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Following cardiac surgery, patients are encouraged to exercise regularly (assume that regular exercise is defined as exercising on 3 or more days per week). A physician suspects that patients exercise regularly immediately following cardiac surgery but tend to reduce, even stop exercising completely, over time. An investigation is planned to estimate the mean number of weeks that patients exercise regularly following cardiac surgery. Assume that the standard deviation (s.d) in the number of weeks cardiac patients exercise regularly following surgery is 6.3 weeks.

(a) If a sample of 40 cardiac patients is followed, and the number of weeks in which each patient exercised regularly is recorded, what is the probability that the sample mean will be no more than 1 week higher than the true mean?

Under central limit therom, 

standard error = s.d./sqrt(n) = 6.3/sqrt(40) = 0.99
X = true mean + 1
Z = (X - true mean) / 0.99 = (true mean + 1 = true mean) / 0.99 = 1/0.99

(b) Find the probability that the sample mean is at least two weeks less than the true mean for a sample of 40 cardiac patients.

X = true mean - 2
Z = (X - true mean) / 0.99 = (true mean -2 - true mean) / 0.99 = -2/0.99

(c) If the sample is increased to 100 cardiac patients, what is the probability that the sample mean will be no more than 1 week higher than the true mean?

same as question Can anyone help me check my answers to see whether is correct or not? Please correct and explain if wrong...

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  • $\begingroup$ I posted a hint below, but I agree that this question will get better attention at stats.SE. $\endgroup$
    – pharmine
    Commented Sep 30, 2011 at 9:51
  • $\begingroup$ There seem to be some notational issues and possibly worse: for example if -2/0.99 is supposed to be a probability, then it is wrong. Probabilities lie in the range $[0,1]$ $\endgroup$
    – Henry
    Commented Sep 30, 2011 at 10:27
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    $\begingroup$ I think the OP tried to calculate the $z$-score, but one needs to know how to utilize that $z$-score to obtain the desired probability... $\endgroup$
    – pharmine
    Commented Sep 30, 2011 at 10:38
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    $\begingroup$ The bizarre aspect of HW problems like this is their internal lack of logical consistency. How is it possible to get in a situation where you have to take a sample to estimate a mean when you already know the SD exactly? In reality, either you're doing a prospective power calculation or else the SD will have to be estimated too. That allows for several interpretations of (c) and could completely change its solution. (This makes it perhaps too difficult as an elementary homework problem, because it requires computation of a prediction interval using a nonstandard formula). $\endgroup$
    – whuber ♦
    Commented Sep 30, 2011 at 14:36

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Hint: If each patient's exercise record (the number of weeks he or she exercises regularly following surgery) follows a normal distribution $N(\mu,\sigma^2)$, the sample mean of 40 patients will follow $N(\mu, \sigma^2/40)$ (Why?). Here, $\mu$ is what you call the true mean, and $\sigma$ (not $\sigma^2$) is the standard deviation of $N(\mu,\sigma^2)$ (which is 6.3 weeks). Hope this helps.

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