document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1171229>Question 1171229</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose a company repacks cereal with the use of a machine that is calibrated so that the weight of cereal dispensed in a box is normally distributed with mean 0.40 kg and variance 0.01 kg2, although the weight stated in the cereal box is 0.45 kg. For quality control purposes, suppose a random sample of 15 cereal boxes is taken. What is the probability of selecting a random sample with mean weight less than the advertised weight? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850976 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=850976\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $X$ be the weight of cereal dispensed in a box. We are given that $X$ is normally distributed with mean $\mu = 0.40$ kg and variance $\sigma^2 = 0.01$ kg$^2$. Thus, the standard deviation is $\sigma = \sqrt{0.01} = 0.1$ kg.\r<BR>\n" );
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document.write( "We take a random sample of $n = 15$ cereal boxes. Let $\bar{X}$ be the sample mean. We want to find the probability that the sample mean is less than the advertised weight, which is 0.45 kg.\r<BR>\n" );
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document.write( "The sample mean $\bar{X}$ is also normally distributed with mean $\mu_{\bar{X}} = \mu = 0.40$ kg and variance $\sigma_{\bar{X}}^2 = \frac{\sigma^2}{n} = \frac{0.01}{15}$.<BR>\n" );
document.write( "The standard deviation of the sample mean is $\sigma_{\bar{X}} = \sqrt{\frac{0.01}{15}} = \frac{0.1}{\sqrt{15}} \approx 0.02582$ kg.\r<BR>\n" );
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document.write( "We want to find $P(\bar{X} < 0.45)$. We standardize the sample mean using the z-score formula:\r<BR>\n" );
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document.write( "$$Z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}}$$\r<BR>\n" );
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document.write( "We calculate the z-score for $\bar{X} = 0.45$:\r<BR>\n" );
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document.write( "$$Z = \frac{0.45 - 0.40}{0.1/\sqrt{15}} = \frac{0.05}{0.1/\sqrt{15}} = \frac{0.05 \sqrt{15}}{0.1} = 0.5 \sqrt{15} \approx 1.9365$$\r<BR>\n" );
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document.write( "We want to find $P(Z < 1.9365)$. Using a standard normal distribution table or a calculator, we find:\r<BR>\n" );
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document.write( "$$P(Z < 1.9365) \approx 0.9736$$\r<BR>\n" );
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document.write( "Therefore, the probability of selecting a random sample with mean weight less than the advertised weight is approximately 0.9736.<BR>\n" );
document.write( " </SPAN>\n" );
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