Normal Distribution

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Answer Key & Explanations

Normal Distribution · Grade 11 · Worksheet 3

1. Aroha is studying the distribution of test scores from a large Grade 11 mathematics competition. The scores are normally distributed with a mean of 52 points and a standard deviation of 9 points. On a normal distribution curve, shade the region that represents scores between 43 and 70 points. Using the 68-95-99.7 rule, approximately what percentage of all scores fall in this shaded region? Answer: 81.5 Solution: The mean is μ = 52 and the standard deviation is σ = 9. Calculate the z-score for the lower boundary, 43: z = (43 - 52) / 9 = -9 / 9 = -1. So 43 is 1 standard deviation below the mean.
   Full step-by-step solution

   Step 1: The mean is μ = 52 and the standard deviation is σ = 9. Calculate the z-score for the lower boundary, 43: z = (43 - 52) / 9 = -9 / 9 = -1. So 43 is 1 standard deviation below the mean. Calculate the z-score for the upper boundary, 70: z = (70 - 52) / 9 = 18 / 9 = 2. So 70 is 2 standard deviations above the mean. Step 2: Using the 68-95-99.7 rule, the percentage of data between z = -1 and z = 1 (within 1 standard deviation) is 68%. Therefore, the percentage from z = 0 to z = 1 is half of that, which is 34%. Also, the percentage of data between z = -2 and z = 2 (within 2 standard deviations) is 95%. Therefore, the percentage from z = 0 to z = 2 is half of that, which is 47.5%. Step 3: The region from z = -1 to z = 2 consists of two parts: from z = -1 to z = 0 (which is 34% because it is symmetric to z = 0 to z = 1) and from z = 0 to z = 2 (which is 47.5%). Add these: 34% + 47.5% = 81.5%. The answer is 81.5.

2. A factory produces precision components with lengths that follow a normal distribution. The quality control manager, Liam, finds that 15% of components are shorter than 48.2 mm and 20% are longer than 51.4 mm. Calculate the mean and standard deviation of the component lengths. Answer: Mean = 50.0 mm, Standard Deviation = 1.5 mm Solution: In normal distribution problems, we often use the property that any value x can be expressed as x = μ + zσ, where μ is the mean, σ is the standard deviation, and z is the z-score from the standard normal distribution.
   Full step-by-step solution

   In normal distribution problems, we often use the property that any value x can be expressed as x = μ + zσ, where μ is the mean, σ is the standard deviation, and z is the z-score from the standard normal distribution. When given percentile information, we can determine the corresponding z-scores and create a system of equations to solve for both parameters. This method is commonly used in quality control, test scoring, and other applications where we need to characterize a normally distributed population from limited sample information.

3. Hana is a botanist studying the height of a particular species of sunflower. She finds that the heights follow a normal distribution with a mean of 210 cm and a standard deviation of 15 cm. Using the 68-95-99.7 rule, what percentage of sunflowers are expected to be taller than 240 cm? Answer: 2.5% Solution: The mean is 210 cm and standard deviation is 15 cm. 240 cm is 240 - 210 = 30 cm above the mean. Since 30 / 15 = 2, 240 cm is 2 standard deviations above the mean.
   Full step-by-step solution

   Step 1: The mean is 210 cm and standard deviation is 15 cm. 240 cm is 240 - 210 = 30 cm above the mean. Since 30 / 15 = 2, 240 cm is 2 standard deviations above the mean. Step 2: Using the 68-95-99.7 rule: within 2 standard deviations of the mean (both above and below) lies 95% of the data. That means 100% - 95% = 5% of the data lies outside this range (both tails combined). Step 3: Because the normal distribution is symmetric, half of that 5% lies in the upper tail (above +2 standard deviations). So 5% / 2 = 2.5% of sunflowers are taller than 240 cm. The answer is 2.5%.

