document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1161078>Question 1161078</A>: <I><SPAN STYLE=\"word-break: break-all;\"> PLEASE HELP ME! I REALLY APPRECIATE ANY SUPPORT!<BR>\n" );
document.write( "1) Suppose that the distribution for the number of points scored in a game by Steph Curry can be described by an approximate normal curve, with a mean of 30 and a standard deviation of 8.  About how many points are needed to ensure there is only approximately a 2.5% chance that Steph exceeds this amount? \r<BR>\n" );
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document.write( "2) Lastly, assume the scoring distribution of Knicks phenom RJ Barret is<BR>\n" );
document.write( "approximately normal in shape. Suppose there is approximately 0.15% chance that R.J scores more than 26 points, and a 2.5% chance that he scores less than 6 points. What is his mean and standard deviation of points scored? <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 #784530 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=784530\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the 97.5th percentile is 45.68 or 46 points<BR>\n" );
document.write( "calculator 2 VARS 3 invnorm (.975,30,8)\r<BR>\n" );
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document.write( "99.85 percentile is 2.967 sd<BR>\n" );
document.write( "97.5 percentile is 1.96 sd\r<BR>\n" );
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document.write( "z=(x-mean)/sd<BR>\n" );
document.write( "so 2.967=(x-mean)/sd<BR>\n" );
document.write( "and -1.96=(x-mean)/sd<BR>\n" );
document.write( "2.97 sd=26-mean <BR>\n" );
document.write( "-1.96 sd=6-mean<BR>\n" );
document.write( "1.96 sd=-6+mean<BR>\n" );
document.write( "4.93 sd=20<BR>\n" );
document.write( "sd=4.056 or 4.06 points\r<BR>\n" );
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document.write( "(26-mean)=4.06*2.967=12.05<BR>\n" );
document.write( "mean=13.95 points\r<BR>\n" );
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document.write( "check<BR>\n" );
document.write( "26-13.95 divided by 4.06=12.05/4.06 or 2.97 points standard deviation.<BR>\n" );
document.write( "and<BR>\n" );
document.write( "-7.95/4.06=-1.96 sd \r<BR>\n" );
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document.write( "can check using normalcdf (6,26,13.95,4.06), which should be everything but 0.15% on the left and 2.5% on the right or 100-2.65=97.35%, and it is 97.34%<BR>\n" );
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