document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1086776>Question 1086776</A>: <I><SPAN STYLE=\"word-break: break-all;\"> At a particular hospital, records shows that each day, on average,  only 80% of the people keep their appointments at the outpatient clinic. Find the probability that when 200 appointments have been booked: more than 170 patients keep their appointments  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #701021 by </B><A HREF=/tutors/aboutme.mpl?userid=mathmate><B>mathmate(429)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mathmate><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=mathmate><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=701021\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Question:<BR>\n" );
document.write( "At a particular hospital, records shows that each day, on average,  only 80% of the people keep their appointments at the outpatient clinic. Find the probability that when 200 appointments have been booked: more than 170 patients keep their appointments <BR>\n" );
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document.write( "Solution:<BR>\n" );
document.write( "The situation calls for the binomial distribution.  Before proceeding, we need to go through the checklist to qualify the use of the distribution.<BR>\n" );
document.write( "1. Bernoulli trials, i.e. exactly two possible outcomes (success = keep or failure=not keep appointments)<BR>\n" );
document.write( "2. Number of trials is known before and constant throughout the experiment, <BR>\n" );
document.write( "i.e. independent of outcomes. (n=200)<BR>\n" );
document.write( "3. All trials are independent of each other. (assumed from context)<BR>\n" );
document.write( "4. Probability of success is known, and remain constant throughout trials. (historical information, p=0.8)<BR>\n" );
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document.write( "Since all criteria are satisfied, we can model with binomial distribution, where the probability of x successes out of N trials each with probability of success p is given by<BR>\n" );
document.write( "P(x)=C(n,x)(p^x)(1-p)^(n-x)<BR>\n" );
document.write( "and,<BR>\n" );
document.write( "C(N,x) is number of combinations of selecting x objects out of N.<BR>\n" );
document.write( " <BR>\n" );
document.write( "Substituting values, <BR>\n" );
document.write( "n=200<BR>\n" );
document.write( "p=0.8<BR>\n" );
document.write( "x>170, or x=171, 172,173,...200.\r<BR>\n" );
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document.write( "P(x>170)<BR>\n" );
document.write( "=1-P(x<=170)<BR>\n" );
document.write( "=1-(sum P(x=i), i=0 to 170)<BR>\n" );
document.write( "=1-0.97172  [obtained from software, such as R, pbinom(170,200,0.8)=0.9717234]<BR>\n" );
document.write( "=0.02828<BR>\n" );
document.write( " <BR>\n" );
document.write( "Alternatively, for large values of n, we can use normal approximation as long as the binomial distribution is not too skewed np(1-p)>5 (preferably 10), or 200(.8)(.2)=32 > 10.<BR>\n" );
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document.write( "Steps:<BR>\n" );
document.write( "1. find mean=np=200*0.8=160<BR>\n" );
document.write( "2. find standard deviation, &sigma;&sup2;=np(1-p)=32, &sigma;=&radic;32=4&radic;2=5.656<BR>\n" );
document.write( "3. apply continuity correction, x=170+0.5=170.5<BR>\n" );
document.write( "find Z=(170.5-160)/5.656=1.85616<BR>\n" );
document.write( "4. find P(z<Z)=0.96828 [from tables or calculator]<BR>\n" );
document.write( "5. P(x>170)=1-0.96828=0.03172<BR>\n" );
document.write( " <BR>\n" );
document.write( "Note that error of approximation is greater than 12%, compared to the exact value using binomial distribution.  Values of p closer to 0.5 would give much better approximations.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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