document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1167889>Question 1167889</A>: <I><SPAN STYLE=\"word-break: break-all;\">  The Democrat and Chronicle reported that 25% of the flights arriving at the San Diego airport during the<BR>\n" );
document.write( "first five months of 2001 were late (Democrat and Chronicle, July 23, 2001). Assume the population<BR>\n" );
document.write( "proportion is p .25.<BR>\n" );
document.write( "a. Show the sampling distribution of , the proportion of late flights in a sample of 1000 flights.<BR>\n" );
document.write( "b. What is the probability that the sample proportion will be within .03 of the population proportion if a<BR>\n" );
document.write( "sample of size 1000 is selected?<BR>\n" );
document.write( "c. Answer part (b) for a sample of 500 flights. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852083 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</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=852083\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> ```python?code_reference&code_event_index=2<BR>\n" );
document.write( "import math<BR>\n" );
document.write( "from scipy.stats import norm\r<BR>\n" );
document.write( "\n" );
document.write( "# Population proportion<BR>\n" );
document.write( "p = 0.25\r<BR>\n" );
document.write( "\n" );
document.write( "# --- Part a: Sampling distribution for n = 1000 ---<BR>\n" );
document.write( "n_a = 1000<BR>\n" );
document.write( "mean_p_bar_a = p<BR>\n" );
document.write( "std_dev_p_bar_a = math.sqrt((p * (1 - p)) / n_a)\r<BR>\n" );
document.write( "\n" );
document.write( "# Check for normality assumption<BR>\n" );
document.write( "np_a = n_a * p<BR>\n" );
document.write( "n_1_minus_p_a = n_a * (1 - p)\r<BR>\n" );
document.write( "\n" );
document.write( "print(f\"Part a (n={n_a}):\")<BR>\n" );
document.write( "print(f\"  Mean of p_bar: {mean_p_bar_a}\")<BR>\n" );
document.write( "print(f\"  Standard deviation of p_bar: {std_dev_p_bar_a}\")<BR>\n" );
document.write( "print(f\"  np >= 5: {np_a} >= 5 (True)\")<BR>\n" );
document.write( "print(f\"  n(1-p) >= 5: {n_1_minus_p_a} >= 5 (True)\")\r<BR>\n" );
document.write( "\n" );
document.write( "# --- Part b: Probability for n = 1000 ---<BR>\n" );
document.write( "diff = 0.03<BR>\n" );
document.write( "lower_bound_b = p - diff<BR>\n" );
document.write( "upper_bound_b = p + diff\r<BR>\n" );
document.write( "\n" );
document.write( "# Calculate Z-scores<BR>\n" );
document.write( "z_lower_b = (lower_bound_b - p) / std_dev_p_bar_a<BR>\n" );
document.write( "z_upper_b = (upper_bound_b - p) / std_dev_p_bar_a\r<BR>\n" );
document.write( "\n" );
document.write( "# Calculate probability using CDF<BR>\n" );
document.write( "probability_b = norm.cdf(z_upper_b) - norm.cdf(z_lower_b)\r<BR>\n" );
document.write( "\n" );
document.write( "print(f\"\nPart b (n={n_a}):\")<BR>\n" );
document.write( "print(f\"  Range: [{lower_bound_b}, {upper_bound_b}]\")<BR>\n" );
document.write( "print(f\"  Z-score for lower bound: {z_lower_b}\")<BR>\n" );
document.write( "print(f\"  Z-score for upper bound: {z_upper_b}\")<BR>\n" );
document.write( "print(f\"  Probability: {probability_b}\")\r<BR>\n" );
document.write( "\n" );
document.write( "# --- Part c: Probability for n = 500 ---<BR>\n" );
document.write( "n_c = 500<BR>\n" );
document.write( "std_dev_p_bar_c = math.sqrt((p * (1 - p)) / n_c)\r<BR>\n" );
document.write( "\n" );
document.write( "# Check for normality assumption<BR>\n" );
document.write( "np_c = n_c * p<BR>\n" );
document.write( "n_1_minus_p_c = n_c * (1 - p)\r<BR>\n" );
