Statistics Quiz
Practice Using Normal Distributions To Estimate Populations in Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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A machine fills cereal boxes, and the fill weights are approximately normal. The mean fill weight is μ=500 g with standard deviation σ=10 g. Approximately what percent of boxes have fill weights between 490 g and 510 g? (Use the empirical rule.)
This quiz focuses on Using Normal Distributions To Estimate Populations, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics.
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A machine fills cereal boxes, and the fill weights are approximately normal. The mean fill weight is μ=500 g with standard deviation σ=10 g. Approximately what percent of boxes have fill weights between 490 g and 510 g? (Use the empirical rule.)
Explanation: We're using a normal model to estimate the percentage of cereal boxes with fill weights between 490 g and 510 g. These cutoffs are 10 g below and above the mean 500 g, so with SD 10 g, they are -1 SD and +1 SD. The empirical rule indicates about 68% of data within 1 SD of the mean. Therefore, approximately 68% of boxes are in this range. This correct answer reflects the central area between symmetric cutoffs. A common error is mistaking it for 95% (2 SD), but count the SDs carefully—it's 1 SD here. To visualize, sketch mean at 500, tick -1 SD at 490 and +1 SD at 510, identify the middle region, and estimate 68%.
Scores on a standardized exam are designed to be approximately normal. Suppose scores have mean μ=70 and standard deviation σ=10. Approximately what percent of students score between 60 and 80? (Use the empirical rule.)
Explanation: This problem involves using a normal model to estimate the percentage of exam scores between 60 and 80. The cutoffs are 60 (10 below mean 70) and 80 (10 above), so with SD 10, these are -1 SD and +1 SD from the mean. The empirical rule states that about 68% of data fall within 1 standard deviation of the mean. Thus, the area between -1 SD and +1 SD is approximately 68%. This matches the correct answer for the percentage between these symmetric cutoffs around the mean. A misconception might be confusing it with 95% for 2 SD, but here it's only 1 SD. For strategy, sketch the mean at 70, mark -1 SD at 60 and +1 SD at 80, note the central area between them, and estimate as 68%.
A factory produces metal rods whose lengths are roughly symmetric and bell-shaped, so a normal model is appropriate. Rod lengths have mean μ=50 cm and standard deviation σ=2 cm. Approximately what percent of rods are longer than 54 cm? (Use the empirical rule approximation.)
Explanation: We're using a normal model to estimate the percentage of metal rods longer than 54 cm, given the bell-shaped distribution. The cutoff of 54 cm is 4 cm above the mean of 50 cm, and with a standard deviation of 2 cm, this is 4/2 = 2 standard deviations above the mean. According to the empirical rule, about 95% of the data fall within 2 standard deviations of the mean, leaving 5% in the tails, or 2.5% in each tail. This area translates to approximately 2.5% of rods being longer than 54 cm. The correct answer matches because it focuses on the upper tail beyond +2 SD, where longer rods are. A common misconception is thinking the entire 5% outside 2 SD applies to one tail, but it's split equally for symmetric normals. To apply this, sketch the mean at 50, mark ticks at +1 SD (52) and +2 SD (54), identify the right tail, and estimate the area as 2.5%.
A quality-control measurement is approximately normal. The measurement has mean μ=200 units and standard deviation σ=5 units. Approximately what percent of measurements are below 190 units? (Use the empirical rule.)
Explanation: We're using a normal model to estimate the percentage of measurements below 190 units. Cutoff 190 is 10 below mean 200, with SD 5, so 10/5 = -2 SD. Empirical rule: 95% within 2 SD, leaving 2.5% below -2 SD. Thus, about 2.5% are below 190. This fits the lower tail direction for below. Common error: using 5% for both tails combined, but it's 2.5% per tail. Sketch mean at 200, mark -1 SD 195 and -2 at 190, focus left tail, estimate 2.5%.
The diameters of ball bearings from a well-controlled process are roughly symmetric and bell-shaped, so assume a normal model. Diameters have mean μ=10.0 mm and standard deviation σ=0.2 mm. Approximately what percent of bearings have diameter greater than 10.4 mm? (Use the empirical rule.)
Explanation: This involves a normal model to estimate the percentage of ball bearings with diameters greater than 10.4 mm. The cutoff 10.4 mm is 0.4 mm above mean 10.0 mm, and with SD 0.2 mm, that's 0.4/0.2 = 2 SD above. Empirical rule: 95% within 2 SD, so 2.5% above +2 SD. This gives about 2.5% greater than 10.4 mm. The answer matches the upper tail direction for greater diameters. Misconception: some might use 5% for the tail, but it's half of the 5% outside. Strategy: sketch mean at 10.0, mark +1 SD at 10.2 and +2 at 10.4, target the right tail, estimate 2.5%.
