document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1162676>Question 1162676</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A random variable X is exponentially distributed with an expected value of 67.<BR>\n" );
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document.write( "a-1. What is the rate parameter λ? (Round your answer to 3 decimal places.)<BR>\n" );
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document.write( "a-2. What is the standard deviation of X?<BR>\n" );
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document.write( "b. Compute P(58 ≤ X ≤ 76). (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)<BR>\n" );
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document.write( "c. Compute P(55 ≤ X ≤ 79). (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)<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 #854397 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2261)</B></A><img src=\"/images/stars/stars-13.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=854397\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> For an exponentially distributed random variable $X$, we use the properties of the exponential distribution where the mean is $\mu = \frac{1}{\lambda}$.\r<BR>\n" );
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document.write( "### a-1. What is the rate parameter $\lambda$?<BR>\n" );
document.write( "The relationship between the expected value and the rate parameter is $E[X] = \frac{1}{\lambda}$.<BR>\n" );
document.write( "$$\lambda = \frac{1}{67} \approx 0.014925$$<BR>\n" );
document.write( "Rounding to 3 decimal places:<BR>\n" );
document.write( "**$\lambda = 0.015$**\r<BR>\n" );
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document.write( "### a-2. What is the standard deviation of $X$?<BR>\n" );
document.write( "For the exponential distribution, the standard deviation ($\sigma$) is equal to the mean ($E[X]$).<BR>\n" );
document.write( "$$\sigma = 67$$<BR>\n" );
document.write( "**Standard deviation = 67**\r<BR>\n" );
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document.write( "### b. Compute $P(58 \leq X \leq 76)$<BR>\n" );
document.write( "The cumulative distribution function (CDF) for an exponential distribution is $F(x) = 1 - e^{-\lambda x}$. To find the probability between two values, we use $P(a \leq X \leq b) = e^{-\lambda a} - e^{-\lambda b}$.\r<BR>\n" );
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document.write( "Using the precise $\lambda = \frac{1}{67} \approx 0.014925$:<BR>\n" );
document.write( "* $e^{-(0.014925 \times 58)} = e^{-0.865671} \approx 0.4208$<BR>\n" );
document.write( "* $e^{-(0.014925 \times 76)} = e^{-1.134328} \approx 0.3216$\r<BR>\n" );
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document.write( "$$P(58 \leq X \leq 76) = 0.42077 - 0.32164 = 0.09913$$<BR>\n" );
document.write( "Rounding to 4 decimal places:<BR>\n" );
document.write( "**$P(58 \leq X \leq 76) = 0.0991$**\r<BR>\n" );
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document.write( "### c. Compute $P(55 \leq X \leq 79)$<BR>\n" );
document.write( "Using the same method:<BR>\n" );
document.write( "* $e^{-(0.014925 \times 55)} = e^{-0.820895} \approx 0.4401$<BR>\n" );
document.write( "* $e^{-(0.014925 \times 79)} = e^{-1.179104} \approx 0.3076$\r<BR>\n" );
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document.write( "$$P(55 \leq X \leq 79) = 0.44005 - 0.30757 = 0.13248$$<BR>\n" );
document.write( "Rounding to 4 decimal places:<BR>\n" );
document.write( "**$P(55 \leq X \leq 79) = 0.1325$** </SPAN>\n" );
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