The Normal Distribution - AP Statistics

A manufacturer reports that the lifetimes (in hours) of a certain type of light bulb are approximately Normal for the population of all bulbs produced, with mean $\mu=800$ hours and standard deviation $\sigma=50$ hours. The value 850 hours is marked on the normal curve. Which statement about the marked value is correct?
Answer: The marked value is 1 standard deviation above the mean, so it is somewhat above average but still fairly typical. This question involves interpreting a value's position in a normal distribution. With μ = 800 hours and σ = 50 hours, we need to determine where 850 hours falls. Calculate the number of standard deviations from the mean: (850 - 800)/50 = 50/50 = 1. So 850 hours is exactly 1 standard deviation above the mean. In a normal distribution, values within 1 standard deviation of the mean are considered fairly typical, as about 68% of all values fall in the interval μ ± 1σ. Being 1σ above the mean means this bulb lasts somewhat longer than average, but it's not unusual or rare. About 16% of bulbs would last even longer than this. Understanding that values within 1σ of the mean are "typical" while those beyond 2σ are "unusual" is key to interpreting normal distributions.
The distribution of diameters (in mm) of ball bearings produced by a factory is approximately Normal for the population of all ball bearings, with mean $\mu=10.00$ mm and standard deviation $\sigma=0.03$ mm. The interval from 9.94 mm to 10.06 mm is marked on the normal curve. Which statement about the marked interval is correct?
Answer: The interval is within 2 standard deviations of the mean, so it contains almost all ball bearings. This question involves interpreting an interval on a normal distribution. With μ = 10.00 mm and σ = 0.03 mm, we analyze the interval from 9.94 mm to 10.06 mm. First, standardize the endpoints: (9.94 - 10.00)/0.03 = -0.06/0.03 = -2 and (10.06 - 10.00)/0.03 = 0.06/0.03 = 2. The interval spans from 2 standard deviations below to 2 standard deviations above the mean. According to the empirical rule for normal distributions, approximately 95% of all values fall within 2 standard deviations of the mean. This means "almost all" ball bearings (about 95%) have diameters in this range. Understanding that μ ± 2σ captures about 95% of the distribution is crucial for quality control applications. The remaining 5% would be split between the two tails, representing unusually small or large ball bearings.
The weights of packages shipped by a warehouse are approximately Normal with mean $\mu=12$ lb and standard deviation $\sigma=3$ lb. The normal curve below marks the interval 9 to 15 lb. Which statement about the marked interval is correct?
Answer: The interval 9 to 15 is about $\mu\pm 1\sigma$ , so it is centered at the mean and includes many packages. This question assesses interpreting an interval in the normal distribution of package weights with μ = 12 lb and σ = 3 lb. The interval 9 to 15 is μ ± 3, but since σ = 3, it's exactly μ ± σ, centered at the mean and including many (about 68%) packages. This range captures typical weights. Choice B distracts by calling it μ ± 2σ, possibly from misjudging the deviation as 6 lb each way. Mini-lesson: In normal models, intervals symmetric around μ use multiples of σ to estimate proportions; μ ± σ for ~68%, helping visualize data concentration without exact probabilities.
In a region, the distribution of monthly household electricity use (in kWh) is approximately Normal for the population of all households. The mean is $\mu=900$ kWh with standard deviation $\sigma=150$ kWh. The value 1050 kWh is marked on the normal curve. Which statement about the marked value is correct?
Answer: The marked use is 1 standard deviation above the mean, so it is higher than average but still fairly typical. This problem requires locating an electricity usage value on a normal distribution. Given μ = 900 kWh and σ = 150 kWh, we need to find where 1050 kWh falls. Calculate the number of standard deviations from the mean: (1050 - 900)/150 = 150/150 = 1. So 1050 kWh is exactly 1 standard deviation above the mean. In a normal distribution, values within 1 standard deviation of the mean are fairly typical, as approximately 68% of all values fall within μ ± 1σ. Being 1σ above the mean indicates this household uses somewhat more electricity than average, but this usage level is not unusual or rare. About 16% of households would use even more electricity than this. The key is recognizing that values within 1σ of μ are common, while those beyond 2σ are uncommon.