IQ scores are often modeled as approximately normal. Suppose IQ has mean μ=100 and standard deviation σ=15. Approximately what percent of people have IQ between 85 and 115? (Use the empirical rule.)
Explanation: This problem uses a normal model to estimate the percentage of people with IQ between 85 and 115. Cutoffs: 85 is 15 below mean 100, 115 is 15 above, with SD 15, so -1 SD to +1 SD. Empirical rule: 68% within 1 SD. Thus, about 68% have IQ in this range. The answer fits the between direction for central symmetric bounds. Misconception: adding tails incorrectly, like thinking 84% for one side, but it's central. Strategy: sketch mean 100, mark -1 SD 85 and +1 SD 115, focus on area between, estimate 68%.
Adult heights in a large population are roughly bell-shaped, so using a normal model is reasonable. Suppose heights have mean μ=170 cm and standard deviation σ=6 cm. Approximately what percent of adults are shorter than 164 cm? (Use the empirical rule.)
Explanation: Here, we're applying a normal model to estimate the percentage of adults shorter than 164 cm in a bell-shaped height distribution. The cutoff 164 cm is 6 cm below the mean 170 cm, and with SD 6 cm, that's exactly -1 standard deviation. By the empirical rule, 68% are within 1 SD, so half of that (34%) is between the mean and -1 SD, leaving 50% below the mean total, but subtracting 34% gives 16% below -1 SD. This approximates to 16% shorter than 164 cm. The answer aligns with the lower tail below -1 SD. People might miscount by thinking it's 32% or forgetting to subtract from 50%, but it's correctly 16%. Sketch the mean at 170, mark -1 SD at 164, focus on the left tail, and estimate the area as 16%.
Scores on a standardized math assessment are modeled well by a normal distribution (approximately symmetric and bell-shaped). The scores have mean μ=70 and standard deviation σ=10. Approximately what percent of students score between 60 and 80? (Use the empirical rule.)
Explanation: This problem involves using a normal model to estimate the percentage of students scoring between 60 and 80 on a math assessment with mean 70 and SD 10. The cutoffs are 60 (70 - 10 = -1 SD) and 80 (70 + 10 = +1 SD), so we're looking between -1 SD and +1 SD from the mean. The empirical rule states that approximately 68% of data lie within 1 SD of the mean. This area directly translates to about 68% of students in that range. The correct choice reflects the between direction for a symmetric 1-SD interval around the mean. Students often miscount SDs or confuse it with the 2-SD rule of 95%, leading to wrong answers. For similar problems, sketch the mean, mark SD intervals like 60, 70, 80, identify the central area between tails, and estimate 68%.
The time it takes a printer to complete a certain job is approximately normal (based on many runs, the distribution is roughly symmetric and bell-shaped). The times have mean μ=30 seconds and standard deviation σ=2 seconds. Approximately what percent of jobs take between 28 and 32 seconds? (Use the empirical rule.)
Explanation: We're estimating with a normal model the percentage of printer jobs taking between 28 and 32 seconds, mean 30 seconds, SD 2 seconds. The interval is from 28 (-1 SD) to 32 (+1 SD). The empirical rule says 68% are within 1 SD of the mean. So, approximately 68% of jobs fall in this range. This fits the between direction for a 1-SD symmetric interval. Common errors include miscounting to 2 SDs for 95% or focusing on one tail like 16%. To apply elsewhere, sketch mean at 30, mark SD at 28 and 32, identify the central area, and estimate 68%.
A set of exam scores is modeled as approximately normal (roughly symmetric and bell-shaped) with mean μ=500 and standard deviation σ=100. Approximately what percent of scores are below 700? (Use the empirical rule.)
Explanation: Using a normal model, estimate percentage of exam scores below 700, mean 500, SD 100. Cutoff 700 is +2 SD (200 / 100 = 2). Area below +2 SD: 50% below mean plus 47.5% from mean to +2 SD (34% to +1 SD + 13.5% to +2 SD), totaling 97.5%. Approximately 97.5% are below 700. This fits the below direction up to +2 SD. Misconception: thinking it's a tail like 2.5% instead of cumulative. Sketch mean 500, mark +1 SD 600, +2 700, add left half and right up to cutoff, estimate 97.5%.
