document.write( "Question 1167859: Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P81, the 81-percentile. This is the temperature reading separating the bottom 81% from the top 19%.
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document.write( "p81=? \n" );
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Algebra.Com's Answer #852085 by CPhill(1959)![]() ![]() You can put this solution on YOUR website! ```python?code_reference&code_event_index=2 \n" ); document.write( "from scipy.stats import norm\r \n" ); document.write( "\n" ); document.write( "# Given parameters \n" ); document.write( "mean = 0 # degrees Celsius \n" ); document.write( "std_dev = 1.00 # degrees Celsius \n" ); document.write( "percentile = 81 # as a percentage\r \n" ); document.write( "\n" ); document.write( "# Convert percentile to a decimal for the PPF function \n" ); document.write( "probability = percentile / 100\r \n" ); document.write( "\n" ); document.write( "# Find the Z-score corresponding to the 81st percentile \n" ); document.write( "# norm.ppf(q, loc=0, scale=1) returns the inverse of the CDF, i.e., the value x \n" ); document.write( "# such that P(X <= x) = q, for a normal distribution with mean=loc and std_dev=scale. \n" ); document.write( "z_score_p81 = norm.ppf(probability)\r \n" ); document.write( "\n" ); document.write( "# Convert the Z-score back to the temperature reading (x) \n" ); document.write( "# x = mean + Z * std_dev \n" ); document.write( "p81_temperature = mean + z_score_p81 * std_dev\r \n" ); document.write( "\n" ); document.write( "print(f\"The Z-score for the 81st percentile is: {z_score_p81:.3f}\") \n" ); document.write( "print(f\"P81, the 81-percentile temperature reading is: {p81_temperature:.2f}°C\") \n" ); document.write( "```\r \n" ); document.write( "\n" ); document.write( "```text?code_stdout&code_event_index=2 \n" ); document.write( "The Z-score for the 81st percentile is: 0.878 \n" ); document.write( "P81, the 81-percentile temperature reading is: 0.88°C\r \n" ); document.write( "\n" ); document.write( "```\r \n" ); document.write( "\n" ); document.write( "To find $P\_{81}$, the 81st percentile temperature reading, we use the properties of the normal distribution.\r \n" ); document.write( "\n" ); document.write( "Given:\r \n" ); document.write( "\n" ); document.write( " * Mean ($\\mu$) = $0^\\circ \\text{C}$ \n" ); document.write( " * Standard Deviation ($\\sigma$) = $1.00^\\circ \\text{C}$ \n" ); document.write( " * Percentile = 81st percentile, which means we are looking for the temperature reading ($x$) such that $81%$ of the readings are below it.\r \n" ); document.write( "\n" ); document.write( "-- end list -->\r \n" ); document.write( "\n" ); document.write( "1. **Find the Z-score corresponding to the 81st percentile:** \n" ); document.write( " The Z-score is the value on the standard normal distribution (mean = 0, standard deviation = 1) that corresponds to a cumulative probability of 0.81. \n" ); document.write( " Using a Z-table or statistical software, the Z-score for $P(Z \\le z) = 0.81$ is approximately $0.878$.\r \n" ); document.write( "\n" ); document.write( "2. **Convert the Z-score back to the temperature reading:** \n" ); document.write( " We use the formula: $x = \\mu + Z\\sigma$ \n" ); document.write( " $x = 0 + (0.878) \\times 1.00$ \n" ); document.write( " $x = 0.878$\r \n" ); document.write( "\n" ); document.write( "Rounding to two decimal places, $P\_{81} = 0.88^\\circ \\text{C}$.\r \n" ); document.write( "\n" ); document.write( "$P\_{81} = \\boxed{0.88}$ \n" ); document.write( " |