A company finds that lifetimes of a certain brand of lightbulb are approximately Normal for the population produced this month, with mean $\mu=1200$ hours and standard deviation $\sigma=150$ hours. A customer reports a bulb lasting 900 hours. Which statement about the marked value is correct?
Answer: 900 hours is about 2 standard deviations below the mean, so it is shorter than most lifetimes and somewhat unusual. This problem tests understanding of negative z-scores in a normal distribution. Given μ = 1200 hours and σ = 150 hours, we find (900 - 1200)/150 = -300/150 = -2, meaning 900 hours is 2 standard deviations below the mean. The distractor about "standard errors" confuses population parameters with sampling distributions. In a normal model, approximately 95% of bulbs last between 900 and 1500 hours (μ ± 2σ). A bulb lasting only 900 hours falls in the bottom 2.5% of the distribution, making it somewhat unusual and potentially indicating a quality issue.
The diameters of ball bearings produced by a factory are approximately Normal with mean $\mu=10.00$ mm and standard deviation $\sigma=0.04$ mm. The normal curve below marks the interval 9.92 to 10.08 mm. Which statement about the marked interval is correct?
Answer: The interval 9.92 to 10.08 is about $\mu\pm 2\sigma$ , so it contains most diameters and is centered at the mean. The skill here is interpreting an interval in the normal distribution of diameters with μ = 10.00 mm and σ = 0.04 mm. The interval 9.92 to 10.08 is from 10 - 0.08 to 10 + 0.08, and since 0.08 = 2 × 0.04, it's μ ± 2σ, containing most (about 95%) diameters and centered at the mean. This is a wide, typical range. Choice A distracts by saying μ ± 1σ, perhaps halving the width incorrectly. Mini-lesson: Empirical rule approximates coverage—95% within 2σ—useful for quality control in manufacturing normal data without detailed calculations.
For the population of newborn weights at a hospital, birth weights are approximately Normal with mean $\mu=3.4$ kg and standard deviation $\sigma=0.5$ kg. The marked value is $x=4.1$ kg. Which statement about the marked value is correct?
Answer: A weight of 4.1 kg is about 1.4 standard deviations above the mean, so it is heavier than average and somewhat uncommon. This question tests z-score calculation for newborn weights. With μ = 3.4 kg and σ = 0.5 kg, the z-score for x = 4.1 kg is: z = (4.1 - 3.4)/0.5 = 0.7/0.5 = 1.4. This means 4.1 kg is 1.4 standard deviations above the mean, making it heavier than average and somewhat uncommon. The distractors include confusing standard deviations with standard errors, reversing the direction (below instead of above), or miscalculating the z-score. In a normal distribution, values between 1 and 2 standard deviations from the mean are somewhat unusual, occurring in roughly 8% of cases on each side.
The time (in minutes) it takes the population of customers to complete an online checkout is approximately Normal with mean $\mu=8.0$ minutes and standard deviation $\sigma=1.5$ minutes. The marked value is $x=5.0$ minutes. Which statement about the marked value is correct?
Answer: A checkout time of 5.0 minutes is about 2 standard deviations below the mean, so it is relatively fast compared with most customers. This problem involves calculating the position of a checkout time relative to the mean. Given μ = 8.0 minutes and σ = 1.5 minutes, we find the z-score for x = 5.0 minutes: z = (5.0 - 8.0)/1.5 = -3.0/1.5 = -2. A checkout time of 5.0 minutes is 2 standard deviations below the mean, making it relatively fast. The distractors include reversing the interpretation (calling it slow when it's fast), confusing standard deviations with standard errors, or miscalculating the z-score. Values 2 standard deviations below the mean occur in only about 2.5% of cases, making this a relatively unusual fast checkout.
