Using Normal Distributions to Estimate Populations
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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.)
About 95%
About 68%
About 16%
About 5%
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 $\mu=70$ and standard deviation $\sigma=10$. Approximately what percent of students score between 60 and 80? (Use the empirical rule.)
About 68%
About 34%
About 95%
About 50%
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 $\mu=50$ cm and standard deviation $\sigma=2$ cm. Approximately what percent of rods are longer than 54 cm? (Use the empirical rule approximation.)
About 97.5%
About 16%
About 2.5%
About 50%
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 $\mu=200$ units and standard deviation $\sigma=5$ units. Approximately what percent of measurements are below 190 units? (Use the empirical rule.)
About 5%
About 2.5%
About 97.5%
About 16%
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 $\mu=10.0$ mm and standard deviation $\sigma=0.2$ mm. Approximately what percent of bearings have diameter greater than 10.4 mm? (Use the empirical rule.)
About 95%
About 97.5%
About 5%
About 2.5%
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 $\mu=100$ and standard deviation $\sigma=15$. Approximately what percent of people have IQ between 85 and 115? (Use the empirical rule.)
About 84%
About 34%
About 95%
About 68%
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 $\mu=170$ cm and standard deviation $\sigma=6$ cm. Approximately what percent of adults are shorter than 164 cm? (Use the empirical rule.)
About 97.5%
About 84%
About 16%
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%.
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.)
5%
32%
68%
95%
Explanation
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%.
Scores on a standardized math assessment are modeled well by a normal distribution (approximately symmetric and bell-shaped). The scores have mean $\mu = 70$ and standard deviation $\sigma = 10$. Approximately what percent of students score between 60 and 80? (Use the empirical rule.)
16%
32%
68%
95%
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 $\mu = 30$ seconds and standard deviation $\sigma = 2$ seconds. Approximately what percent of jobs take between 28 and 32 seconds? (Use the empirical rule.)
16%
32%
68%
95%
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%.