According to Health Affairs (Oct. 28, 2004), 50% of all patients wait more than
ID: 3334895 • Letter: A
Question
According to Health Affairs (Oct. 28, 2004), 50% of all patients wait more than 30 minutes in a typical U.S. emergency room. Suppose 150 patients visit the emergency room on a particular day. Using normal aproximation to work out the following probabilities.
a.) Let X be the number of the patients out of the 150 who wait more than 30 minutes. What is the distribution of X?
b.) What is the probability that more than half of the patients wait more than 30 minutes?
c.) What is the probability that more than 85 of the patients wait more than 30 minutes?
d.) What is the probability that more than 60 but fewer than 90 will wait more than 30 minutews that day?
Explanation / Answer
a) distribution of X is binomial with N=150 and p=0.5
normal approximation of above:
mean=np =150*0.5 =75
and std deviation =(np(1-p))1/2 =6.1237
b)probability that more than half of the patients wait more than 30 minutes =P(X>75)=1-P(X<=75)
=1-P(Z<(75.5-75)/6.1237)=1-P(Z<0.0816)=1-0.5325 =0.4675
c) probability that more than 85 of the patients wait more than 30 minutes=P(X>85)=1-P(X<=85)
=1-P(Z<(85.5-75)/6.1237)=1-P(Z<1.7146)=1-0.9568 =0.0432
d) probability that more than 60 but fewer than 90 will wait more than 30 minutews that day
=P(60<X<90)=P((60.5-75)/6.1237<Z<(89.5-75)/6.1237)=P(-2.3678<Z<2.3678)=0.9911-0.0089 =0.9821