Using Normal Distributions to Estimate Populations - Statistics and Probability

The results of an online placement test are approximately normal (roughly symmetric and bell-shaped). Scores are modeled as $N(\mu=100,\,\sigma=15)$ . Approximately what percent of test-takers score above $115$ ? (Use the empirical rule.)
Answer: 16% This problem asks for the percentage of test-takers who score above 115. The cutoff of 115 is exactly 1 standard deviation above the mean since (115 - 100)/15 = 1. According to the empirical rule, about 68% of values fall within 1 SD of the mean, leaving 32% in the two tails combined. Since the normal distribution is symmetric, half of this 32% (which is 16%) falls in each tail. Therefore, approximately 16% of test-takers score above 115. A common error is thinking 84% score above this value, but that's actually the percentage below 115. When solving these problems, always identify whether you need the upper tail, lower tail, or middle region.
The fill weights of cereal boxes from a machine are checked and the distribution is approximately bell-shaped, so it is reasonable to use a normal model. Assume fill weights are normally distributed with mean $\mu = 368\text{ g}$ and standard deviation $\sigma = 4\text{ g}$ . Approximately what percent of boxes have fill weight below $360\text{ g}$ ?
Answer: About 2.5% This question uses a normal model to estimate the percentage of cereal boxes with fill weights below 360 g, given mean 368 g and SD 4 g. The cutoff 360 g is 8 g below the mean, equaling 2 SD units (8 / 4 = 2). The empirical rule states that about 95% of data is within 2 SDs, leaving 5% in the tails, or 2.5% below μ - 2σ. Thus, approximately 2.5% of boxes weigh less than 360 g. This matches the correct answer as it's the lower tail beyond 2 SDs, indicating rarity. A misconception is assuming it's 5% by combining both tails, but we only want the lower one. To generalize, sketch the mean, mark SD intervals, pinpoint the lower tail, and apply the empirical rule for the outer 0.3% or inner percentages accordingly.
Times (in minutes) to complete a certain assembly task are roughly symmetric and bell-shaped, so a normal model is appropriate. Completion times are modeled as $N(\mu=40,\,\sigma=4)$ . Approximately what percent of tasks take more than $48$ minutes? (Use the empirical rule.)
Answer: 2.5% We need to find what percentage of assembly tasks take more than 48 minutes. The cutoff of 48 minutes is exactly 2 standard deviations above the mean since (48 - 40)/4 = 2. The empirical rule tells us that approximately 95% of values fall within 2 SDs of the mean, leaving 5% in the two tails combined. Since the normal curve is symmetric, half of this 5% (which is 2.5%) falls in each tail. Therefore, approximately 2.5% of tasks take more than 48 minutes. Don't confuse this with 16%, which would be the percentage above 44 minutes (one SD above mean). To transfer this strategy: sketch the normal curve, mark the mean at 40, add SD ticks at 4-minute intervals, then identify that 48 is in the far right tail.
The weights of apples in a large shipment are roughly bell-shaped, so assume a normal model. Apple weights have mean $\mu=150$ g and standard deviation $\sigma=20$ g. Approximately what percent of apples weigh more than 170 g? (Use the empirical rule.)
Answer: About 16% This uses a normal model to estimate the percentage of apples weighing more than 170 g. Cutoff 170 is 20 g above mean 150, with SD 20, so +1 SD. Empirical rule: 68% within 1 SD, so 16% above +1 SD (50% above mean minus 34%). Approximately 16% weigh more than 170 g. The answer corresponds to the upper tail above +1 SD. Misconception: confusing with 2 SD (2.5%), but it's 1 SD here. Strategy: sketch mean 150, mark +1 SD at 170, target right tail, estimate 16%.
A company's customer service ratings (on an internal scale) are approximately normal for a large set of calls. Ratings have mean $\mu=100$ and standard deviation $\sigma=15$ . Approximately what percent of calls have a rating below 85? (Use the empirical rule.)
Answer: 16% 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.
A factory produces metal rods whose lengths are roughly symmetric and bell-shaped, so a normal model is appropriate. Rod lengths have mean $\mu=50$ cm and standard deviation $\sigma=2$ cm. Approximately what percent of rods are longer than 54 cm? (Use the empirical rule approximation.)
Answer: About 2.5% 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 factory makes metal rods whose lengths are roughly symmetric and bell-shaped, so a normal model is appropriate. The lengths are modeled as $N(\mu=50\text{ cm},\,\sigma=2\text{ cm})$ . Approximately what percent of rods are longer than $52$ cm? (Use the empirical rule.)
Answer: 16% This problem asks us to use a normal model to estimate what percentage of rods are longer than 52 cm. The cutoff of 52 cm is exactly 1 standard deviation above the mean since (52 - 50)/2 = 1. According to the empirical rule, about 68% of values fall within 1 SD of the mean, leaving 32% in the two tails combined. Since the normal distribution is symmetric, half of this 32% (which is 16%) falls in each tail. Therefore, approximately 16% of rods are longer than 52 cm. A common mistake is thinking that 84% are above this value, but that would be the percentage below 52 cm (everything except the upper 16%).
