Question

Question Help The standard deviation of the lengths of hospital stay on the intervention ward is 78 days. Come parts through (c) below a. For the variable length of hospital stay." determine the sampling distribution of the sample mear for samples of patients The standard deviation of the sample mean is days. (Round to four decimal places as needed.) b. The distribution of the length of hospital stayis right skewed Does this invalidate your result in part (a? Explain your answer O A. No, because it is normally distributed, then x must be normally distributed OB. No, because the sample sizes are sufficiently large that will be approximately normally distributed, regardless of the distribution of OC. Yes, because it is only normally distributed it is normally distributed OD. Yos, because the sample sizes are not sufficiently large that will be approximately normally distributed, regardless of the distribution of c. Obtain the probability that the sampling error made in the population in length of stay by the mean length of stay of a sample of patients will be at most 2 days The probability is approximately (Round to three decimal places as needed)

Answer 1

Facts : 1) IfX, Z be the normal and standard normal distibuted random variables respectively. To find the probabilites for X we convertX to Z by using Z = X-M Where u = mean(X) o = standard deviation(X) o X = The length (in days) at hospital. Given that X follows an unknown distribution with standard deviation(o) = 7.8. a) Sample size = n = 68 7.8 = Ox 0.9459 Vn 768 .: 0x = 0.9459

b) X is sample mean(Sample average) random variable. According to central limit theorem, X follows normal distribution irrespective of the distribution of X if sample size (n) > 30 In our problem, sample size = n = 68 > 30 Therefore, sample mean follows normal distribution. .. Option(B) is the answer. c) Now we have to find P(X - Mg <2). (X-4X 2 P( - < 2) = P OX 0.9459 = P(Z < 2.11) = 0.9826 :: P(X - ug < 2) = 0.9826

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