The standard deviation of the lengths of hospital stay on the intervention ward is 6.4days. Complete parts(a) through(c) below.

a. For the variable"length of hospitalstay," determine the sampling distribution of the sample mean for samples of 82 patients.

The standard deviation of the sample mean is x=

0.7068

0.7068 days.

(Round to four decimal places asneeded.)

b. The distribution of the length of hospital stay is right skewed. Does this invalidate your result in part(a)? Explain your answer.

A.

No, because the sample sizes are sufficiently large that x will be approximately normallydistributed, regardless of the distribution of x.

Your answer is correct.

B.

Yes, because the sample sizes are not sufficiently large that x will be approximately normallydistributed, regardless of the distribution of x.

C.

Yes, because x is only normally distributed if x is normally distributed.

D.

No, because if x is normallydistributed, then x must be normally distributed.

c. Obtain the probability that the sampling error made in estimating the population mean length of stay by the mean length of stay of a sample of 82 patients will be at most 2 days.

The probability is approximately

nothing

.

(Round to three decimal places asneeded.)