The standard deviation of the lengths of hospital stay on the intervention ward is 7.9 days. Complete parts (a) through (c) below. . . . a. For the variable "length of hospital stay," determine the sampling distribution of the sample mean for samples of 77 patients. The sampling distribution will be approximately normal with mean u and standard deviation of = days. (Round to four decimal places as needed.) b. The distribution of the length of hospital stay is right skewed. Does this invalidate your result in part (a)? Explain your answer. O A. Yes, because the sample sizes are not sufficiently large that x will be approximately normally distributed, regardless of the distribution of x. O B. Yes, because x is only normally distributed if x is normally distributed. O C. No, because the sample size is sufficiently large so that x will be approximately normally distributed, regardless of the distribution of x. O D. No, because if x is normally distributed, 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 77 patients will be at most 2 days. The probability is approximately (Round to three decimal places as needed.)