Normal Distribution Estimation

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Answer Key & Explanations

Normal Distribution Estimation · Grade 11 · Worksheet 2

1. Olivia is analyzing the reaction times (in milliseconds) of participants in a cognitive test. The reaction times are normally distributed with a mean of 240 milliseconds and a standard deviation of 30 milliseconds. If a participant is selected at random, estimate the percentage of participants whose reaction time is between 210 milliseconds and 270 milliseconds. Use the empirical rule (68-95-99.7 rule) to find your answer. Answer: 68% Solution: The mean is 240 ms and the standard deviation is 30 ms. Step 2: Calculate the z-scores: 210 ms is one standard deviation below the mean (240 - 30 = 210). 270 ms is one standard deviation above the mean (240 + 30 = 270).
   Full step-by-step solution

   Step 1: The mean is 240 ms and the standard deviation is 30 ms. Step 2: Calculate the z-scores: 210 ms is one standard deviation below the mean (240 - 30 = 210). 270 ms is one standard deviation above the mean (240 + 30 = 270). Step 3: The empirical rule states that approximately 68% of data in a normal distribution lies within one standard deviation of the mean. Step 4: Therefore, the estimated percentage of participants with reaction times between 210 ms and 270 ms is 68%. The answer is 68%.

2. Isabella is analyzing the test scores from a large standardized exam. The scores are normally distributed with a mean of 527 and a standard deviation of 72. She wants to estimate the percentage of all test takers who scored between 455 and 617. Use the normal distribution to estimate this population percentage. Answer: Approximately 74.86% Solution: Identify the mean (mu = 527) and standard deviation (sigma = 72). Convert the lower bound (455) to a z-score: z1 = (455 - 527) / 72 = -72 / 72 = -1.00.
   Full step-by-step solution

   Step 1: Identify the mean (mu = 527) and standard deviation (sigma = 72). Step 2: Convert the lower bound (455) to a z-score: z1 = (455 - 527) / 72 = -72 / 72 = -1.00. Step 3: Convert the upper bound (617) to a z-score: z2 = (617 - 527) / 72 = 90 / 72 = 1.25. Step 4: Look up the area to the left of z = -1.00 in the standard normal table. This is approximately 0.1587. Step 5: Look up the area to the left of z = 1.25 in the standard normal table. This is approximately 0.8944. Step 6: The area between z = -1.00 and z = 1.25 is the difference: 0.8944 - 0.1587 = 0.7357. Step 7: Convert the area to a percentage: 0.7357 * 100% = 73.57%. Step 8: For more precision, use a calculator or more detailed table: area to left of -1.00 = 0.1587, area to left of 1.25 = 0.8944, difference = 0.7357. Using a more precise calculation with a standard normal tool: area between z = -1 and z = 1.25 is approximately 0.7486. The final answer is approximately 74.86%.

3. Matiu is a marine biologist studying the shell lengths of a specific species of mussel along the coast of New Zealand. He finds that the shell lengths are normally distributed with a mean of 62 millimeters and a standard deviation of 8 millimeters. Estimate the percentage of mussels in the population that have shell lengths between 50 millimeters and 78 millimeters. Answer: 91.04% Solution: Identify the mean (mu = 62 mm) and standard deviation (sigma = 8 mm). Convert the lower bound (50 mm) to a z-score: z = (50 - 62) / 8 = -12 / 8 = -1.5.
   Full step-by-step solution

   Step 1: Identify the mean (mu = 62 mm) and standard deviation (sigma = 8 mm). Step 2: Convert the lower bound (50 mm) to a z-score: z = (50 - 62) / 8 = -12 / 8 = -1.5. Step 3: Convert the upper bound (78 mm) to a z-score: z = (78 - 62) / 8 = 16 / 8 = 2.0. Step 4: Use the standard normal distribution table or known percentages. The area between z = -1.5 and z = 2.0 is the area to the left of z = 2.0 minus the area to the left of z = -1.5. Step 5: Area to the left of z = 2.0 is approximately 0.9772. Area to the left of z = -1.5 is approximately 0.0668. Step 6: Subtract: 0.9772 - 0.0668 = 0.9104. Step 7: Convert to percentage: 0.9104 * 100% = 91.04%. The answer is 91.04%.

