document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1197679>Question 1197679</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 9. A population has a mean of 200 and a standard deviation of 50. A simple random sample of <BR>\n" );
document.write( "size 100 will be taken and the sample mean 𝑥 will be used to estimate the population mean.<BR>\n" );
document.write( "(a) What is the expected value of 𝑥?<BR>\n" );
document.write( "(b) What is the standard deviation of 𝑥?<BR>\n" );
document.write( "(c) Show the sampling distribution of 𝑥?<BR>\n" );
document.write( "(d) What does the sampling distribution of 𝑥 show?<BR>\n" );
document.write( "(e) What is the probability that the sample mean will be within ±5 of the population mean?<BR>\n" );
document.write( "(f) What is the probability that the sample mean will be within ±10 of the population mean </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #848353 by </B><A HREF=/tutors/aboutme.mpl?userid=onyulee><B>onyulee(41)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=onyulee><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=848353\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> **a) Expected Value of 𝑥**\r<BR>\n" );
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document.write( "* The expected value of the sample mean (𝑥) is equal to the population mean (μ). \r<BR>\n" );
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document.write( "* **E(𝑥) = μ = 200**\r<BR>\n" );
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document.write( "**b) Standard Deviation of 𝑥 (Standard Error)**\r<BR>\n" );
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document.write( "* The standard deviation of the sample mean is given by:\r<BR>\n" );
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document.write( "   * σ_𝑥 = σ / √n \r<BR>\n" );
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document.write( "   * where:<BR>\n" );
document.write( "      * σ is the population standard deviation (50)<BR>\n" );
document.write( "      * n is the sample size (100)\r<BR>\n" );
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document.write( "   * σ_𝑥 = 50 / √100 = 50 / 10 = 5\r<BR>\n" );
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document.write( "**c) Sampling Distribution of 𝑥**\r<BR>\n" );
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document.write( "* **According to the Central Limit Theorem:**<BR>\n" );
document.write( "    * If the sample size (n) is sufficiently large (generally considered to be n ≥ 30), the sampling distribution of the sample mean (𝑥) will be approximately normally distributed, regardless of the shape of the population distribution.\r<BR>\n" );
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document.write( "* **In this case:**<BR>\n" );
document.write( "    * The sample size (n = 100) is large.<BR>\n" );
document.write( "    * Therefore, the sampling distribution of 𝑥 will be approximately normally distributed.\r<BR>\n" );
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document.write( "**d) What does the sampling distribution of 𝑥 show?**\r<BR>\n" );
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document.write( "* The sampling distribution of 𝑥 shows the probability distribution of all possible sample means that could be obtained from repeated random samples of size 100 from the population. <BR>\n" );
document.write( "* It illustrates how the sample means are likely to vary around the population mean.\r<BR>\n" );
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document.write( "**e) Probability that the sample mean will be within ±5 of the population mean**\r<BR>\n" );
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document.write( "1. **Calculate the z-scores:**\r<BR>\n" );
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document.write( "   * z1 = (μ - 5 - μ) / σ_𝑥 = -5 / 5 = -1<BR>\n" );
document.write( "   * z2 = (μ + 5 - μ) / σ_𝑥 = 5 / 5 = 1\r<BR>\n" );
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document.write( "2. **Find the probability using a standard normal distribution table (z-table):**\r<BR>\n" );
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document.write( "   * P(-1 < z < 1) = P(z < 1) - P(z < -1) <BR>\n" );
document.write( "   * P(-1 < z < 1) = 0.8413 - 0.1587 = 0.6826\r<BR>\n" );
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document.write( "* **The probability that the sample mean will be within ±5 of the population mean is approximately 0.6826 or 68.26%.**\r<BR>\n" );
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document.write( "**f) Probability that the sample mean will be within ±10 of the population mean**\r<BR>\n" );
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document.write( "1. **Calculate the z-scores:**\r<BR>\n" );
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document.write( "   * z1 = (μ - 10 - μ) / σ_𝑥 = -10 / 5 = -2<BR>\n" );
document.write( "   * z2 = (μ + 10 - μ) / σ_𝑥 = 10 / 5 = 2\r<BR>\n" );
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document.write( "2. **Find the probability using a standard normal distribution table (z-table):**\r<BR>\n" );
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document.write( "   * P(-2 < z < 2) = P(z < 2) - P(z < -2) <BR>\n" );
document.write( "   * P(-2 < z < 2) = 0.9772 - 0.0228 = 0.9544\r<BR>\n" );
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document.write( "* **The probability that the sample mean will be within ±10 of the population mean is approximately 0.9544 or 95.44%.**<BR>\n" );
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