document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1160812>Question 1160812</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A gym knows that each member, on average, spends 50 minutes at the gym per week, with a standard deviation of 10 minutes. Assume the amount of time each customer spends at the gym is normally distributed.\r<BR>\n" );
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document.write( "a. Suppose the gym surveys a random sample of 30 members about the amount of time they spend at the gym each week. What are the expected value and standard deviation (standard error) of the sample mean of the time spent at the gym? <BR>\n" );
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document.write( "b. If 30 members are randomly selected, what is the probability that the average time spent at the gym exceeds 55 minutes? <BR>\n" );
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document.write( "c. If 30 members are randomly selected, what is the probability that a randomly selected customer spends between 45 minutes to 55 minutes at the gym? <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #784210 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=784210\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a. E(X)=50 minutes, same as individual average<BR>\n" );
document.write( "SEM=10/sqrt(30)=1.826 minutes\r<BR>\n" );
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document.write( "z=(x-mean)/sd/sqrt(n)<BR>\n" );
document.write( "> 5*sqrt(30)10=2.74<BR>\n" );
document.write( "prob. z>2.74 is 0.0031\r<BR>\n" );
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document.write( "this would be z between -.5 and +.5 which is 0.3829\r<BR>\n" );
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document.write( "Much more difficult for the average of 30 to differ greatly from the mean than a single person.<BR>\n" );
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