document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210220>Question 1210220</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 	Let us assume that from a population with mean μ=100 and standard deviation σ=15 a sample random variable of n=900 is selected. \r<BR>\n" );
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document.write( "	What is the probability P(X ̅<101.1)? \r<BR>\n" );
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document.write( "	What is the probability P(X ̅>101.5)? \r<BR>\n" );
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document.write( "	What is the probability P(99.3<X ̅<100.5)? \r<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 #851660 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</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=851660\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $\bar{X}$ be the sample mean of a random sample of size $n=900$ drawn from a population with mean $\mu=100$ and standard deviation $\sigma=15$.\r<BR>\n" );
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document.write( "According to the Central Limit Theorem, the sampling distribution of the sample mean $\bar{X}$ will be approximately normal with mean $\mu_{\bar{X}} = \mu$ and standard deviation $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$.\r<BR>\n" );
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document.write( "In this case:<BR>\n" );
document.write( "$\mu_{\bar{X}} = 100$<BR>\n" );
document.write( "$\sigma_{\bar{X}} = \frac{15}{\sqrt{900}} = \frac{15}{30} = 0.5$\r<BR>\n" );
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document.write( "Now we can answer each part of the question by converting the sample mean to a z-score:<BR>\n" );
document.write( "$z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}}$\r<BR>\n" );
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document.write( "**1. What is the probability P($\bar{X} < 101.1$)?**\r<BR>\n" );
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document.write( "First, calculate the z-score for $\bar{X} = 101.1$:<BR>\n" );
document.write( "$z = \frac{101.1 - 100}{0.5} = \frac{1.1}{0.5} = 2.2$\r<BR>\n" );
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document.write( "Now, we need to find the probability $P(Z < 2.2)$, where $Z$ is a standard normal random variable. Looking up the value in a standard normal distribution table or using a calculator, we find:<BR>\n" );
document.write( "$P(Z < 2.2) \approx 0.9861$\r<BR>\n" );
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document.write( "So, $P(\bar{X} < 101.1) \approx 0.9861$.\r<BR>\n" );
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document.write( "**2. What is the probability P($\bar{X} > 101.5$)?**\r<BR>\n" );
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document.write( "First, calculate the z-score for $\bar{X} = 101.5$:<BR>\n" );
document.write( "$z = \frac{101.5 - 100}{0.5} = \frac{1.5}{0.5} = 3$\r<BR>\n" );
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document.write( "Now, we need to find the probability $P(Z > 3)$. This is equal to $1 - P(Z \le 3)$. Looking up the value in a standard normal distribution table or using a calculator, we find:<BR>\n" );
document.write( "$P(Z \le 3) \approx 0.9987$\r<BR>\n" );
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document.write( "So, $P(Z > 3) = 1 - 0.9987 = 0.0013$\r<BR>\n" );
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document.write( "Thus, $P(\bar{X} > 101.5) \approx 0.0013$.\r<BR>\n" );
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document.write( "**3. What is the probability P($99.3 < \bar{X} < 100.8$)?**\r<BR>\n" );
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document.write( "First, calculate the z-scores for $\bar{X} = 99.3$ and $\bar{X} = 100.8$:<BR>\n" );
document.write( "For $\bar{X} = 99.3$:<BR>\n" );
document.write( "$z_1 = \frac{99.3 - 100}{0.5} = \frac{-0.7}{0.5} = -1.4$\r<BR>\n" );
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document.write( "For $\bar{X} = 100.8$:<BR>\n" );
document.write( "$z_2 = \frac{100.8 - 100}{0.5} = \frac{0.8}{0.5} = 1.6$\r<BR>\n" );
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document.write( "Now, we need to find the probability $P(-1.4 < Z < 1.6)$, which is equal to $P(Z < 1.6) - P(Z \le -1.4)$. Looking up the values in a standard normal distribution table or using a calculator, we find:<BR>\n" );
document.write( "$P(Z < 1.6) \approx 0.9452$<BR>\n" );
document.write( "$P(Z \le -1.4) \approx 0.0808$\r<BR>\n" );
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document.write( "So, $P(-1.4 < Z < 1.6) = 0.9452 - 0.0808 = 0.8644$\r<BR>\n" );
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document.write( "Thus, $P(99.3 < \bar{X} < 100.8) \approx 0.8644$.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{P(\bar{X} < 101.1) \approx 0.9861, P(\bar{X} > 101.5) \approx 0.0013, P(99.3 < \bar{X} < 100.8) \approx 0.8644}$ </SPAN>\n" );
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