A company reports that the diameters of its ball bearings are approximately normal (roughly symmetric and bell-shaped). The diameters have mean μ=10.00 mm and standard deviation σ=0.02 mm. Approximately what percent of ball bearings have diameter above 10.02 mm? (Use the empirical rule.)
Explanation: We're using a normal model to estimate the percentage of ball bearings with diameters above 10.02 mm, given a mean of 10.00 mm and standard deviation of 0.02 mm. The cutoff of 10.02 mm is 0.02 mm above the mean, which is exactly 1 standard deviation unit (since 0.02 / 0.02 = 1). According to the empirical rule, about 68% of data fall within 1 SD of the mean, leaving 32% outside, or 16% in each tail. This 16% in the upper tail translates to approximately 16% of ball bearings having diameters above 10.02 mm. The correct answer matches the upper tail direction, estimating the percentage above the mean plus 1 SD. A common misconception is thinking this is the lower tail or miscounting it as 2 SDs, which would incorrectly suggest 2.5%. To apply this generally, sketch the mean at 10.00, mark SD ticks at 9.98, 10.02, etc., decide it's the upper tail, and estimate the area as 16%.
A large set of reading assessment scores is approximately normal (bell-shaped). Scores have mean μ=500 and standard deviation σ=100. Approximately what percent of students score between 300 and 700? (Use the empirical rule.)
Explanation: We're finding what percent of students score between 300 and 700. With mean 500 and SD = 100, let's convert to z-scores: 300 is (300-500)/100 = -2 SDs below the mean, and 700 is (700-500)/100 = +2 SDs above the mean. The empirical rule tells us that approximately 95% of data in a normal distribution falls within 2 standard deviations of the mean. Therefore, about 95% of students score between 300 and 700. This leaves 5% total in the tails (2.5% below 300 and 2.5% above 700). Students might mistakenly choose 99.7% thinking of 3 SDs, but we're only going out 2 SDs from the mean. Remember the key benchmarks: 68% within 1 SD, 95% within 2 SD, and 99.7% within 3 SD.
The time (in minutes) it takes trained workers to complete a routine assembly task is roughly bell-shaped, so a normal model is reasonable. Completion times have mean μ=30 and standard deviation σ=4. Approximately what percent of workers finish the task between 26 and 34 minutes? (Use the empirical rule.)
Explanation: We're finding what percent of workers finish between 26 and 34 minutes using a normal model. With mean 30 minutes and SD = 4 minutes, let's convert to z-scores: 26 is (26-30)/4 = -1 SD below the mean, and 34 is (34-30)/4 = +1 SD above the mean. The empirical rule tells us that approximately 68% of data in a normal distribution falls within 1 standard deviation of the mean. Therefore, about 68% of workers complete the task between 26 and 34 minutes. Some students might confuse this with 34% (which is just one side from mean to 1 SD) or 95% (which is within 2 SDs). To solve these efficiently, memorize the key percentages: 68% within 1 SD, 95% within 2 SD, and 99.7% within 3 SD.
The diameters of ball bearings produced by a machine are approximately normally distributed. The diameters have mean μ=10.00 mm and standard deviation σ=0.05 mm. Approximately what percent of ball bearings have diameter above 10.10 mm? (Use the empirical rule.)
Explanation: We need to find what percent of ball bearings have diameter above 10.10 mm. With mean 10.00 mm and SD = 0.05 mm, let's find how many SDs away 10.10 is: (10.10-10.00)/0.05 = 2, so 10.10 mm is exactly 2 SDs above the mean. The empirical rule states that 95% of data falls within 2 SDs of the mean, leaving 5% in both tails combined. Since we want only the upper tail (above 10.10), and the distribution is symmetric, we take half of 5%, which gives us 2.5%. Therefore, approximately 2.5% of ball bearings have diameter above 10.10 mm. Students often mistake this for 5% by forgetting that the 5% outside 2 SDs is split between both tails. Always identify whether you need one tail or both when solving these problems.
The fill weights of a snack package are monitored and are known to be approximately normal (bell-shaped with no strong outliers). The weights have mean μ=50 g and standard deviation σ=2 g. Approximately what percent of packages have a fill weight between 48 g and 52 g? (Use the empirical rule.)
Explanation: We need to find what percent of packages weigh between 48g and 52g using a normal model. The mean is 50g with SD = 2g, so 48g is (48-50)/2 = -1 SD below the mean, and 52g is (52-50)/2 = +1 SD above the mean. The empirical rule states that approximately 68% of data in a normal distribution falls within 1 standard deviation of the mean. This means about 68% of packages have fill weights between 48g and 52g. Some students might confuse this with 95% (which is within 2 SDs) or 32% (which is outside 1 SD). When solving, always convert your values to z-scores first, then apply the empirical rule. Remember: within 1 SD = 68%, within 2 SD = 95%, within 3 SD = 99.7%.