document.write( "\n" );
document.write( "# Calculate Z-scores<BR>\n" );
document.write( "z_lower_c = (lower_bound_b - p) / std_dev_p_bar_c<BR>\n" );
document.write( "z_upper_c = (upper_bound_b - p) / std_dev_p_bar_c\r<BR>\n" );
document.write( "\n" );
document.write( "# Calculate probability using CDF<BR>\n" );
document.write( "probability_c = norm.cdf(z_upper_c) - norm.cdf(z_lower_c)\r<BR>\n" );
document.write( "\n" );
document.write( "print(f\"\nPart c (n={n_c}):\")<BR>\n" );
document.write( "print(f\"  Standard deviation of p_bar: {std_dev_p_bar_c}\")<BR>\n" );
document.write( "print(f\"  np >= 5: {np_c} >= 5 (True)\")<BR>\n" );
document.write( "print(f\"  n(1-p) >= 5: {n_1_minus_p_c} >= 5 (True)\")<BR>\n" );
document.write( "print(f\"  Z-score for lower bound: {z_lower_c}\")<BR>\n" );
document.write( "print(f\"  Z-score for upper bound: {z_upper_c}\")<BR>\n" );
document.write( "print(f\"  Probability: {probability_c}\")<BR>\n" );
document.write( "```\r<BR>\n" );
document.write( "\n" );
document.write( "```text?code_stdout&code_event_index=2<BR>\n" );
document.write( "Part a (n=1000):<BR>\n" );
document.write( "  Mean of p_bar: 0.25<BR>\n" );
document.write( "  Standard deviation of p_bar: 0.013693063937629153<BR>\n" );
document.write( "  np >= 5: 250.0 >= 5 (True)<BR>\n" );
document.write( "  n(1-p) >= 5: 750.0 >= 5 (True)\r<BR>\n" );
document.write( "\n" );
document.write( "Part b (n=1000):<BR>\n" );
document.write( "  Range: [0.22, 0.28]<BR>\n" );
document.write( "  Z-score for lower bound: -2.1908902300206643<BR>\n" );
document.write( "  Z-score for upper bound: 2.1908902300206665<BR>\n" );
document.write( "  Probability: 0.9715402630836895\r<BR>\n" );
document.write( "\n" );
document.write( "Part c (n=500):<BR>\n" );
document.write( "  Standard deviation of p_bar: 0.019364916731037084<BR>\n" );
document.write( "  np >= 5: 125.0 >= 5 (True)<BR>\n" );
document.write( "  n(1-p) >= 5: 375.0 >= 5 (True)<BR>\n" );
document.write( "  Z-score for lower bound: -1.5491933384829668<BR>\n" );
document.write( "  Z-score for upper bound: 1.5491933384829681<BR>\n" );
document.write( "  Probability: 0.878664749641518\r<BR>\n" );
document.write( "\n" );
document.write( "```\r<BR>\n" );
document.write( "\n" );
document.write( "The population proportion of late flights is $p = 0.25$.\r<BR>\n" );
document.write( "\n" );
document.write( "**a. Show the sampling distribution of $\\bar{p}$, the proportion of late flights in a sample of 1000 flights.**\r<BR>\n" );
document.write( "\n" );
document.write( "For a sample of size $n=1000$:\r<BR>\n" );
document.write( "\n" );
document.write( "  * **Mean of the sampling distribution ($E(\\bar{p})$):**<BR>\n" );
document.write( "    $E(\\bar{p}) = p = 0.25$\r<BR>\n" );
document.write( "\n" );
document.write( "  * **Standard deviation of the sampling distribution ($\\sigma\_{\\bar{p}}$):**<BR>\n" );
document.write( "    $\\sigma\_{\\bar{p}} = \\sqrt{\\frac{p(1-p)}{n}} = \\sqrt{\\frac{0.25(1-0.25)}{1000}} = \\sqrt{\\frac{0.25 \\times 0.75}{1000}} = \\sqrt{\\frac{0.1875}{1000}} = \\sqrt{0.0001875} \\approx 0.01369$\r<BR>\n" );
document.write( "\n" );
document.write( "  * **Shape of the sampling distribution:**<BR>\n" );
document.write( "    To determine the shape, we check if $np \\ge 5$ and $n(1-p) \\ge 5$.<BR>\n" );
document.write( "    $np = 1000 \\times 0.25 = 250$<BR>\n" );
document.write( "    $n(1-p) = 1000 \\times 0.75 = 750$<BR>\n" );
document.write( "    Since both $250 \\ge 5$ and $750 \\ge 5$, the sampling distribution of $\\bar{p}$ is **approximately normal**.\r<BR>\n" );