A company measures the diameters of ball bearings it produces. The diameters are roughly symmetric and bell-shaped, so a normal model is appropriate. Assume diameters are normally distributed with mean $\mu = 10.00\text{ mm}$ and standard deviation $\sigma = 0.02\text{ mm}$ . Approximately what percent of ball bearings have a diameter between $9.96\text{ mm}$ and $10.04\text{ mm}$ ?
Answer: About 95% This problem involves using a normal model to estimate the percentage of ball bearings with diameters between two values. The cutoffs are 9.96 mm and 10.04 mm, which are 0.04 mm below and above the mean of 10.00 mm; since the standard deviation is 0.02 mm, these are 2 SD units away (0.04 / 0.02 = 2). Using the empirical rule, about 95% of the data falls within 2 standard deviations of the mean in a normal distribution. This area translates to approximately 95% of the ball bearings having diameters in that range. The correct answer matches because it's the percentage between μ - 2σ and μ + 2σ, covering the central bulk of the distribution. A common misconception is thinking it's within 1 SD, which would be 68%, but here it's 2 SDs, so double-check the distance from the mean. To apply this generally, sketch the mean on a number line, mark ticks at each SD interval, decide it's a central interval, and estimate the area using the empirical rule (68-95-99.7).
A machine fills cereal boxes, and the fill weights are approximately normal. The mean fill weight is $\mu=500$ g with standard deviation $\sigma=10$ g. Approximately what percent of boxes have fill weights between 490 g and 510 g? (Use the empirical rule.)
Answer: About 68% 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%.
A large company reports that employee typing speeds (in words per minute) are roughly symmetric and bell-shaped, so a normal model is appropriate. Assume typing speeds are normally distributed with mean $\mu=70$ wpm and standard deviation $\sigma=10$ wpm. Approximately what percent of employees type above 80 wpm? (Use the empirical rule.)
Answer: 16% This problem asks us to use a normal model to estimate the percentage of employees typing above 80 wpm. First, we need to find how far 80 wpm is from the mean of 70 wpm in standard deviation units: (80 - 70) / 10 = 1 SD above the mean. According to the empirical rule, about 68% of values fall within 1 SD of the mean (between 60 and 80 wpm), leaving 32% in the two tails combined. Since the normal distribution is symmetric, half of this 32% is above 80 wpm, giving us 16%. The correct answer is A (16%) because we're looking for the upper tail beyond 1 SD. A common mistake would be to report 84%, which is the percentage below 80 wpm, not above. To solve these problems, always sketch the curve, mark the mean and SD boundaries, identify which tail(s) you need, and apply the empirical rule percentages.
A factory monitors the thickness of a coating, which is approximately normal (the histogram is roughly symmetric and bell-shaped). Thickness has mean $\mu = 1.20$ mm and standard deviation $\sigma = 0.05$ mm. Approximately what percent of items have thickness below 1.15 mm? (Use the empirical rule.)
Answer: 16% This problem uses a normal model to estimate the percentage of items with thickness below 1.15 mm, mean 1.20 mm, SD 0.05 mm. The cutoff 1.15 is 0.05 below mean, or -1 SD (0.05 / 0.05 = 1). Empirical rule: 68% within 1 SD, so 16% in lower tail. Approximately 16% are below 1.15 mm. It matches the below direction for the lower tail. A frequent misconception is using the upper tail or counting 2 SDs for 2.5%. To transfer, sketch mean 1.20, mark -1 SD at 1.15, decide lower tail, estimate 16%.
The lifetimes of a certain brand of light bulb are approximately normal (based on quality-control data, the distribution is roughly bell-shaped). Lifetimes have mean $\mu = 1000$ hours and standard deviation $\sigma = 100$ hours. Approximately what percent of bulbs last between 900 and 1100 hours? (Use the empirical rule.)
Answer: 68% We're applying a normal model to find the percentage of light bulbs lasting between 900 and 1100 hours, mean 1000, SD 100. Cutoffs: 900 (-1 SD), 1100 (+1 SD). Empirical rule gives 68% within 1 SD. So, about 68% last in this range. This aligns with the between direction for symmetric 1-SD bounds. Errors often involve confusing with 95% for 2 SD or tail percentages. Generally, sketch mean 1000, mark 900 and 1100 as ±1 SD, target central area, estimate 68%.
A factory produces bolts whose lengths are roughly symmetric and bell-shaped, so a normal model is appropriate. Bolt lengths are modeled as $N(\mu=50\text{ mm},\,\sigma=2\text{ mm})$ . Approximately what percent of bolts are between 48 mm and 52 mm? (Use the empirical rule.)
