There is a clinical trial in place. Let X be the number of patients, out of grou

Question

There is a clinical trial in place.

Let X be the number of patients, out of group size n, who present a successful recoevery under the treatment in question.

Assume patients recover independently of one another. Suppose also, that the probability that a particular patient recover is Uniformly distributed over

[0,1} that the probability of successful recovery is p.

a) What is the distribution of X/P=p?

b) Find expected number of patients, out of this group, who recover successfully.

c) Find variance of patients, out of this group, who recover successfully.

Explain all parts of the questions in detail please. Thank You.

Explanation / Answer

Part a

Answer:

We know that the sampling distribution of any sample statistics follows an approximate normal distribution. For the given scenario, the sampling distribution of proportion p = X/P or sample proportion will follow an approximate normal distribution.

Part b

Answer:

We have to find the expected number of patients who recover successfully. We are given the sample size or group size as n and the population proportion p or proportion of successful recovery.

We know, the mean or expected number of patients who recover successfully is given as below:

Mean = E(X) = n*p

Part c

Answer:

For the above scenario, the variance of patients who recover successful recovery is given as below:

Variance = 2 = n*p*q

Where, q = 1 – p

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.