document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1177613>Question 1177613</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose that Y is a continuous random variable whose pdf is given by<BR>\n" );
document.write( "f (y) = <BR>\n" );
document.write( "K(4y − 2y^2), 0 < y < 2<BR>\n" );
document.write( "0, elsewhere.\r<BR>\n" );
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document.write( "(a) What is the value of K?<BR>\n" );
document.write( "(b) Find P(Y > 1).<BR>\n" );
document.write( "(c) Find F (y)\r<BR>\n" );
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document.write( "thank you </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850434 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850434\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Absolutely, let's break down the problem step by step.\r<BR>\n" );
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document.write( "**(a) Finding the value of K**\r<BR>\n" );
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document.write( "To find the value of K, we'll use the property that the integral of a probability density function (PDF) over its entire range must equal 1. \r<BR>\n" );
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document.write( "Therefore, we need to solve the following equation for K:\r<BR>\n" );
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document.write( "```<BR>\n" );
document.write( "∫[0,2] K(4y - 2y^2) dy = 1<BR>\n" );
document.write( "```\r<BR>\n" );
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document.write( "Evaluating the integral:\r<BR>\n" );
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document.write( "```<BR>\n" );
document.write( "K [2y^2 - (2/3)y^3] from 0 to 2 = 1<BR>\n" );
document.write( "K (8 - (16/3)) = 1<BR>\n" );
document.write( "K (8/3) = 1<BR>\n" );
document.write( "```\r<BR>\n" );
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document.write( "Solving for K:\r<BR>\n" );
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document.write( "```<BR>\n" );
document.write( "K = 3/8<BR>\n" );
document.write( "```\r<BR>\n" );
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document.write( "**(b) Finding P(Y > 1)**\r<BR>\n" );
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document.write( "To find P(Y > 1), we'll integrate the PDF from 1 to 2:\r<BR>\n" );
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document.write( "```<BR>\n" );
document.write( "P(Y > 1) = ∫[1,2] (3/8)(4y - 2y^2) dy<BR>\n" );
document.write( "          = (3/8) [2y^2 - (2/3)y^3] from 1 to 2<BR>\n" );
document.write( "          = (3/8) [(8 - (16/3)) - (2 - (2/3))]<BR>\n" );
document.write( "          = (3/8) (2/3) = 1/4<BR>\n" );
document.write( "```\r<BR>\n" );
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document.write( "Therefore, P(Y > 1) = 1/4.\r<BR>\n" );
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document.write( "**(c) Finding F(y)**\r<BR>\n" );
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document.write( "The cumulative distribution function (CDF), F(y), is the integral of the PDF from negative infinity to y.  We can define it piecewise:\r<BR>\n" );
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document.write( "For y ≤ 0:<BR>\n" );
document.write( "```<BR>\n" );
document.write( "F(y) = 0<BR>\n" );
document.write( "```\r<BR>\n" );
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document.write( "For 0 < y < 2:<BR>\n" );
document.write( "```<BR>\n" );
document.write( "F(y) = ∫[0,y] (3/8)(4t - 2t^2) dt<BR>\n" );
document.write( "     = (3/8) [2t^2 - (2/3)t^3] from 0 to y<BR>\n" );
document.write( "     = (3/8) (2y^2 - (2/3)y^3)<BR>\n" );
document.write( "     = (3/4)y^2 - (1/4)y^3<BR>\n" );
document.write( "```\r<BR>\n" );
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document.write( "For y ≥ 2:<BR>\n" );
document.write( "```<BR>\n" );
document.write( "F(y) = 1<BR>\n" );
document.write( "```\r<BR>\n" );
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document.write( "So, the complete CDF is:\r<BR>\n" );
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document.write( "```<BR>\n" );
document.write( "F(y) = <BR>\n" );
document.write( "   0,                 y ≤ 0<BR>\n" );
document.write( "   (3/4)y^2 - (1/4)y^3,  0 < y < 2<BR>\n" );
document.write( "   1,                 y ≥ 2<BR>\n" );
document.write( "``` </SPAN>\n" );
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