Answer: 68% This problem asks us to use a normal model to estimate what percentage of bolts fall between 48 mm and 52 mm. The mean is 50 mm and the standard deviation is 2 mm, so we need to find how far 48 and 52 are from the mean in SD units: 48 is (48-50)/2 = -1 SD below the mean, and 52 is (52-50)/2 = +1 SD above the mean. According to the empirical rule, approximately 68% of values in a normal distribution fall within 1 standard deviation of the mean (between μ-σ and μ+σ). Therefore, about 68% of bolts have lengths between 48 mm and 52 mm, making B the correct answer. A common mistake is confusing this with 95% (which is for ±2 SD) or 16% (which might come from looking at just one tail). To solve problems like this, always sketch the normal curve, mark the mean and SD boundaries, identify which region you need, and apply the empirical rule percentages.
A bottling plant's fill volumes are closely monitored and are approximately normal (roughly symmetric and bell-shaped). Assume fill volumes are normally distributed with mean $\mu = 1.00\text{ L}$ and standard deviation $\sigma = 0.03\text{ L}$ . Approximately what percent of bottles have fill volume between $0.97\text{ L}$ and $1.03\text{ L}$ ?
Answer: About 68% This involves a normal model to estimate the percentage of bottles with fill volumes between 0.97 L and 1.03 L, mean 1.00 L, SD 0.03 L. The cutoffs are 0.03 L from the mean, each 1 SD (0.03 / 0.03 = 1). The empirical rule states about 68% falls within 1 SD. So, approximately 68% of bottles are in this volume range. The choice is accurate for the central interval within 1 SD. Some might think it's 2 SDs if they double the SD value by mistake, leading to 95%, but confirm the math. To transfer, draw the mean, add SD marks, define the between area, and apply empirical rule percentages for approximation.
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 $\mu=500$ and standard deviation $\sigma=50$ . Approximately what percent of days have output between 450 and 550 units? (Use the empirical rule.)
Answer: 68% 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.
Adult male heights in a large population are roughly normal (approximately symmetric and bell-shaped). Suppose heights have mean $\mu=70$ in and standard deviation $\sigma=3$ in. Approximately what percent of men are below 64 inches? (Use the empirical rule.)
Answer: 2.5% We're finding what percent of men are below 64 inches using a normal model. With mean 70 inches and SD = 3 inches, let's calculate how far 64 is from the mean: (64-70)/3 = -2, so 64 inches is exactly 2 SDs below the mean. The empirical rule tells us that 95% of data falls within 2 SDs of the mean (between 64 and 76 inches), leaving 5% in the two tails combined. Since the normal curve is symmetric, 2.5% is below 64 inches and 2.5% is above 76 inches. Therefore, approximately 2.5% of men are below 64 inches. A common error is forgetting to divide the tail area by 2, leading to an answer of 5%. To avoid mistakes, always sketch the distribution and clearly mark which tail(s) you need.
The diameters of ball bearings from a well-controlled process are roughly symmetric and bell-shaped, so assume a normal model. Diameters have mean $\mu=10.0$ mm and standard deviation $\sigma=0.2$ mm. Approximately what percent of bearings have diameter greater than 10.4 mm? (Use the empirical rule.)
Answer: About 2.5% 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%.
A calibration error (in units) for a sensor is approximately normal, centered at 0. The mean error is $\mu=0$ and the standard deviation is $\sigma=3$ . Approximately what percent of errors are between $-6$ and $6$ ? (Use the empirical rule.)
Answer: About 95% We're applying a normal model to estimate the percentage of sensor errors between -6 and 6. With mean 0 and SD 3, -6 is -2 SD and +6 is +2 SD. Empirical rule: about 95% within 2 SD of mean. So, approximately 95% are between -6 and 6. This matches the between direction for ±2 SD. A common mistake is using 99.7% for 3 SD, but count SDs—it's 2 here. Sketch mean at 0, mark -2 SD at -6 and +2 at 6, identify central area, estimate 95%.
Scores on a standardized exam are designed to be approximately normal. Suppose scores have mean $\mu=70$ and standard deviation $\sigma=10$ . Approximately what percent of students score between 60 and 80? (Use the empirical rule.)
Answer: About 68% 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%.
The fill amounts of a cereal machine are monitored and are roughly symmetric and bell-shaped, so a normal model is appropriate. Fill amounts are modeled as $N(\mu=500\text{ g},\,\sigma=10\text{ g})$ . Approximately what percent of boxes have more than 520 g? (Use the empirical rule.)
Answer: 2.5% We need to find what percentage of cereal boxes contain more than 520 g when fill amounts follow N(500, 10). First, calculate how far 520 g is from the mean in SD units: (520-500)/10 = 2 standard deviations above the mean. The empirical rule states that approximately 95% of values fall within 2 standard deviations of the mean (between μ-2σ and μ+2σ), leaving 5% in the two tails combined. Since we want only the upper tail (boxes with more than 520 g) and the distribution is symmetric, we take half of 5%, which is 2.5%. Therefore, approximately 2.5% of boxes have more than 520 g, making C the correct answer. Students might mistakenly choose 5% by forgetting to divide for just one tail, or 97.5% by looking at the wrong side of the distribution. When solving these problems, always draw the curve, mark your boundary in SD units, identify which tail you need, and apply the empirical rule carefully.

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