The weights of packages shipped by a warehouse are roughly normal for a certain product line. Package weights have mean μ=20 lb and standard deviation σ=3 lb. Approximately what percent of packages weigh below 17 lb? (Use the empirical rule.)
Explanation: We need to find what percent of packages weigh below 17 lb. With mean 20 lb and SD = 3 lb, let's calculate the z-score: (17-20)/3 = -1, so 17 lb is exactly 1 SD below the mean. The empirical rule states that 68% of data falls within 1 SD of the mean (between 17 and 23 lb), leaving 32% in both tails combined. Since the normal distribution is symmetric, half of that 32% is below 17 lb and half is above 23 lb. Therefore, approximately 16% of packages weigh below 17 lb. Some students might think 84% are below 17 lb, but that's actually the percent above 17 lb (everything except the lower tail). To solve these problems accurately, always sketch the distribution and shade the region you're calculating.
Daily output from a manufacturing line (in units) is roughly symmetric and bell-shaped over long periods, so a normal model is appropriate. Output has mean μ=500 and standard deviation σ=50. Approximately what percent of days have output between 450 and 550 units? (Use the empirical rule.)
Explanation: We're finding what percent of days have output between 450 and 550 units. With mean 500 and SD = 50, let's convert to standard units: 450 is (450-500)/50 = -1 SD below the mean, and 550 is (550-500)/50 = +1 SD above the mean. The empirical rule tells us that approximately 68% of data in a normal distribution falls within 1 standard deviation of the mean. Therefore, about 68% of days have output between 450 and 550 units. This is a straightforward application of the 68-95-99.7 rule. Students sometimes confuse the percentages or miscalculate the number of SDs from the mean. To master these problems, practice converting raw scores to z-scores and memorize the empirical rule percentages.
Scores on a large district’s standardized math test are roughly symmetric and bell-shaped, so a normal model is appropriate. The scores have mean μ=70 and standard deviation σ=10. Approximately what percent of students score above 80? (Use the empirical rule.)
Explanation: We're using a normal model to estimate what percent of students score above 80. First, let's find how far 80 is from the mean of 70 in standard deviation units: (80-70)/10 = 1, so 80 is exactly 1 SD above the mean. The empirical rule tells us that about 68% of data falls within 1 SD of the mean (between 60 and 80), leaving 32% in the two tails combined. Since the normal distribution is symmetric, half of that 32% is above 80 and half is below 60. Therefore, approximately 16% of students score above 80. A common mistake is thinking 84% score above 80, but that's actually the percent scoring below 80 (the entire left side plus the middle 68%). To solve these problems, always sketch the curve, mark the mean and SD boundaries, then identify which area you need.
A company’s customer service ratings (on an internal scale) are approximately normal for a large set of calls. Ratings have mean μ=100 and standard deviation σ=15. Approximately what percent of calls have a rating below 85? (Use the empirical rule.)
Explanation: We need to find what percent of calls have a rating below 85. With mean 100 and SD = 15, let's calculate the z-score: (85-100)/15 = -1, so 85 is exactly 1 SD below the mean. The empirical rule states that 68% of data falls within 1 SD of the mean, leaving 32% in both tails combined. Since the distribution is symmetric, 16% is below 85 (the lower tail) and 16% is above 115 (the upper tail). Therefore, approximately 16% of calls have a rating below 85. A common mistake is thinking 84% are below 85, but that would be the case if we asked for ratings below 115 (1 SD above the mean). Always be careful about which tail you're calculating - sketch the curve to avoid confusion.
Scores on a standardized test are designed to be approximately normal. Suppose scores are normally distributed with mean μ=500 and standard deviation σ=100. Approximately what percent of students score below 300? (Use the empirical rule.)
Explanation: This problem asks for the percentage of students scoring below 300 on a test with mean 500 and SD 100. First, let's find how many standard deviations 300 is from the mean: (300 - 500) / 100 = -2 SD below the mean. According to the empirical rule, approximately 95% of values fall within 2 SD of the mean (between 300 and 700), leaving 5% in the two tails combined. Since the normal distribution is symmetric, half of this 5% is below 300, giving us 2.5%. The correct answer is A (2.5%) because we want only the lower tail beyond -2 SD. A common mistake would be to report 97.5%, which represents the percentage above 300, not below. To avoid errors, always determine whether you need the area in the tail (extreme values) or the area closer to the mean.