document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1185911>Question 1185911</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Question: Use Table A4 to compute the probability for any Normal random variable to take a value within 1.5 interquartile ranges from population quartiles.\r<BR>\n" );
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document.write( "P.S: Table A4 is a Standard Normal distribution Table.\r<BR>\n" );
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document.write( "I don't know how to approach this question because I didn't understand how to take a random variable within 1.5 IQR from population quartiles.\r<BR>\n" );
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document.write( "I would appreciate a more elaborate explanation and maybe a solution. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849510 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=849510\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Absolutely! Let's break down this problem step by step.\r<BR>\n" );
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document.write( "**Understanding the Interquartile Range (IQR)**\r<BR>\n" );
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document.write( "The interquartile range (IQR) is a measure of statistical dispersion, or how spread out the data is. It's the range between the 25th and 75th percentiles of the data.\r<BR>\n" );
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document.write( "*   **Q1 (First Quartile):** The value that separates the lowest 25% of the data.<BR>\n" );
document.write( "*   **Q3 (Third Quartile):** The value that separates the highest 25% of the data.<BR>\n" );
document.write( "*   **IQR = Q3 - Q1**\r<BR>\n" );
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document.write( "**Relating IQR to the Standard Normal Distribution**\r<BR>\n" );
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document.write( "When dealing with a standard normal distribution (mean = 0, standard deviation = 1), we can use a Z-table (like Table A4) to find the Z-scores corresponding to Q1 and Q3.\r<BR>\n" );
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document.write( "*   The Z-score for Q1 is approximately -0.674<BR>\n" );
document.write( "*   The Z-score for Q3 is approximately 0.674\r<BR>\n" );
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document.write( "This means that the IQR of a standard normal distribution is 0.674 - (-0.674) = 1.348.\r<BR>\n" );
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document.write( "**Calculating the Probability**\r<BR>\n" );
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document.write( "1.  **Determine the range:** We want to find the probability of a random variable falling within 1.5 IQRs from the population quartiles. Since the IQR is centered around the median (which is also the mean in a normal distribution), we need to calculate the range as follows:\r<BR>\n" );
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document.write( "    *   Lower bound = Q1 - 1.5 * IQR = -0.674 - 1.5 * 1.348 = -2.696<BR>\n" );
document.write( "    *   Upper bound = Q3 + 1.5 * IQR = 0.674 + 1.5 * 1.348 = 2.696\r<BR>\n" );
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document.write( "2.  **Use the Z-table:** Look up the cumulative probabilities for the upper and lower bounds in Table A4.\r<BR>\n" );
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document.write( "    *   P(Z < 2.696) ≈ 0.9965<BR>\n" );
document.write( "    *   P(Z < -2.696) ≈ 0.0035\r<BR>\n" );
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document.write( "3.  **Calculate the probability:** Subtract the smaller probability from the larger probability to find the probability of the random variable falling within the desired range.\r<BR>\n" );
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document.write( "    P(-2.696 < Z < 2.696) = P(Z < 2.696) - P(Z < -2.696)<BR>\n" );
document.write( "    P(-2.696 < Z < 2.696) = 0.9965 - 0.0035 = 0.993\r<BR>\n" );
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document.write( "**Therefore, the probability of a normal random variable falling within 1.5 interquartile ranges from the population quartiles is approximately 0.993.**\r<BR>\n" );
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document.write( "**Key Points**\r<BR>\n" );
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document.write( "*   The IQR is a measure of spread that is less sensitive to outliers than the standard deviation.<BR>\n" );
document.write( "*   In a normal distribution, about 68% of the data falls within one standard deviation of the mean, and about 95% falls within two standard deviations. Our result shows that almost all (99.3%) of the data falls within 1.5 IQRs of the quartiles. This is because the range we are considering is wider than two standard deviations.\r<BR>\n" );
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document.write( "Let me know if you have any other questions!<BR>\n" );
document.write( " </SPAN>\n" );
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