Normal Distribution

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Answer Key & Explanations

Normal Distribution · Grade 11 · Worksheet 3

1. Olivia is a quality control analyst at a battery manufacturing plant. The lifespan of the batteries produced follows a normal distribution with a mean of 450 hours and a standard deviation of 15 hours. The company guarantees that batteries will last at least 420 hours. Using the 68-95-99.7 rule, what percentage of batteries will fail to meet this guarantee (i.e., have a lifespan less than 420 hours)? Answer: 2.5% Solution: Find the z-score. z = (420 - 450) / 15 = -30 / 15 = -2. So 420 hours is 2 standard deviations below the mean.
   Full step-by-step solution

   Step 1: Find the z-score. z = (420 - 450) / 15 = -30 / 15 = -2. So 420 hours is 2 standard deviations below the mean. Step 2: Using the 68-95-99.7 rule: 95% of data falls within 2 standard deviations of the mean (between 420 and 480 hours). This means 5% of data falls outside this range. Step 3: The tails are symmetric, so half of that 5% is below 420 hours and half is above 480 hours. Step 4: Percentage below 420 hours = 5% / 2 = 2.5%. Therefore, 2.5% of batteries will fail to meet the guarantee.

2. Charlotte is analyzing the annual snowfall amounts in a mountain town. The snowfall amounts follow a normal distribution with a mean of 172 cm and a standard deviation of 27 cm. In a particularly snowy year, the town recorded 226 cm of snow. Approximately what percentage of years would be expected to have snowfall amounts greater than 226 cm? Use the 68-95-99.7 rule to determine your answer. Answer: 2.5% Solution: Calculate the z-score (number of standard deviations from the mean). z = (x - mu) / sigma = (226 - 172) / 27 = 54 / 27 = 2 The value 226 cm is exactly 2 standard deviations above the mean.
   Full step-by-step solution

   Step 1: Calculate the z-score (number of standard deviations from the mean). z = (x - mu) / sigma = (226 - 172) / 27 = 54 / 27 = 2 The value 226 cm is exactly 2 standard deviations above the mean. Step 2: Apply the 68-95-99.7 rule. This rule states that approximately 95% of the data falls within 2 standard deviations of the mean (between mu - 2sigma and mu + 2sigma). Step 3: Calculate the percentage outside this range. The remaining 5% of data falls outside the range of 2 standard deviations from the mean. Since the normal distribution is symmetric, half of this 5% (2.5%) is above mu + 2sigma, and half (2.5%) is below mu - 2sigma. Step 4: State the answer. Therefore, approximately 2.5% of years are expected to have snowfall amounts greater than 226 cm. The answer is 2.5%.

3. Emma is analyzing the annual salaries of employees at a mid-sized technology firm. The salaries follow a normal distribution with a mean of $63,000 and a standard deviation of $9,000. The company's CEO announces that any employee whose salary is at or above the 84th percentile will receive a special bonus. Using the 68-95-99.7 rule, what is the minimum salary an employee must earn to qualify for the bonus? Answer: $72,000 Solution: Recall the 68-95-99.7 rule for normal distributions. About 68% of data falls within 1 standard deviation of the mean, 95% within 2 standard deviations, and 99.7% within 3 standard deviations.
   Full step-by-step solution

   Step 1: Recall the 68-95-99.7 rule for normal distributions. About 68% of data falls within 1 standard deviation of the mean, 95% within 2 standard deviations, and 99.7% within 3 standard deviations. Step 2: The mean is $63,000 and the standard deviation is $9,000. One standard deviation above the mean is: $63,000 + $9,000 = $72,000. Step 3: In a normal distribution, the area to the left of the mean is 50%. Adding the 34% (half of 68%) that lies between the mean and one standard deviation above gives: 50% + 34% = 84%. So the 84th percentile is exactly one standard deviation above the mean. Step 4: Therefore, the minimum salary to qualify for the bonus is $72,000. The answer is $72,000.

