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The standard deviation of the lengths of hospital stay on the intervention ward is 6.8 days. Complete parts (a) through (c) below. K a. For the variable "length of hospital stay," determine the sampling distribution of the sample mean for samples of 88 patients. The sampling distribution will be approximately normal with mean u and standard deviation a; = 0.7249 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 x is only normally distributed if x is normally distributed. B. No, because the sample size is sufficiently large so that x will be approximately normally distributed, regardless of the distribution of x. O C. No, because if x is normally distributed, then x must be normally distributed. O D. Yes, because the sample sizes are not sufficiently large that x will be approximately normally distributed, regardless of the distribution of x. 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 88 patients will be at most 2 days. The probability is approximately]. (Round to three decimal places as needed.)