document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1137508>Question 1137508</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Q.2 By using Normal Distribution, solve the following:\r<BR>\n" );
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document.write( "a) A certain type of battery lasts, on average, 3.0 years with a standard deviation of 0.5 years. Assuming that the battery lives are normally distributed, find the probability that a given battery will last less than 2.3 years. \r<BR>\n" );
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document.write( "b) An electrical firm manufactures light bulbs that have a life, before burn-out, that is normally distributed with mean equal to 800 hours and a standard deviation of 40 hours. Find the probability that a bulb burns between 778 and 834 hours. <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 #755388 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=755388\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"times\" color=\"black\" size=\"3\"><BR>\n" );
document.write( "Part A\r<BR>\n" );
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document.write( "mu = 3 = mean<BR>\n" );
document.write( "sigma = 0.5 = standard deviation\r<BR>\n" );
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document.write( "Convert the raw score x = 2.3 to its corresponding z score<BR>\n" );
document.write( "z = (x-mu)/sigma<BR>\n" );
document.write( "z = (2.3-3)/0.5<BR>\n" );
document.write( "z = -1.40\r<BR>\n" );
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document.write( "Use <a  rel=nofollow HREF= \"http://users.stat.ufl.edu/~athienit/Tables/Ztable.pdf\">this table</a> (or similar) to find that P(Z < -1.40) = 0.0808<BR>\n" );
document.write( "To find this result, turn to page 1 of that PDF, go to the row that starts with -1.4 and the column that has 0.00 at the top. The row and column intersect at 0.0808<BR>\n" );
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document.write( "The probability is approximately 0.0808<BR>\n" );
document.write( "which is roughly 8.08% in percent form<BR>\n" );
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document.write( "Part B\r<BR>\n" );
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document.write( "mu = 800<BR>\n" );
document.write( "sigma = 40\r<BR>\n" );
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document.write( "Convert the raw score x = 778 to its corresponding z score<BR>\n" );
document.write( "z = (x-mu)/sigma<BR>\n" );
document.write( "z = (778-800)/40<BR>\n" );
document.write( "z = -0.55\r<BR>\n" );
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document.write( "Repeat for x = 834<BR>\n" );
document.write( "z = (x-mu)/sigma<BR>\n" );
document.write( "z = (834-800)/40<BR>\n" );
document.write( "z = 0.85\r<BR>\n" );
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document.write( "So, P(778 < X < 834) is the same as P(-0.55 < Z < 0.85)\r<BR>\n" );
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document.write( "We will use this formula here<BR>\n" );
document.write( "P(A < Z < B) = P(Z < B) - P(Z < A)<BR>\n" );
document.write( "to help compute the area we want between the two z scores\r<BR>\n" );
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document.write( "A = -0.55<BR>\n" );
document.write( "B = 0.85<BR>\n" );
document.write( "P(A < Z < B) = P(Z < B) - P(Z < A)<BR>\n" );
document.write( "P(-0.55 < Z < 0.85) = P(Z < 0.85) - P(Z < -0.55)<BR>\n" );
document.write( "P(-0.55 < Z < 0.85) = 0.8023 - 0.2912 <font color=blue>use the same <a  rel=nofollow HREF= \"http://users.stat.ufl.edu/~athienit/Tables/Ztable.pdf\">table</a> as done in part A</font><BR>\n" );
document.write( "P(-0.55 < Z < 0.85) = 0.5111\r<BR>\n" );
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document.write( "The approximate probability is 0.5111<BR>\n" );
document.write( "which in percent form would be roughly 51.11%\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"http://davidmlane.com/hyperstat/z_table.html\">Here</a> is a handy calculator to help check your work\r<BR>\n" );
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document.write( "</font> </SPAN>\n" );
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