For adult men in a country, heights (in inches) are approximately Normal with mean $\mu=70$ and standard deviation $\sigma=3$ . An interval is marked: from $x=67$ to $x=73$ inches. Which statement about the marked interval is correct?
Answer: The marked interval is within about 1 standard deviation of the mean, so it contains a large majority (roughly two-thirds) of the population. This question tests understanding of intervals in normal distributions. With μ = 70 and σ = 3, the interval from 67 to 73 inches spans from μ - σ to μ + σ, which is μ ± 1σ. In a normal distribution, approximately 68% of the population falls within one standard deviation of the mean, making this a large majority but not "nearly all." The interval is symmetric around the mean, containing the middle portion of the distribution. The distractor about "95% of the population" would apply to an interval of μ ± 2σ (64 to 76 inches). Understanding these benchmark percentages (68-95-99.7 rule) is crucial for interpreting normal models and determining what proportion of a population falls within specific ranges.
For the population of commuters on a certain highway, daily one-way travel times are approximately Normal with mean $\mu=35$ minutes and standard deviation $\sigma=5$ minutes. A commuter's time of 25 minutes is recorded on a particular day. Which statement about the marked value is correct?
Answer: 25 minutes is about 2 standard deviations below the mean, so it is faster than most commutes and somewhat unusual. This question involves interpreting a value below the mean in terms of standard deviations. With μ = 35 minutes and σ = 5 minutes, we calculate (25 - 35)/5 = -10/5 = -2, showing 25 minutes is 2 standard deviations below the mean. The distractor mentioning "standard errors" incorrectly applies sampling concepts to individual observations. In a normal distribution, about 95% of commute times fall between 25 and 45 minutes (μ ± 2σ). A 25-minute commute is faster than approximately 97.5% of all commutes, making it somewhat unusual but not extremely rare.
For all students at a high school, the number of text messages sent per day is approximately Normal with mean $\mu=80$ and standard deviation $\sigma=20$ . The normal curve below marks a value of 40 texts. Which statement about the marked value is correct?
Answer: Sending 40 texts is about 2 standard deviations below the mean, so it is relatively unusual and in the left tail. The skill involves positioning a text message count in a normal distribution with μ = 80 and σ = 20. For 40 texts, z = (40 - 80) / 20 = -2, placing it 2 standard deviations below the mean, unusual in the left tail where only about 2.5% of students send fewer. This indicates significantly low texting activity. Choice C is a distractor, incorrectly stating 2 SDs above, likely from ignoring the negative sign. Mini-lesson: Z-scores quantify distance from μ in σ units; positive for above, negative for below, with tails beyond ±2σ holding rare values, useful for identifying outliers in data.
The time it takes the population of all customers to complete an online checkout is approximately Normal with mean $\mu=4.5$ minutes and standard deviation $\sigma=1.0$ minute. A normal curve is shown with the interval from 3.5 to 5.5 minutes shaded. Normal curve (minutes) | | .-''''-. | .-' #### '-. | .' #### '. | / #### \ |--------|----|----|----|----|----|----> x 1.5 2.5 3.5 4.5 5.5 6.5 7.5 [3.5, 5.5] \mu=4.5 Which statement about the marked interval is correct?
Answer: The interval from 3.5 to 5.5 minutes is about $\pm 1\sigma$ from the mean, so it contains a majority of checkout times. This question tests interval interpretation in normal distributions for AP Statistics. Checkout times are normal with μ = 4.5 minutes and σ = 1.0 minute. The interval 3.5 to 5.5 spans 1 minute on each side, so ±1σ (1 / 1 = 1). The empirical rule states about 68% fall within ±1σ, which is a majority. Distractor choice B claims ±2σ for nearly all, but that's 95% and would be ±2 minutes here, not matching. In normal models, these percentages (68-95-99.7) provide quick estimates of data concentration around μ, aiding in understanding variability.