4. A university's psychology department is studying reaction times to visual stimuli. The reaction times follow a normal distribution. Researchers find that 20% of participants have reaction times shorter than 180 milliseconds, while 15% have reaction times longer than 240 milliseconds. Calculate the mean and standard deviation of the reaction time distribution. Answer: mean = 207.6, standard deviation = 20.8 Solution: Convert the given percentiles to z-scores using the standard normal distribution table. - 20th percentile corresponds to z = -0.8416 - 85th percentile (since 100% - 15% = 85%) corresponds to z = 1.0364 Set up equations using the formula: value = mean + (z-score × standard deviation) For the 20th…
   Full step-by-step solution

   Step 1: Convert the given percentiles to z-scores using the standard normal distribution table. - 20th percentile corresponds to z = -0.8416 - 85th percentile (since 100% - 15% = 85%) corresponds to z = 1.0364 Step 2: Set up equations using the formula: value = mean + (z-score × standard deviation) For the 20th percentile: 180 = μ + (-0.8416)σ For the 85th percentile: 240 = μ + (1.0364)σ Step 3: Solve the system of equations. From the first equation: μ = 180 + 0.8416σ Substitute into the second equation: 240 = (180 + 0.8416σ) + 1.0364σ 240 = 180 + 1.878σ 60 = 1.878σ σ = 60 / 1.878 = 31.96 Step 4: Calculate the mean. μ = 180 + 0.8416 × 31.96 = 180 + 26.90 = 206.90 The mean reaction time is 206.9 milliseconds and the standard deviation is 32.0 milliseconds.

5. Isabella is analyzing the annual rainfall in a coastal city. The annual rainfall amounts follow a normal distribution with a mean of 850 mm and a standard deviation of 35 mm. The city government considers a year to be a 'drought year' if the annual rainfall is below 780 mm. Based on this normal distribution, what percentage of years would be classified as drought years? Use the 68-95-99.7 rule to estimate your answer. Answer: Approximately 2.5% Solution: Identify the mean and standard deviation. Mean (μ) = 850 mm, standard deviation (σ) = 35 mm. Determine how many standard deviations 780 mm is below the mean.
   Full step-by-step solution

   Step 1: Identify the mean and standard deviation. Mean (μ) = 850 mm, standard deviation (σ) = 35 mm. Step 2: Determine how many standard deviations 780 mm is below the mean. Difference = 850 - 780 = 70 mm. Number of standard deviations = 70 / 35 = 2. So, 780 mm is 2 standard deviations below the mean. Step 3: Use the 68-95-99.7 rule. This rule states: - 68% of data falls within 1 standard deviation of the mean. - 95% of data falls within 2 standard deviations of the mean. - 99.7% of data falls within 3 standard deviations of the mean. Step 4: Since 95% of data falls within 2 standard deviations of the mean, this means 5% of data falls outside that range (both tails combined). Because the normal distribution is symmetric, half of that 5% (i.e., 2.5%) lies in the lower tail (below 2 standard deviations below the mean). Step 5: Therefore, the percentage of years with rainfall below 780 mm is approximately 2.5%. Answer: Approximately 2.5%.

6. Noah is analyzing the weights of adult male African elephants in a reserve. The weights follow a normal distribution with a mean of 5000 kg and a standard deviation of 400 kg. Using the 68-95-99.7 rule, what percentage of these elephants weigh between 4200 kg and 5800 kg? Answer: 95% Solution: Identify the mean (μ = 5000 kg) and standard deviation (σ = 400 kg). Find how many standard deviations 4200 kg is from the mean: (5000 - 4200) / 400 = 800 / 400 = 2 standard deviations below the mean.
   Full step-by-step solution

   Step 1: Identify the mean (μ = 5000 kg) and standard deviation (σ = 400 kg). Step 2: Find how many standard deviations 4200 kg is from the mean: (5000 - 4200) / 400 = 800 / 400 = 2 standard deviations below the mean. Step 3: Find how many standard deviations 5800 kg is from the mean: (5800 - 5000) / 400 = 800 / 400 = 2 standard deviations above the mean. Step 4: The 68-95-99.7 rule states that approximately 95% of data in a normal distribution lies within 2 standard deviations of the mean. Step 5: Therefore, about 95% of the elephants weigh between 4200 kg and 5800 kg. Answer: 95%