The Length of Stay (LOS) for an acute patient in a metropolitan hospital
Questions 13 to 15 refer to the following information: The Length of Stay (LOS) for an acute patient in a metropolitan hospital has a mean (µ ) of 5.5 days and a
standard deviation (? ) of 2.5 days. It is known that the distribution of LOS is skewed. Assume that the number of admission (ie. LOS) records in the hospital database
is very large. Question 13 If 49 admission records are randomly selected from the hospital database, what is the probability that the average LOS (for the sample of
49) is greater than 6 days? (a) 0.1793 (b) 0.4207 (c) 0.0808 (d) 0.4192 (e) cannot be determined unless Length of Stay (LOS) is normally distributed Question 14 If 9
admission records are randomly selected from the hospital database, what is the probability that the average LOS (for the sample of 9) is greater than 6 days? (a)
0.1793 (b) 0.4207 (c) 0.2743 (d) 0.2257 (e) cannot be determined unless Length of Stay (LOS) is normally distributed Question 15 If 16 admission records are to be
randomly selected from the hospital database, what is the standard error of the LOS sample mean (for the sample of 16)? (a) 2.5 (b) 0.1563 (c) 1.375 (d) 0.625 (e)
cannot be determined as n is too small (n < 30)