Scores on a professional certification exam for all test takers are approximately Normal with mean $\mu=500$ and standard deviation $\sigma=80$ . A score of $x=660$ is marked. Which statement about the marked value is correct?
Answer: The marked score is about 2 standard deviations above the mean, so it is in the high tail and relatively unusual. This problem tests interpreting standardized scores in a normal distribution. Given μ = 500 and σ = 80, we calculate the z-score for x = 660: (660 - 500) / 80 = 160 / 80 = 2. The marked score is exactly 2 standard deviations above the mean, placing it in the upper tail of the distribution. In a normal model, only about 2.5% of values exceed μ + 2σ, making this score relatively unusual and quite high. The distractor about "16% scoring above 660" incorrectly divides the distance from mean by 10 instead of using the proper normal distribution properties. Remember that in normal distributions, extreme values (beyond 2 standard deviations) are rare, which is why they're often used as benchmarks for identifying unusual observations.
For the population of all adult male heights in a region, heights are approximately Normal with mean $\mu=69$ inches and standard deviation $\sigma=3$ inches. A normal curve is shown with the value $63$ inches marked. Normal curve (height) | | .-''''-. | .-' '-. | .' '. | / \ |----------|----|----|----|----|----|----|----> x 60 63 66 69 72 75 78 63 $\mu=69$ Which statement about the marked value is correct?
Answer: A height of 63 inches is about 2 standard deviations below the mean, so it is unusually short compared with most adult males. This question evaluates z-score application to heights in normal distributions for AP Statistics. Heights are normal with $\mu = 69$ inches and $\sigma = 3$ inches. For 63 inches, $z = \frac{63 - 69}{3} = -2$ , meaning 2 standard deviations below the mean. The empirical rule notes 95% within $\pm 2\sigma$ , so this is unusually short. Choice E is a distractor, wrongly claiming most are shorter, but only about 2.5% are in that lower tail. Normal models help by standardizing values to compare across distributions, emphasizing how $\sigma$ quantifies deviation from $\mu$ .
The weights (in grams) of all packages of a certain snack are approximately Normal with mean $\mu=250$ and standard deviation $\sigma=8$ . A package weight of $x=242$ grams is marked. Which statement about the marked value is correct?
Answer: The marked weight is about 1 standard deviation below the mean, so it is somewhat light but not extremely unusual. This question involves determining the position of a value in a normal distribution. With μ = 250 and σ = 8, we find where x = 242 falls: (242 - 250) / 8 = -8 / 8 = -1. The marked weight is exactly 1 standard deviation below the mean, making it somewhat light but not extremely unusual. In a normal distribution, about 68% of values fall within 1 standard deviation of the mean, so values at exactly 1 standard deviation away occur reasonably often (about 16% on each side). The distractor mentioning "standard error" confuses the population parameter σ with the standard error used in sampling distributions. When interpreting normal models, values within 1 standard deviation are considered fairly typical, while those beyond 2 standard deviations are unusual.
The lifetimes of a certain brand of LED bulbs (in thousands of hours) are approximately Normal with mean $\mu=30$ and standard deviation $\sigma=4$ . The normal curve below marks the interval 26 to 34 thousand hours. Which statement about the marked interval is correct?
Answer: The interval 26 to 34 is about $\mu\pm 1\sigma$ , so it contains about two-thirds of bulb lifetimes and is centered at the mean. This question evaluates understanding intervals in a normal distribution for bulb lifetimes with μ = 30 thousand hours and σ = 4. The interval 26 to 34 is from 30 - 4 to 30 + 4, exactly μ ± σ, which contains about two-thirds (68%) of lifetimes and is centered at the mean. This makes it a typical range for most bulbs. A distractor is choice C, which claims it's μ ± 2σ, perhaps from thinking the width is 16 instead of 8. Mini-lesson on interpreting normal models: Use empirical rule for rough proportions—68% within 1σ, 95% within 2σ, 99.7% within 3σ—to estimate coverage without tables, aiding quick assessments of intervals.
In a state, the systolic blood pressure (in mmHg) for all adults is approximately Normal with mean $\mu=120$ and standard deviation $\sigma=15$ . A value is marked at $x=150$ mmHg. Which statement about the marked value is correct?
Answer: The marked blood pressure is about 2 standard deviations above the mean, so it is in the high tail and relatively unusual. This problem requires calculating the standardized position of a blood pressure reading. Given μ = 120 and σ = 15, we find: (150 - 120) / 15 = 30 / 15 = 2. The marked value is 2 standard deviations above the mean, placing it in the upper tail of the distribution where only about 2.5% of values occur. This makes the blood pressure reading relatively unusual and concerning from a medical standpoint. The distractor claiming "30% above 150" incorrectly assumes that being 30 units above the mean means 30% are higher, ignoring the normal distribution's shape. In medical contexts, values beyond 2 standard deviations often indicate conditions requiring attention, as they represent the extreme 5% of the population.
Scores on a statewide AP Statistics practice exam for all 11th-grade students in a large district are approximately Normal with mean $\mu=72$ points and standard deviation $\sigma=10$ points. The normal curve below marks a score of 92 points. Which statement about the marked value is correct? (Interpret the marked value in terms of position and spread, not an exact probability.)
Answer: A score of 92 is about 2 standard deviations above the mean, so it is relatively unusual and in the right tail. This question assesses the skill of interpreting a value's position in a normal distribution using the mean and standard deviation. The scores follow a normal distribution with μ = 72 and σ = 10, so to find the position of 92, calculate the z-score: (92 - 72) / 10 = 2, meaning it's 2 standard deviations above the mean. This places it in the right tail, making it relatively unusual as values beyond 2 SDs occur in about 5% of the data combined in both tails. A common distractor is choice A, which incorrectly states it's about 1 SD above the mean, perhaps from miscalculating the difference as 10 instead of 20. In normal models, we interpret positions by standardizing to z-scores: values with |z| < 1 are typical and near the center, |z| around 2 are unusual in the tails, and |z| > 3 are extreme. Understanding this helps visualize where data points fall relative to the bulk of the distribution without needing exact probabilities.
For the population of daily high temperatures in July in a certain desert town, the distribution is approximately Normal with mean $\mu=102^\circ\text{F}$ and standard deviation $\sigma=4^\circ\text{F}$ . The marked interval is from $98^\circ\text{F}$ to $106^\circ\text{F}$ . Which statement about the marked interval is correct?
Answer: The interval $98$ to $106$ is within about 1 standard deviation of the mean, so it contains about two-thirds of July days. This question tests understanding of intervals in normal distributions and the empirical rule. With μ = 102°F and σ = 4°F, the interval 98°F to 106°F extends from (98-102)/4 = -1 standard deviation to (106-102)/4 = +1 standard deviation from the mean. The empirical rule states that approximately 68% (about two-thirds) of data falls within one standard deviation of the mean in a normal distribution. The distractor mentioning "standard error" incorrectly applies a sampling concept to population data. The interval is symmetric around the mean, not shifted, so it captures typical values, not unusually cool days.
For the population of adult women in a city, resting heart rates are approximately Normal with mean $\mu=72$ beats per minute and standard deviation $\sigma=12$ beats per minute. A clinic flags anyone with a resting heart rate below 48 bpm for follow-up. Which statement about the marked value is correct?
Answer: 48 bpm is about 2 standard deviations below the mean, so it is unusually low compared with most women. This question requires interpreting a value below the mean in a normal distribution. With μ = 72 and σ = 12, we calculate (48 - 72)/12 = -24/12 = -2, showing that 48 bpm is 2 standard deviations below the mean. The distractor mentioning "standard errors" incorrectly applies sampling distribution concepts to individual values. In a normal distribution, values 2 standard deviations below the mean are relatively rare, occurring in only about 2.5% of the population. This makes 48 bpm unusually low for resting heart rate, warranting the clinic's follow-up protocol.

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