The standard deviation of the lengths of hospital stay on the intervention ward is 7 7 days. Complete parts (a) through (c) below. b. The distribution of the length of hospital stay is right skewed. Does this invalidate your result in part (a)? Explain your answer Of A. No, because the sample size is sufficiently large so that x will be approximately normally distributed, regardless of the distribution of x. Q B. Yes, because the sample sizes are not sufficiently large that x will be approximately normally distributed regardless of the distribution of x. O C. Yes, because x is only normally distributed if x is normally distributed. 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 85 patients will be at most 2 days. The probability is approximately (Round to three decimal places as needed. )