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he standard deviation of the lengths of hospital stay on the intervention ward is 7.3 days. Complete parts (a) through (c) below. . For the variable "length of hospital stay," determine the sampling distribution of the sample mean for samples of 85 patients. he standard deviation of the sample mean is o; = days. Round to four decimal places as needed.) . The distribution of the length of hospital stay is right skewed. Does this invalidate your result in part (a)? Explain your answer. OA. No, because if x is normally distributed, then x must be normally distributed. O B. Yes, because x is only normally distributed if x is normally distributed. O C. Yes, because the sample sizes are not sufficiently large that x will be approximately normally distributed, regardless of the distribution of x. O D. No, because the sample sizes are sufficiently large that x will be approximately normally distributed, regardless of the distribution of x. . 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. he probability is approximately. Round to three decimal places as needed.)

he standard deviation of the lengths of hospital stay on the intervention ward is 7.3 days. Complete parts (a) through (c) below. . For the variable "length of hospital stay," determine the sampling distribution of the sample mean for samples of 85 patients. he standard deviation of the sample mean is o; = days. Round to four decimal places as needed.) . The distribution of the length of hospital stay is right skewed. Does this invalidate your result in part (a)? Explain your answer. OA. No, because if x is normally distributed, then x must be normally distributed. O B. Yes, because x is only normally distributed if x is normally distributed. O C. Yes, because the sample sizes are not sufficiently large that x will be approximately normally distributed, regardless of the distribution of x. O D. No, because the sample sizes are sufficiently large that x will be approximately normally distributed, regardless of the distribution of x. . 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. he probability is approximately. Round to three decimal places as needed.)

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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The standard deviation of the lengths of hospital stay on the intervention ward is 7.3 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 85 patients. The standard deviation of the sample mean is \( \sigma_{\bar{x}} = \square \) 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. - ⬜ A. No, because if \(\bar{x}\) is normally distributed, then x must be normally distributed. - ⬜ B. Yes, because \(\bar{x}\) is only normally distributed if x is normally distributed. - ⬜ C. Yes, because the sample sizes are not sufficiently large that \(\bar{x}\) will be approximately normally distributed, regardless of the distribution of x. - ⬜ D. No, because the sample sizes are sufficiently large that \(\bar{x}\) 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 85 patients will be at most 2 days. The probability is approximately \(\square\). (Round to three decimal places as needed.)
Transcribed Image Text:The standard deviation of the lengths of hospital stay on the intervention ward is 7.3 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 85 patients. The standard deviation of the sample mean is \( \sigma_{\bar{x}} = \square \) 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. - ⬜ A. No, because if \(\bar{x}\) is normally distributed, then x must be normally distributed. - ⬜ B. Yes, because \(\bar{x}\) is only normally distributed if x is normally distributed. - ⬜ C. Yes, because the sample sizes are not sufficiently large that \(\bar{x}\) will be approximately normally distributed, regardless of the distribution of x. - ⬜ D. No, because the sample sizes are sufficiently large that \(\bar{x}\) 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 85 patients will be at most 2 days. The probability is approximately \(\square\). (Round to three decimal places as needed.)
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