Normal Distribution Estimation

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Answer Key & Explanations

Normal Distribution Estimation · Grade 11 · Worksheet 1

1. log₂(32) - log₂(2) = ? Answer: 4 Solution: Step 1: Apply the logarithmic subtraction property: log₂(32) - log₂(2) = log₂(32/2) Step 2: Simplify the division: 32/2 = 16 Step 3: Evaluate log₂(16): 2^4 = 16 Step 4: Therefore, log₂(16) = 4 The answer is 4.
   Full step-by-step solution

   Step 1: Apply the logarithmic subtraction property: log₂(32) - log₂(2) = log₂(32/2) Step 2: Simplify the division: 32/2 = 16 Step 3: Evaluate log₂(16): 2^4 = 16 Step 4: Therefore, log₂(16) = 4 The answer is 4.

2. Charlotte's reaction times are normally distributed with μ=272 ms and σ=17 ms. Estimate the percentage of reaction times between 238 ms and 306 ms. Answer: 95 Solution: Calculate the z-score for 238 ms: z = (238 - 272)/17 = -34/17 = -2. Step 2: Calculate the z-score for 306 ms: z = (306 - 272)/17 = 34/17 = 2.
   Full step-by-step solution

   Step 1: Calculate the z-score for 238 ms: z = (238 - 272)/17 = -34/17 = -2. Step 2: Calculate the z-score for 306 ms: z = (306 - 272)/17 = 34/17 = 2. Step 3: According to the empirical rule, approximately 95% of data in a normal distribution lies within 2 standard deviations of the mean (between z = -2 and z = 2). Step 4: Therefore, approximately 95% of Charlotte's reaction times are between 238 ms and 306 ms. The answer is 95.

3. Hana is a marine ecologist studying the shell widths of a particular species of limpet found on rocky shores. She measures a large population and finds that the shell widths are normally distributed with a mean of 28 millimeters and a standard deviation of 5 millimeters. Estimate the percentage of limpets in this population that have a shell width greater than 33 millimeters. Answer: About 15.9% Solution: Identify the mean (μ = 28 mm) and standard deviation (σ = 5 mm). Calculate the z-score for 33 mm: z = (33 - 28) / 5 = 5 / 5 = 1.
   Full step-by-step solution

   Step 1: Identify the mean (μ = 28 mm) and standard deviation (σ = 5 mm). Step 2: Calculate the z-score for 33 mm: z = (33 - 28) / 5 = 5 / 5 = 1. Step 3: The percentage of limpets with shell widths greater than 33 mm corresponds to the area under the normal curve to the right of z = 1. Step 4: From the standard normal table, the area to the left of z = 1 is approximately 0.8413 (84.13%). Step 5: The area to the right is 1 - 0.8413 = 0.1587. Step 6: Convert to a percentage: 0.1587 × 100% = 15.87%. Step 7: Rounded to one decimal place, the estimated percentage is about 15.9%. The answer is about 15.9%.

4. Isabella's reaction times are normally distributed with μ=220 milliseconds and σ=25 milliseconds. Estimate the percentage of reaction times that are greater than 270 milliseconds. Answer: 2.28 Solution: Calculate the z-score for 270 milliseconds: z = (270 - 220)/25 = 50/25 = 2.00 The z-score of 2.00 means 270 is 2 standard deviations above the mean.
   Full step-by-step solution

   Step 1: Calculate the z-score for 270 milliseconds: z = (270 - 220)/25 = 50/25 = 2.00 Step 2: The z-score of 2.00 means 270 is 2 standard deviations above the mean. Step 3: Using the standard normal distribution table, the area to the left of z = 2.00 is 0.9772. Step 4: The area to the right (greater than 270) is 1 - 0.9772 = 0.0228. Step 5: Convert to a percentage: 0.0228 × 100 = 2.28%. The answer is 2.28.

