document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1182183>Question 1182183</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of time in hours before a mobile requires charging has a normal distribution with a mean of<BR>\n" );
document.write( "100 hours and a standard deviation of 15 hours.<BR>\n" );
document.write( "a) Find the probability that the time before charging is greater than 127 hours.<BR>\n" );
document.write( "b) Find the 10th percentile<BR>\n" );
document.write( "c) You are about to go on a 6 hour trip. Given you last charged your phone 127 hours ago, what is<BR>\n" );
document.write( "the probability your mobile will not need charging until you complete the trip? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849941 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</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=849941\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to solve this problem:\r<BR>\n" );
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document.write( "**a) Probability of time before charging being greater than 127 hours:**\r<BR>\n" );
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document.write( "1. **Calculate the z-score:**<BR>\n" );
document.write( "   z = (x - μ) / σ<BR>\n" );
document.write( "   z = (127 - 100) / 15<BR>\n" );
document.write( "   z = 1.8\r<BR>\n" );
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document.write( "2. **Find the probability:** Use a standard normal distribution table (z-table) or a calculator to find the probability associated with a z-score of 1.8.  We want the probability of the time being *greater* than 127 hours, so we look for the area to the *right* of z = 1.8.<BR>\n" );
document.write( "   P(x > 127) = P(z > 1.8) = 1 - P(z < 1.8) ≈ 1 - 0.9641 ≈ 0.0359\r<BR>\n" );
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document.write( "**b) Find the 10th percentile:**\r<BR>\n" );
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document.write( "1. **Find the z-score:** The 10th percentile is the value below which 10% of the data falls. Look up 0.10 in the *body* of the z-table (or use a calculator) to find the corresponding z-score. The z-score corresponding to 0.10 is approximately -1.28.\r<BR>\n" );
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document.write( "2. **Convert the z-score to hours:**<BR>\n" );
document.write( "   x = μ + zσ<BR>\n" );
document.write( "   x = 100 + (-1.28 * 15)<BR>\n" );
document.write( "   x ≈ 80.8 hours\r<BR>\n" );
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document.write( "**c) Probability of not needing charging during a 6-hour trip, given it was charged 127 hours ago:**\r<BR>\n" );
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document.write( "This is a conditional probability problem.  We know the phone lasted 127 hours, and we want to know the probability it will last *at least* an additional 6 hours (127 + 6 = 133 hours).\r<BR>\n" );
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document.write( "1. **Calculate the conditional probability:** We want P(x > 133 | x > 127). This can be written as:\r<BR>\n" );
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document.write( "   P(x > 133 and x > 127) / P(x > 127)\r<BR>\n" );
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document.write( "   Since if x>133, it is automatically true that x>127, then P(x > 133 and x > 127) simplifies to P(x>133)\r<BR>\n" );
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document.write( "   So we have P(x > 133) / P(x > 127)\r<BR>\n" );
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document.write( "2. **Calculate P(x > 133):**<BR>\n" );
document.write( "   z = (133 - 100) / 15<BR>\n" );
document.write( "   z = 2.2<BR>\n" );
document.write( "   P(x > 133) = P(z > 2.2) = 1 - P(z < 2.2) ≈ 1 - 0.9861 ≈ 0.0139\r<BR>\n" );
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document.write( "3. **Calculate the conditional probability:**<BR>\n" );
document.write( "   P(x > 133 | x > 127) = P(x > 133) / P(x > 127) ≈ 0.0139 / 0.0359 ≈ 0.387\r<BR>\n" );
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document.write( "**Answers:**\r<BR>\n" );
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document.write( "*   a) The probability that the time before charging is greater than 127 hours is approximately 0.0359 or 3.59%.<BR>\n" );
document.write( "*   b) The 10th percentile is approximately 80.8 hours.<BR>\n" );
document.write( "*   c) The probability that the mobile will not need charging during the 6-hour trip, given it was charged 127 hours ago, is approximately 0.387 or 38.7%.<BR>\n" );
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