document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1112917>Question 1112917</A>: <I><SPAN STYLE=\"word-break: break-all;\">  The diameter of a brand of&#8203; Ping-Pong balls is approximately normally&#8203; distributed, with a mean of 1.31 inches and a standard deviation of 0.04 inch. A random sample of 4 &#8203;Ping-Pong balls is selected. What is the probability that the sample is between 1.26 and 1.3 &#8203;inches? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #727942 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=727942\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The diameter of a brand of&#8203; Ping-Pong balls is approximately normally&#8203; distributed, with a mean of 1.31 inches and a standard deviation of 0.04 inch. A random sample of 4 &#8203;Ping-Pong balls is selected. What is the probability that the sample is between 1.26 and 1.3 &#8203;inches? <BR>\n" );
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document.write( "z(1.26) = (1.26-1.31)/(0.04/sqrt(4)) = -0.05*2/0.04 = -0.10/0.04 = -2.5<BR>\n" );
document.write( "z(1.3) = (1.3-1.31)/(0.04/sqrt(4)) = -0.01*2/0.04 = -0.02/0.04 = - 1/2<BR>\n" );
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document.write( "P(1.26 < x-bar < 1.3) = p(-2.5 < z < -0.5) = normalcdf(-2.5,-0.5) = 0.3023<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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