document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1192227>Question 1192227</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A certain industrial material is packed in 2 kilogram packets. Packets selected at random have a mean mass of mean mass of 2025 with a standard deviation of 15 grams. If the masses are normally distributed, determine the percentage of packets having a mass<BR>\n" );
document.write( "i. less than 2 kilograms<BR>\n" );
document.write( "ii. More than 2030 grams<BR>\n" );
document.write( "iii. The manufacturer decides to allow 10% of the packets to be sold underweight. If he is able to maintain the same standard deviation, what mass should he adjust his mean?<BR>\n" );
document.write( "iv. A government inspector allows this but insists that only 10 % of the packets may be less than 1995 grams. To what must the manufacturer now adjust his standard deviation to meet this new requirement, if the mean is not altered? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849068 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=849068\">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( "**i. Packets less than 2 kilograms (2000 grams):**\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 = (2000 - 2025) / 15<BR>\n" );
document.write( "   z = -1.67\r<BR>\n" );
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document.write( "2. **Find the probability:**<BR>\n" );
document.write( "   Using a z-table or calculator, the probability of a z-score less than -1.67 is approximately 0.0475.\r<BR>\n" );
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document.write( "3. **Percentage:**<BR>\n" );
document.write( "   0.0475 * 100% = 4.75%\r<BR>\n" );
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document.write( "**ii. Packets more than 2030 grams:**\r<BR>\n" );
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document.write( "1. **Calculate the z-score:**<BR>\n" );
document.write( "   z = (2030 - 2025) / 15<BR>\n" );
document.write( "   z = 0.33\r<BR>\n" );
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document.write( "2. **Find the probability:**<BR>\n" );
document.write( "   The probability of a z-score greater than 0.33 is 1 - P(z < 0.33).  Using a z-table, P(z < 0.33) is approximately 0.6293. So, the probability of a z-score greater than 0.33 is 1 - 0.6293 = 0.3707.\r<BR>\n" );
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document.write( "3. **Percentage:**<BR>\n" );
document.write( "   0.3707 * 100% = 37.07%\r<BR>\n" );
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document.write( "**iii. Adjusting the mean for 10% underweight:**\r<BR>\n" );
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document.write( "1. **Find the z-score for the 10th percentile:**<BR>\n" );
document.write( "   Using a z-table or calculator, the z-score corresponding to 10% (0.10) is approximately -1.28.\r<BR>\n" );
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document.write( "2. **Use the z-score formula to find the new mean (μ_new):**<BR>\n" );
document.write( "   x = μ_new + zσ<BR>\n" );
document.write( "   2000 = μ_new + (-1.28)(15)<BR>\n" );
document.write( "   2000 = μ_new - 19.2<BR>\n" );
document.write( "   μ_new = 2000 + 19.2<BR>\n" );
document.write( "   μ_new = 2019.2 grams\r<BR>\n" );
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document.write( "**iv. Adjusting the standard deviation for 10% less than 1995 grams:**\r<BR>\n" );
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document.write( "1. **Find the z-score for the 10th percentile:**<BR>\n" );
document.write( "   As before, the z-score for 10% is approximately -1.28.\r<BR>\n" );
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document.write( "2. **Use the z-score formula to find the new standard deviation (σ_new):**<BR>\n" );
document.write( "   x = μ + zσ_new<BR>\n" );
document.write( "   1995 = 2025 + (-1.28)σ_new<BR>\n" );
document.write( "   -30 = -1.28σ_new<BR>\n" );
document.write( "   σ_new = 30 / 1.28<BR>\n" );
document.write( "   σ_new ≈ 23.44 grams\r<BR>\n" );
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document.write( "**Summary of Answers:**\r<BR>\n" );
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document.write( "* i. Approximately 4.75% of packets are less than 2 kg.<BR>\n" );
document.write( "* ii. Approximately 37.07% of packets are more than 2030 grams.<BR>\n" );
document.write( "* iii. The manufacturer should adjust the mean to approximately 2019.2 grams.<BR>\n" );
document.write( "* iv. The manufacturer should adjust the standard deviation to approximately 23.44 grams.<BR>\n" );
document.write( " </SPAN>\n" );
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