Question

For Problems 13 to 28, use the normal approximation to the binomial to calculate the indicated probabilities, or explain why it

is not appropriate to use the normal approximation to the binomia

A certain doctor’s office has a tendency to make people wait longer than usual. In fact, 65% of all patients of this

doctor have to wait in the waiting rooms for more than 30 minutes beyond their scheduled appointment time. If the

office has 36 patients scheduled for that day, what is the probability that less than half of the patients that day have to

wait for more than 30 minutes?

Answer 1

Given that 65%. Of all patient of the doctor have to wait in the waiting rooms for more than 30 minutes beyond their scheduled appoint ment time. Hence the poobability that a randomly selected patient wait for more than 30 minutes = 0-65 . Number of patients on the given day = 36 Let x be the number of patients who have to wait for more than 30 minutes on that day. Then xv Binomial (n=36, p=0.65). As both n*p= 36*0-65- 23.4 and * (1-P)=36* 0.35= 12-6 are greater than 5; we can use a nomal appoorimation here. Mean of this Binomial distribution is pe nip=36*0-65=23-4 and standard deviation n*p*(1-P) = 36*0*65*0.35 -J8.19 # 2. 8618 Hence x will be approximately nommal with Me - 23.4 and 6 = 2.8618 Now; we need to calculate the probability that less than half of the patients that day have to wait for more than 30 mérites i.e pcx < 36)=P(x<18). Before using the normal appaocimation ; as binomial is a discrete distribution and normal is a continuous deterbution we need to Qupply continudy correction. Continuity correction for p(x<a)=P(X <Q-0.5 :: P ( x <18) = P(x <18-0.5) - P(x< 17.5), 7-score for 17.5 = 175- :: PC x <17-5) P(Z < -2.06). Area to the left-of -2.06 under the standard normal tables 0.0197 (toom standard normal cumulatine distribution table). Hence podbability that less than half of the patient that day have to wait for more than 30 minutes = 0.0197 0.02 - 17-5- 23.4 2.8618 2.06