All questions
Question 1
Normal data have mean 500 and SD 100. What score is the 90th percentile?
- 628 (correct answer)
- 600
- 650
- 664
Explanation: The 90th percentile of a normal distribution is about 1.28 standard deviations above the mean. So add 1.28 times 100 to 500: 500 + 128 = 628. The tempting answer 600 is only 1 standard deviation above the mean, which falls around the 84th percentile.
Question 2
A normal data set has mean 100 and SD 15. Estimate the percentage of values from 85 to 130.
- About 68%
- About 81.5% (correct answer)
- About 95%
- About 99.7%
Explanation: 85 is one standard deviation below the mean and 130 is two standard deviations above it. The interval 85 to 115 holds about 68%, and 115 to 130 adds about 13.5%, so the total is about 81.5%. The tempting 95% mistake would apply to two standard deviations on both sides, from 70 to 130, not from one below to two above.
Question 3
Which clue most strongly suggests a normal model is inappropriate?
- Mean is far above median (correct answer)
- Sample size is exactly 40
- Standard deviation is small
- All values are positive
Explanation: In a normal model, the distribution is symmetric, so the mean and median should be roughly equal. A mean far above the median indicates right skew, which directly violates normality. A small standard deviation only affects spread, not shape, so it is not evidence against a normal model.
Question 4
In a normal distribution, the area below z=1.28 is 0.90. What percentage is between z=-1.28 and z=1.28?
- 95%
- 90%
- 80% (correct answer)
- 10%
Explanation: The area below z=1.28 is 0.90, so the area above it is 0.10. By symmetry, the area below z=-1.28 is also 0.10. Subtract both tails from the total: 1 - 0.10 - 0.10 = 0.80, or 80%. The tempting 90% is wrong because 0.90 is the area below z=1.28, not the area between the two z-values.
Question 5
For normal data, mean 68 and SD 4. Estimate the percentile of 75.
- 84th percentile
- 93rd percentile
- 95th percentile
- 96th percentile (correct answer)
Explanation: 75 is 1.75 standard deviations above the mean of 68, since 75 minus 68 is 7 and 7 divided by 4 is 1.75. In a normal distribution, 1.75 SD above the mean corresponds to about the 96th percentile. The tempting 95th percentile is wrong because it matches only about 1.64 SD above the mean, and 75 is higher than that.
Question 6
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.)
- About 95%
- About 5%
- About 16%
- About 68% (correct answer)
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%.
Question 7
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.)
- About 68% (correct answer)
- About 34%
- About 50%
- About 95%
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%.
Question 8
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.)
- About 97.5%
- About 50%
- About 16%
- About 2.5% (correct answer)
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%.
Question 9
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.)
- About 5%
- About 97.5%
- About 16%
- About 2.5% (correct answer)
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%.
Question 10
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.)
- About 95%
- About 97.5%
- About 5%
- About 2.5% (correct answer)
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%.
Question 11
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.)
- About 34%
- About 68% (correct answer)
- About 84%
- About 95%
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%.
Question 12
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.)
- About 97.5%
- About 16% (correct answer)
- About 84%
- About 2.5%
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%.
Question 13
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.)
- 95%
- 68% (correct answer)
- 32%
- 16%
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%.
Question 14
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.)
- 32%
- 68% (correct answer)
- 95%
- 16%
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%.
Question 15
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.)
- 97.5% (correct answer)
- 84%
- 2.5%
- 16%
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%.
Question 16
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.)
- 16% (correct answer)
- 50%
- 2.5%
- 84%
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%.
Question 17
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.)
- 5%
- 95% (correct answer)
- 68%
- 99.7%
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.
Question 18
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.)
- 68% (correct answer)
- 95%
- 34%
- 32%
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.
Question 19
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.)
- 16%
- 2.5% (correct answer)
- 5%
- 95%
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.
Question 20
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.)
- 95%
- 32%
- 50%
- 68% (correct answer)
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%.