document.write( "Question 1195851: Historically, 93% of the deliveries of an overnight mail service arrive before 10:30 the following morning. If a random sample of 500 deliveries is selected, what is the probability the sample will have more than 95% of the deliveries arriving before 10:30 the following morning? \n" ); document.write( "
Algebra.Com's Answer #828469 by Boreal(15235)\"\" \"About 
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This is calculating a z-value which is here z=(phat-p)/sqrt (p*(1-p)/n)
\n" ); document.write( "=0.02/sqrt(0.93*0.07/500)\r
\n" ); document.write( "\n" ); document.write( "=0.02/0.0114
\n" ); document.write( "=1.75
\n" ); document.write( "Probability z > 1.75 is 0.0398
\n" ); document.write( "-
\n" ); document.write( "Can check with a normal approximation
\n" ); document.write( "mean is 0.93*500=465
\n" ); document.write( "variance is that *0.07=32.55
\n" ); document.write( "sd is sqrt of that=5.71
\n" ); document.write( "95% of 500 is 475
\n" ); document.write( "so z=(475.5-465)/5.71, using the continuity correction factor
\n" ); document.write( "=1.84 and the probability of z> 1.84 is 0.0329.
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