4. Charlotte is studying the distribution of diameters of mature maple trees in a large forest. The diameters are normally distributed with a mean of 28 inches and a standard deviation of 7 inches. On a normal distribution curve, shade the region that represents trees with diameters between 14 inches and 35 inches. Using the 68-95-99.7 rule, approximately what percentage of all trees fall in this shaded region? Answer: 81.5 Solution: The mean is μ = 28 and the standard deviation is σ = 7. Calculate the z-score for the lower boundary, 14: z = (14 - 28) / 7 = -14 / 7 = -2. So 14 is 2 standard deviations below the mean.
   Full step-by-step solution

   Step 1: The mean is μ = 28 and the standard deviation is σ = 7. Calculate the z-score for the lower boundary, 14: z = (14 - 28) / 7 = -14 / 7 = -2. So 14 is 2 standard deviations below the mean. Calculate the z-score for the upper boundary, 35: z = (35 - 28) / 7 = 7 / 7 = 1. So 35 is 1 standard deviation above the mean. Step 2: Using the 68-95-99.7 rule, the percentage of data between z = -2 and z = 2 (within 2 standard deviations) is 95%. Therefore, the percentage from z = -2 to z = 0 is half of that, which is 47.5%. Also, the percentage of data between z = -1 and z = 1 (within 1 standard deviation) is 68%. Therefore, the percentage from z = 0 to z = 1 is half of that, which is 34%. Step 3: The region from z = -2 to z = 1 consists of two parts: from z = -2 to z = 0 (which is 47.5%) and from z = 0 to z = 1 (which is 34%). Add these: 47.5% + 34% = 81.5%. The answer is 81.5.

5. Liam is a quality control engineer at a factory that produces steel rods. The lengths of the rods follow a normal distribution with a mean of 125 cm and a standard deviation of 7 cm. The factory has a quality standard that rods must be between 118 cm and 139 cm in length to be accepted. Using the 68-95-99.7 rule, what percentage of the rods produced will be accepted? Answer: 81.5% Solution: The mean is 125 cm and the standard deviation is 7 cm. Find the z-scores for the boundaries. For 118 cm: (125 - 118) / 7 = 7 / 7 = 1 standard deviation below the mean.
   Full step-by-step solution

   Step 1: The mean is 125 cm and the standard deviation is 7 cm. Step 2: Find the z-scores for the boundaries. For 118 cm: (125 - 118) / 7 = 7 / 7 = 1 standard deviation below the mean. For 139 cm: (139 - 125) / 7 = 14 / 7 = 2 standard deviations above the mean. Step 3: Apply the 68-95-99.7 rule. Within 1 standard deviation of the mean (118 cm to 132 cm) is 68% of the data. Within 2 standard deviations of the mean (111 cm to 139 cm) is 95% of the data. Step 4: The interval from 118 cm to 139 cm includes all data within 1 standard deviation (68%) plus the data between 1 and 2 standard deviations above the mean. The data between 1 and 2 standard deviations above the mean is half of (95% - 68%) = half of 27% = 13.5%. Step 5: Total percentage accepted = 68% + 13.5% = 81.5%. The answer is 81.5%.

6. Noah is analyzing the distribution of scores from a large Grade 11 mathematics contest. The scores are normally distributed with a mean of 61 and a standard deviation of 11. A histogram of the scores shows a symmetric bell curve. Shade the region under the normal distribution curve that represents scores between 50 and 72. What percentage of the scores fall in this shaded region? Use the 68-95-99.7 rule and give your answer as a whole number. Answer: 68 Solution: The mean is 61 and the standard deviation is 11. Find how far 50 is from the mean: 61 - 50 = 11. This is exactly 1 standard deviation below the mean (since 11 / 11 = 1).
   Full step-by-step solution

   Step 1: The mean is 61 and the standard deviation is 11. Step 2: Find how far 50 is from the mean: 61 - 50 = 11. This is exactly 1 standard deviation below the mean (since 11 / 11 = 1). Step 3: Find how far 72 is from the mean: 72 - 61 = 11. This is exactly 1 standard deviation above the mean (since 11 / 11 = 1). Step 4: The interval from 50 to 72 represents all data within 1 standard deviation of the mean (from mu - 1 sigma to mu + 1 sigma). Step 5: According to the 68-95-99.7 rule for normal distributions, approximately 68% of the data falls within 1 standard deviation of the mean. Step 6: Therefore, the percentage of scores between 50 and 72 is approximately 68%. The answer is 68.