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The patient recovery time from a particular surgical procedure is normally distributed with a mean of 3 days and a standard deviation of 1.7 days. Let X be the recovery time for a randomly selected patient. Round all answers to 4 decimal places where possible.a. What is the distribution of X? X ~ N(,)b. What is the median recovery time? daysc. What is the Z-score for a patient that took 4.1 days to recover? d. What is the probability of spending more than 2.4 days in recovery? e. What is the probability of spending between 2.7 and 3.4 days in recovery? f. The 80th percentile for recovery times is days.

Respuesta :

Given

mean = 3 days

standard deviation = 1.7 days

Find

a. What is the distribution of X?

b. What is the median recovery time?

c. What is the Z-score for a patient that took 4.1 days to recover?

d. What is the probability of spending more than 2.4 days in recovery?

e. What is the probability of spending between 2.7 and 3.4 days in recovery?

f. The 80th percentile for recovery times

Explanation

a) Distribution of X is given by X ~ N( 3 , 1.7)

b) for the normal distibution ,the median is the same as the mean .

so , the median recovery time is 3 days

c) z - score for the patient that took 4.1 days to recover is

[tex]\begin{gathered} z=\frac{X-\mu}{\sigma} \\ \\ z=\frac{4.1-3}{1.7} \\ \\ z=0.64705882352\approx0.6471 \end{gathered}[/tex]

d) probability of spending more than 2.4 days in recovery

[tex]\begin{gathered} P(X>2.4)=P(\frac{X-\mu}{\sigma}>\frac{2.4-3}{1.7}) \\ \\ P(X>2.4)=P(Z>-0.3529) \\ P(X>2.4)=P(Z<0.3529) \\ \\ P(X>2.4)=0.6379 \end{gathered}[/tex]

e) probability of spending between 2.7 and 3.4 days in recovery

[tex]\begin{gathered} P(2.7f) 80th percentile for recovery times = [tex]\begin{gathered} P(XFinal Answer

Hence , the above are the required answers.

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