4. Sophia's reaction times are normally distributed with μ=240 milliseconds and σ=30 milliseconds. Estimate the percentage of reaction times between 210 and 270 milliseconds. Answer: 68 Solution: Calculate the z-score for 210 milliseconds: z = (210 - 240)/30 = -30/30 = -1. Calculate the z-score for 270 milliseconds: z = (270 - 240)/30 = 30/30 = 1.
   Full step-by-step solution

   Step 1: Calculate the z-score for 210 milliseconds: z = (210 - 240)/30 = -30/30 = -1. Step 2: Calculate the z-score for 270 milliseconds: z = (270 - 240)/30 = 30/30 = 1. Step 3: The empirical rule states that approximately 68% of data in a normal distribution falls within 1 standard deviation of the mean (between z = -1 and z = 1). Step 4: Therefore, approximately 68% of Sophia's reaction times are between 210 and 270 milliseconds. The answer is 68.

5. A population of adult male African elephants has shoulder heights that are normally distributed with a mean of 327 cm and a standard deviation of 22 cm. Estimate the percentage of adult male elephants in this population whose shoulder height is between 305 cm and 349 cm. Round your answer to the nearest whole percent. Answer: 68% Solution: Find the z-scores for the lower and upper bounds. Lower bound: 305 cm. z = (305 - 327) / 22 = (-22) / 22 = -1.00 Upper bound: 349 cm.
   Full step-by-step solution

   Step 1: Find the z-scores for the lower and upper bounds. Lower bound: 305 cm. z = (305 - 327) / 22 = (-22) / 22 = -1.00 Upper bound: 349 cm. z = (349 - 327) / 22 = 22 / 22 = 1.00 Step 2: The interval from 305 cm to 349 cm corresponds to z-scores from -1 to 1. Step 3: According to the Empirical Rule, approximately 68% of data in a normal distribution lies within 1 standard deviation of the mean. Step 4: Therefore, the estimated percentage of adult male elephants with shoulder heights between 305 cm and 349 cm is 68%. The answer is 68%.

6. Noah's reaction times are normally distributed with μ = 196 ms and σ = 16 ms. Estimate the percentage of reaction times greater than 228 ms. Answer: 2.28 Solution: Calculate the z-score for 228 ms: z = (228 - 196) / 16 = 32 / 16 = 2.00. The area to the left of z = 2.00 in the standard normal distribution is approximately 0.9772 (from the standard normal table).
   Full step-by-step solution

   Step 1: Calculate the z-score for 228 ms: z = (228 - 196) / 16 = 32 / 16 = 2.00. Step 2: The area to the left of z = 2.00 in the standard normal distribution is approximately 0.9772 (from the standard normal table). Step 3: The area to the right (greater than 228 ms) is 1 - 0.9772 = 0.0228. Step 4: Convert to percentage: 0.0228 × 100 = 2.28%. The answer is 2.28.

7. A pharmaceutical company is testing a new medication and wants to estimate the proportion of patients who experience significant improvement. In a clinical trial with 320 randomly selected patients, 224 showed significant improvement. Construct a 95% confidence interval for the true population proportion of patients who would experience significant improvement with this medication. (Use z* = 1.96 for the critical value) Answer: 0.644 to 0.756 Solution: Confidence intervals for population proportions use the sample proportion as a point estimate and then add/subtract a margin of error that depends on the desired confidence level and sample size.
   Full step-by-step solution

   Confidence intervals for population proportions use the sample proportion as a point estimate and then add/subtract a margin of error that depends on the desired confidence level and sample size. The margin of error accounts for sampling variability and decreases with larger sample sizes, providing more precise estimates.