A study conducted at a large hospital found the 27.8 percent of all patients adm
ID: 3222505 • Letter: A
Question
A study conducted at a large hospital found the 27.8 percent of all patients admitted to the hospital's intensive care unit (ICU) remained in the ICU for less than 24 hours. (a) If 8 patients selected at random from the ICU patients at this hospital, what is the probability that 2 or fewer of them remained in the ICU for less than 24 hours? (b) Suppose that 20 patients are selected at random from the ICU patients at this hospital Calculate the mean and the standard deviation of the number of these patients who remained in the ICU for less than 24 hours. (c) Another result of the study was that for elective (non-emergency) admissions to the hospital's ICU, the length of stay in the ICU had a mean of 18.9 hours and a standard deviation of 3.9 hours. Assuming that the length of stay in the ICU for elective admissions approximately normally distributed, what proportion of elective admissions remained in the ICU for less than 24 hours?Explanation / Answer
a) probability of success, p = 0.278, N=8
using binomial distribution formula P(X=x) = Ncn * pn * (1-p)(N-n)
P(X<=2) = P(X=0) + P(X=1) + P(X=2) = 8C0 * 0.2780 * 0.7228 + 8C1 * 0.2781 * 0.7227 + 8C2 * 0.2782 * 0.7226
= 0.6078
b) N=20, mean = N*p = 20*0.278 = 5.56, std. dev = N*p*(1-p) = 20*0.278*0.722 = 4.01432
c) mean = 18.9, std. dev. = 3.9
z = (x-mean)/std dev = (24-18.9)/3.9 = 1.3077, p(z < 1.3077) = 0.9045 stayed less than 24 hours.
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