document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=576057>Question 576057</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 8.The probability function for the number of insurance policies John will sell to a customer is given by     f(x) .5 - (x/6) for x = 0, 1, or 2 \r<BR>\n" );
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document.write( "a.Is this a valid probability function? Explain your answer.<BR>\n" );
document.write( "b.What is the probability that John will sell exactly 2 policies to a customer?<BR>\n" );
document.write( "c.What is the probability that John will sell at least 2 policies to a customer?<BR>\n" );
document.write( "d.What is the expected number of policies John will sell?<BR>\n" );
document.write( "e.What is the variance of the number of policies John will sell?<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 #369711 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=369711\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a)\r<BR>\n" );
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document.write( "It's only a valid probability function if all the individual probabilities add to one.\r<BR>\n" );
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document.write( "f(0) = 0.5 - 0/6 = 0.5, so f(0) = 0.5 or f(0) = 1/2\r<BR>\n" );
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document.write( "f(1) = 0.5 - 1/6 = 1/3, so f(1) = 1/3\r<BR>\n" );
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document.write( "f(2) = 0.5 - 2/6 = 1/6, so f(2) = 1/6\r<BR>\n" );
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document.write( "Now add up the individual probabilities:\r<BR>\n" );
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document.write( "f(0) + f(1) + f(2) = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = (3+2+1)/6 = 6/6 = 1\r<BR>\n" );
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document.write( "Since they all add to 1, this is a valid probability function\r<BR>\n" );
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document.write( "b)\r<BR>\n" );
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document.write( "P(X = 2) = 1/6 = 0.1667 and this was found in part a)\r<BR>\n" );
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document.write( "c)\r<BR>\n" );
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document.write( "P(At least 2) = 1 - P(None)\r<BR>\n" );
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document.write( "P(At least 2) = 1 - 1/2\r<BR>\n" );
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document.write( "P(At least 2) = 1/2\r<BR>\n" );
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document.write( "P(At least 2) = 0.5\r<BR>\n" );
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document.write( "So the probability of selling at least two policies to a customer is 0.5 ( which is 50% chance)\r<BR>\n" );
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document.write( "d)\r<BR>\n" );
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document.write( "Expected number = Expected value = Sum of values*probabilities = (0)*(1/2)+(1)*(1/3)+(2)*(1/6) = 2/3 = 0.667\r<BR>\n" );
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document.write( "So the expected number is 0.667, which means that he expects to sell somewhere between 0 and one policy (with more weight/chance towards selling 1 policy)\r<BR>\n" );
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document.write( "e)\r<BR>\n" );
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document.write( "E(X^2) = Sum(X^2*probability)\r<BR>\n" );
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document.write( "E(X^2) = (0^2)*(1/2) + (1^2)*(1/3) + (2^2)*(1/6)\r<BR>\n" );
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document.write( "E(X^2) = 0 + 1/3 + 2/3\r<BR>\n" );
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document.write( "E(X^2) = 1\r<BR>\n" );
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document.write( "Variance\r<BR>\n" );
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document.write( "sigma^2 = E(X^2) - (E(X))^2\r<BR>\n" );
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document.write( "sigma^2 = 1 - (2/3)^2\r<BR>\n" );
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document.write( "sigma^2 = 1 - 4/9\r<BR>\n" );
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document.write( "sigma^2 = 5/9\r<BR>\n" );
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document.write( "sigma^2 = 0.556\r<BR>\n" );
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document.write( "So the variance of the number of policies John will sell is roughly 0.556 </SPAN>\n" );
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