5. Charlotte is a botanist studying the growth of a rare species of fern. She measures the heights of a large population of mature ferns and finds that the heights are normally distributed with a mean of 42 centimeters and a standard deviation of 7 centimeters. What percentage of mature ferns in the population are estimated to have a height between 35 centimeters and 56 centimeters? Answer: 81.85% Solution: Identify the parameters. Mean (mu) = 42 cm, standard deviation (sigma) = 7 cm. We want the percentage of ferns with heights between 35 cm and 56 cm.
   Full step-by-step solution

   Step 1: Identify the parameters. Mean (mu) = 42 cm, standard deviation (sigma) = 7 cm. We want the percentage of ferns with heights between 35 cm and 56 cm. Step 2: Convert the lower bound (35 cm) to a z-score: z = (35 - 42)/7 = (-7)/7 = -1.00. Step 3: Convert the upper bound (56 cm) to a z-score: z = (56 - 42)/7 = 14/7 = 2.00. Step 4: Find the area to the left of z = -1.00 using a standard normal table. Area = 0.1587. Step 5: Find the area to the left of z = 2.00. Area = 0.9772. Step 6: Subtract the smaller area from the larger area to get the area between: 0.9772 - 0.1587 = 0.8185. Step 7: Convert to a percentage: 0.8185 * 100% = 81.85%. The estimated percentage of mature ferns with heights between 35 cm and 56 cm is 81.85%.

6. Noah is a botanist studying the heights of a certain species of flowering plant. The heights are normally distributed with a mean of 46 centimeters and a standard deviation of 11 centimeters. Estimate the percentage of these plants that have a height between 36 centimeters and 56 centimeters. Answer: Approximately 63.68% Solution: Identify the mean (μ = 46 cm) and standard deviation (σ = 11 cm). Find the z-score for the lower bound, x = 36 cm: z1 = (36 - 46) / 11 = -10 / 11 ≈ -0.909.
   Full step-by-step solution

   Step 1: Identify the mean (μ = 46 cm) and standard deviation (σ = 11 cm). Step 2: Find the z-score for the lower bound, x = 36 cm: z1 = (36 - 46) / 11 = -10 / 11 ≈ -0.909. Step 3: Find the z-score for the upper bound, x = 56 cm: z2 = (56 - 46) / 11 = 10 / 11 ≈ 0.909. Step 4: Use the standard normal distribution table. The area to the left of z = 0.909 is approximately 0.8186. The area to the left of z = -0.909 is approximately 0.1814. Step 5: Subtract the smaller area from the larger area: 0.8186 - 0.1814 = 0.6372. Step 6: Convert to a percentage: 0.6372 × 100% ≈ 63.72%. Using more precise values (z = 0.9091), the area between is about 0.6368, so approximately 63.68% of the plants have heights between 36 cm and 56 cm. The answer is approximately 63.68%.

7. Emma's exam scores are normally distributed with μ=75 and σ=10. Estimate the percentage of students scoring between 65 and 85. Answer: 68 Solution: Find the z-scores for the boundaries. For 65: z = (65 - 75) / 10 = -10 / 10 = -1. For 85: z = (85 - 75) / 10 = 10 / 10 = 1.
   Full step-by-step solution

   Step 1: Find the z-scores for the boundaries. For 65: z = (65 - 75) / 10 = -10 / 10 = -1. For 85: z = (85 - 75) / 10 = 10 / 10 = 1. Step 2: The empirical rule states that approximately 68% of data in a normal distribution lies within 1 standard deviation of the mean (between z = -1 and z = 1). Step 3: Therefore, the estimated percentage of students scoring between 65 and 85 is 68%. The answer is 68.

8. Mason's test scores are normally distributed with μ=72 and σ=7. Estimate the percentage of students scoring between 58 and 86. Answer: 95 Solution: Calculate the z-score for 58: z = (58 - 72)/7 = -14/7 = -2 Calculate the z-score for 86: z = (86 - 72)/7 = 14/7 = 2 The scores 58 and 86 are 2 standard deviations below and above the mean, respectively.
   Full step-by-step solution

   Step 1: Calculate the z-score for 58: z = (58 - 72)/7 = -14/7 = -2 Step 2: Calculate the z-score for 86: z = (86 - 72)/7 = 14/7 = 2 Step 3: The scores 58 and 86 are 2 standard deviations below and above the mean, respectively. Step 4: According to the empirical rule, approximately 95% of data in a normal distribution falls within 2 standard deviations of the mean. Step 5: Therefore, approximately 95% of students score between 58 and 86. The answer is 95.