document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1162932>Question 1162932</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 #854394 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=854394\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> ```python?code_reference&code_event_index=2<BR>\n" );
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document.write( "# Given<BR>\n" );
document.write( "mean_x = 67\r<BR>\n" );
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document.write( "# a-1. Rate parameter lambda<BR>\n" );
document.write( "lambda_val = 1 / mean_x<BR>\n" );
document.write( "lambda_rounded = round(lambda_val, 3)\r<BR>\n" );
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document.write( "# a-2. Standard deviation<BR>\n" );
document.write( "std_dev = mean_x\r<BR>\n" );
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document.write( "# b. P(58 <= X <= 76)<BR>\n" );
document.write( "# P(a <= X <= b) = F(b) - F(a) = (1 - exp(-lambda*b)) - (1 - exp(-lambda*a)) = exp(-lambda*a) - exp(-lambda*b)<BR>\n" );
document.write( "a_b = 58<BR>\n" );
document.write( "b_b = 76<BR>\n" );
document.write( "prob_b = math.exp(-lambda_val * a_b) - math.exp(-lambda_val * b_b)\r<BR>\n" );
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document.write( "# c. P(55 <= X <= 79)<BR>\n" );
document.write( "a_c = 55<BR>\n" );
document.write( "b_c = 79<BR>\n" );
document.write( "prob_c = math.exp(-lambda_val * a_c) - math.exp(-lambda_val * b_c)\r<BR>\n" );
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document.write( "print(f\"{lambda_val=}\")<BR>\n" );
document.write( "print(f\"{lambda_rounded=}\")<BR>\n" );
document.write( "print(f\"{std_dev=}\")<BR>\n" );
document.write( "print(f\"{prob_b=}\")<BR>\n" );
document.write( "print(f\"{prob_c=}\")\r<BR>\n" );
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document.write( "```<BR>\n" );
document.write( "```text?code_stdout&code_event_index=2<BR>\n" );
document.write( "lambda_val=0.014925373134328358<BR>\n" );
document.write( "lambda_rounded=0.015<BR>\n" );
document.write( "std_dev=67<BR>\n" );
document.write( "prob_b=0.09913077737880754<BR>\n" );
document.write( "prob_c=0.132483377810241\r<BR>\n" );
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document.write( "```<BR>\n" );
document.write( "For an exponentially distributed random variable $X$:\r<BR>\n" );
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document.write( "### a-1. Rate parameter $\lambda$<BR>\n" );
document.write( "The relationship between the expected value (mean) and the rate parameter $\lambda$ 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. Standard deviation of $X$<BR>\n" );
document.write( "For an exponential distribution, the standard deviation is equal to the mean.<BR>\n" );
document.write( "$$\sigma = E[X] = 67$$<BR>\n" );
document.write( "**Standard deviation = 67**\r<BR>\n" );
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document.write( "---\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 $F(x) = 1 - e^{-\lambda x}$.<BR>\n" );
document.write( "The probability $P(a \leq X \leq b)$ is calculated as $F(b) - F(a) = e^{-\lambda a} - e^{-\lambda b}$.\r<BR>\n" );
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document.write( "Using $\lambda = \frac{1}{67} \approx 0.014925$:<BR>\n" );
document.write( "* $e^{-(0.014925 \times 58)} \approx e^{-0.8657} \approx 0.4208$<BR>\n" );
document.write( "* $e^{-(0.014925 \times 76)} \approx e^{-1.1343} \approx 0.3216$\r<BR>\n" );
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document.write( "$$P(58 \leq X \leq 76) = 0.4208 - 0.3216 = 0.0992$$<BR>\n" );
document.write( "*(Using more precise intermediate values: $0.42077 - 0.32164 = 0.09913$)*\r<BR>\n" );
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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( "---\r<BR>\n" );
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document.write( "### c. Compute $P(55 \leq X \leq 79)$<BR>\n" );
document.write( "Using the same formula: $P(55 \leq X \leq 79) = e^{-\lambda(55)} - e^{-\lambda(79)}$\r<BR>\n" );
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document.write( "* $e^{-(0.014925 \times 55)} \approx e^{-0.8209} \approx 0.4400$<BR>\n" );
document.write( "* $e^{-(0.014925 \times 79)} \approx e^{-1.1791} \approx 0.3076$\r<BR>\n" );
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document.write( "$$P(55 \leq X \leq 79) = 0.4400 - 0.3076 = 0.1324$$<BR>\n" );
document.write( "*(Using more precise intermediate values: $0.44005 - 0.30757 = 0.13248$)*\r<BR>\n" );
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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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