find the probability that the mean benefit for a random sample of 27 patients is more than $4290.
[tex]\begin{gathered} X_i=\text{ 4290} \\ \mu\text{ = 4064} \\ \sigma\text{ = 460} \\ \text{n = 27} \end{gathered}[/tex][tex]Z=\frac{4290\text{ - 4064}}{\frac{460}{\sqrt[]{27}}}[/tex][tex]\begin{gathered} Z\text{ = }\frac{226}{88.52704128} \\ Z\text{ = }2.552892277 \\ Z\text{ }\approx\text{ 2.55 ( 2 decimal places)} \end{gathered}[/tex][tex]\text{ P( x > Z ) = P ( x > 2. 55 ) = ?}[/tex]From the Probability table, we have that:
Given Z = 2.55,
P(x>Z) = 0.0053861
[tex]P\text{ ( x > Z ) = 0.0054 ( 4 decimal places)}[/tex]