document.write( "\n" );
document.write( "**Summary of Sampling Distribution for $n=1000$:**\r<BR>\n" );
document.write( "\n" );
document.write( "  * **Mean:** $0.25$<BR>\n" );
document.write( "  * **Standard Deviation:** Approximately $0.01369$<BR>\n" );
document.write( "  * **Shape:** Approximately Normal\r<BR>\n" );
document.write( "\n" );
document.write( "**b. What is the probability that the sample proportion will be within .03 of the population proportion if a sample of size 1000 is selected?**\r<BR>\n" );
document.write( "\n" );
document.write( "We want to find $P(p - 0.03 \\le \\bar{p} \\le p + 0.03)$, which is $P(0.25 - 0.03 \\le \\bar{p} \\le 0.25 + 0.03) = P(0.22 \\le \\bar{p} \\le 0.28)$.\r<BR>\n" );
document.write( "\n" );
document.write( "We use the Z-score formula: $Z = \\frac{\\bar{p} - p}{\\sigma\_{\\bar{p}}}$\r<BR>\n" );
document.write( "\n" );
document.write( "For $\\bar{p} = 0.22$:<BR>\n" );
document.write( "$Z\_{lower} = \\frac{0.22 - 0.25}{0.01369} = \\frac{-0.03}{0.01369} \\approx -2.191$\r<BR>\n" );
document.write( "\n" );
document.write( "For $\\bar{p} = 0.28$:<BR>\n" );
document.write( "$Z\_{upper} = \\frac{0.28 - 0.25}{0.01369} = \\frac{0.03}{0.01369} \\approx 2.191$\r<BR>\n" );
document.write( "\n" );
document.write( "Now, we find the probability $P(-2.191 \\le Z \\le 2.191)$:<BR>\n" );
document.write( "$P(Z \\le 2.191) - P(Z \\le -2.191)$<BR>\n" );
document.write( "Using a standard normal table or calculator:<BR>\n" );
document.write( "$P(Z \\le 2.191) \\approx 0.9857$<BR>\n" );
document.write( "$P(Z \\le -2.191) \\approx 0.0143$<BR>\n" );
document.write( "Probability $= 0.9857 - 0.0143 = 0.9714$.\r<BR>\n" );
document.write( "\n" );
document.write( "The probability that the sample proportion will be within $0.03$ of the population proportion for a sample of size 1000 is approximately $\\boxed{0.9715}$.\r<BR>\n" );
document.write( "\n" );
document.write( "**c. Answer part (b) for a sample of 500 flights.**\r<BR>\n" );
document.write( "\n" );
document.write( "For a sample of size $n=500$:\r<BR>\n" );
document.write( "\n" );
document.write( "  * **Standard deviation of the sampling distribution ($\\sigma\_{\\bar{p}}$):**<BR>\n" );
document.write( "    $\\sigma\_{\\bar{p}} = \\sqrt{\\frac{p(1-p)}{n}} = \\sqrt{\\frac{0.25(1-0.25)}{500}} = \\sqrt{\\frac{0.1875}{500}} = \\sqrt{0.000375} \\approx 0.01936$\r<BR>\n" );
document.write( "\n" );
document.write( "We want to find $P(0.22 \\le \\bar{p} \\le 0.28)$ with the new standard deviation.\r<BR>\n" );
document.write( "\n" );
document.write( "For $\\bar{p} = 0.22$:<BR>\n" );
document.write( "$Z\_{lower} = \\frac{0.22 - 0.25}{0.01936} = \\frac{-0.03}{0.01936} \\approx -1.549$\r<BR>\n" );
document.write( "\n" );
document.write( "For $\\bar{p} = 0.28$:<BR>\n" );
document.write( "$Z\_{upper} = \\frac{0.28 - 0.25}{0.01936} = \\frac{0.03}{0.01936} \\approx 1.549$\r<BR>\n" );
document.write( "\n" );
document.write( "Now, we find the probability $P(-1.549 \\le Z \\le 1.549)$:<BR>\n" );
document.write( "$P(Z \\le 1.549) - P(Z \\le -1.549)$<BR>\n" );
document.write( "Using a standard normal table or calculator:<BR>\n" );
document.write( "$P(Z \\le 1.549) \\approx 0.9393$<BR>\n" );
document.write( "$P(Z \\le -1.549) \\approx 0.0607$<BR>\n" );
document.write( "Probability $= 0.9393 - 0.0607 = 0.8786$.\r<BR>\n" );
document.write( "\n" );
document.write( "The probability that the sample proportion will be within $0.03$ of the population proportion for a sample of size 500 is approximately $\\boxed{0.8787}$. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );