AP Statistics Quiz: The Normal Distribution

What this quiz covers

This quiz focuses on The Normal Distribution, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

  Normal curve (height)
  |
  |                 .-''''-.
  |              .-'        '-.
  |            .'              '.
  |           /                  \
  |----------|----|----|----|----|----|----|----> x
            60   63   66   69   72   75   78
                 63      $\mu=69$

1. A height of 63 inches is about 2 standard deviations below the mean, so it is unusually short compared with most adult males. (correct answer)
2. A height of 63 inches is about 2 standard deviations above the mean, so it is unusually tall compared with most adult males.
3. A height of 63 inches is about 1 standard deviation below the mean, so it is very typical.
4. A height of 63 inches is about 2 standard errors below the mean, so it is unusually short.
5. A height of 63 inches is about 2 standard deviations below the mean, so most adult males are shorter than 63 inches.

Explanation: 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 $\mu$.

Question 2

  Normal curve (hours)
  |
  |                 .-''''-.
  |              .-'        '-.
  |            .'              '.
  |           /                  \
  |----------|----|----|----|----|----|----|----> x
           750  900  1050 1200 1350 1500 1650
                     <u>1050</u>  \mu=1200

1. A lifetime of 1050 hours is about 1 standard deviation below the mean, so it is somewhat shorter than typical but not unusual. (correct answer)
2. A lifetime of 1050 hours is about 2 standard deviations below the mean, so it is unusually short.
3. A lifetime of 1050 hours is about 1 standard deviation above the mean, so it is somewhat longer than typical.
4. A lifetime of 1050 hours is about 1 standard error below the mean, so it is somewhat shorter than typical.
5. A lifetime of 1050 hours is about 1 standard deviation below the mean, so about 95% of bulbs last less than 1050 hours.

Explanation: This question assesses interpreting lifetimes in normal distributions via AP Statistics. Bulb lifetimes are normal with μ = 1200 hours and σ = 150 hours. The z-score for 1050 hours is (1050 - 1200) / 150 = -1, or 1 standard deviation below the mean. The empirical rule shows 68% within ±1σ, so this is somewhat short but typical, not unusual. Distractor choice E incorrectly states 95% last less than 1050, confusing the rule—actually, about 16% are shorter in the lower tail. In normal models, we reference μ as the center and σ for spread to understand proportions and typical ranges.

Question 3

1. A length of 2.6 minutes is about 1 standard deviation below the mean, so it is slightly short but not unusual.
2. A length of 2.6 minutes is about 2 standard deviations below the mean, so it is relatively unusual and in the left tail. (correct answer)
3. A length of 2.6 minutes is about 2 standard deviations above the mean, so it is relatively unusual and in the right tail.
4. A length of 2.6 minutes is about 2 standard errors below the mean, so it suggests the mean song length is less than 3.8.
5. A length of 2.6 minutes is close to the mean because it differs from 3.8 by only 1.2 minutes.

Explanation: This question tests positioning a song length in a normal distribution with μ = 3.8 minutes and σ = 0.6. For 2.6, z = (2.6 - 3.8) / 0.6 = -2, 2 standard deviations below the mean, unusual in the left tail with about 2.5% shorter songs. This marks it as notably short. Choice C distracts by saying 2 SDs above, ignoring the subtraction direction. Mini-lesson: Normal curves are bell-shaped with most data near μ; use z = (x - μ) / σ to classify: central for |z| < 1, tails for |z| > 2, for intuitive rarity judgments.

Question 4

1. A blood pressure of 135 mmHg is about 1 standard deviation above the mean, so it is higher than average but not especially unusual. (correct answer)
2. A blood pressure of 135 mmHg is about 1 standard deviation below the mean, so it is lower than average but not especially unusual.
3. A blood pressure of 135 mmHg is about 15 standard deviations above the mean, so it is extremely unusual.
4. A blood pressure of 135 mmHg is about 1 standard error above the mean, so it is higher than average but not especially unusual.
5. A blood pressure of 135 mmHg is about 2 standard deviations above the mean, so it is extremely unusual.

Explanation: This question tests z-score calculation for blood pressure readings. Given μ = 120 mmHg and σ = 15 mmHg, we find the z-score for x = 135 mmHg: z = (135 - 120)/15 = 15/15 = 1. A blood pressure of 135 mmHg is exactly 1 standard deviation above the mean, making it higher than average but not especially unusual. The distractors include confusing the direction, mixing up standard deviations with standard errors, or drastically miscalculating the z-score. In a normal distribution, about 16% of values fall more than 1 standard deviation above the mean, so this reading is elevated but not rare.

Question 5

1. A heart rate of 44 bpm is about 2 standard deviations below the mean, so it is relatively low compared with most runners. (correct answer)
2. A heart rate of 44 bpm is about 2 standard deviations above the mean, so it is relatively high compared with most runners.
3. A heart rate of 44 bpm is about 2 standard errors below the mean, so it is relatively low compared with most runners.
4. A heart rate of 44 bpm is about 1 standard deviation below the mean, so it is extremely low compared with most runners.
5. A heart rate of 44 bpm is about 0.5 standard deviations below the mean, so it is very typical compared with most runners.

Explanation: This problem involves interpreting a heart rate value in a normal distribution. Given μ = 60 bpm and σ = 8 bpm, we calculate the z-score for x = 44 bpm: z = (44 - 60)/8 = -16/8 = -2. A heart rate of 44 bpm is 2 standard deviations below the mean, making it relatively low for runners. The distractors include confusing the direction (above vs. below), mixing up standard deviations with standard errors, or miscalculating how many standard deviations away the value is. Values 2 standard deviations below the mean are relatively unusual, occurring in only about 2.5% of the population.

Question 6

1. The interval is within 1 standard deviation of the mean, so it contains most apples.
2. The interval is within 2 standard deviations of the mean, so it contains almost all apples. (correct answer)
3. The interval is within 3 standard deviations of the mean, so it contains nearly all apples.
4. The interval is within 2 standard errors of the mean, so it contains about 95% of sample means regardless of sample size.
5. The interval is centered away from the mean, so it contains about half of apples.

Explanation: This question tests understanding of intervals in normal distributions. With μ = 150 g and σ = 10 g, we analyze the interval from 130 g to 170 g. First, standardize the endpoints: (130 - 150)/10 = -20/10 = -2 and (170 - 150)/10 = 20/10 = 2. The interval spans from 2 standard deviations below to 2 standard deviations above the mean. According to the empirical rule, approximately 95% of values in a normal distribution fall within 2 standard deviations of the mean. This means "almost all" apples (about 95%) have weights in this range. The key insight is recognizing that μ ± 2σ captures approximately 95% of the distribution, while μ ± 1σ captures only about 68%. Understanding these benchmarks helps interpret what proportion of the population falls within various intervals.

Question 7

1. A lifetime of 900 hours is about 2 standard deviations below the mean, so it is shorter than typical and relatively uncommon. (correct answer)
2. A lifetime of 900 hours is about 2 standard deviations above the mean, so it is longer than typical and relatively uncommon.
3. A lifetime of 900 hours is about 0.2 standard deviations below the mean, so it is very typical.
4. A lifetime of 900 hours is about 2 standard errors below the mean, so it is shorter than typical and relatively uncommon.
5. A lifetime of 900 hours is about 3 standard deviations below the mean, so it is extremely short compared with most bulbs.

Explanation: This problem requires calculating how many standard deviations a lightbulb lifetime is from the mean. With μ = 1200 hours and σ = 150 hours, the z-score for x = 900 hours is: z = (900 - 1200)/150 = -300/150 = -2. A lifetime of 900 hours is 2 standard deviations below the mean, making it shorter than typical and relatively uncommon. The distractors include reversing the interpretation (longer instead of shorter), 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 bulb's lifetime relatively unusual.

Question 8

1. The interval $9.92$ to $10.08$ is within about 2 standard deviations of the mean, so it should include most (but not all) bearings. (correct answer)
2. The interval $9.92$ to $10.08$ is within about 1 standard deviation of the mean, so it should include nearly all bearings.
3. The interval $9.92$ to $10.08$ is within about 4 standard deviations of the mean, so it should include only about two-thirds of bearings.
4. The interval $9.92$ to $10.08$ is within about 2 standard errors of the mean, so it should include most bearings.
5. The interval $9.92$ to $10.08$ is centered below the mean, so it captures mostly undersized bearings.

Explanation: This problem involves interpreting an interval in a normal distribution of bearing diameters. With μ = 10.00 mm and σ = 0.04 mm, the interval 9.92 to 10.08 mm extends from (9.92-10.00)/0.04 = -2 to (10.08-10.00)/0.04 = +2 standard deviations from the mean. According to the empirical rule, approximately 95% of values fall within 2 standard deviations of the mean, which means most (but not all) bearings. The distractors incorrectly calculate the number of standard deviations, confuse standard deviation with standard error, or misinterpret the interval's position. The interval is symmetric around the mean, not shifted below it.

Question 9

1. A height of 76 inches is about 2 standard deviations above the mean, so it is relatively unusual compared with most adult males. (correct answer)
2. A height of 76 inches is about 2 standard errors above the mean, so it is relatively unusual compared with most adult males.
3. A height of 76 inches is about 2 standard deviations below the mean, so it is relatively unusual compared with most adult males.
4. A height of 76 inches is about 6 standard deviations above the mean, so it is extremely unusual compared with most adult males.
5. A height of 76 inches is about 1 standard deviation above the mean, so it is very typical compared with most adult males.

Explanation: This question tests understanding of interpreting values in a normal distribution using z-scores. Given μ = 70 inches and σ = 3 inches, we need to find how many standard deviations 76 inches is from the mean. The z-score is (76 - 70)/3 = 6/3 = 2, meaning 76 inches is 2 standard deviations above the mean. The common distractor is confusing "standard deviation" with "standard error" - standard error relates to sampling distributions, not individual values. In a normal distribution, about 95% of values fall within 2 standard deviations of the mean, so a value 2 standard deviations away is relatively unusual, occurring in only about 2.5% of the population.

Question 10

1. The marked reaction time is 2 standard deviations below the mean, so it is in the lower tail and would be relatively uncommon (very fast). (correct answer)
2. The marked reaction time is 2 standard deviations above the mean, so it is in the upper tail and would be relatively uncommon (very slow).
3. The marked reaction time is 1 standard deviation below the mean, so it is typical.
4. The marked reaction time is 2 standard errors below the mean, so it indicates an unusually low sample mean.
5. The marked reaction time is near the mean, so about half of attempts are faster than this.

Explanation: This question tests understanding of extreme values in a normal distribution. With μ = 300 ms and σ = 40 ms, we need to determine where 220 ms falls. Calculate the standardized value: (220 - 300)/40 = -80/40 = -2. This means 220 ms is 2 standard deviations below the mean. In a normal distribution, values that are 2 or more standard deviations from the mean are considered unusual, occurring in only about 2.5% of cases in each tail. Since this is 2σ below the mean, it represents an unusually fast reaction time. In the context of video games, lower reaction times are better, so this would be an exceptionally good performance. Understanding that values beyond 2σ from the mean are rare helps identify which outcomes are typical versus exceptional.

Question 11

1. The interval is within 1 standard deviation of the mean, so it contains a majority of bottles but not nearly all of them. (correct answer)
2. The interval is within 1 standard error of the mean, so it contains about 68% of sample means for any sample size.
3. The interval is within 2 standard deviations of the mean, so it contains almost all bottles.
4. The interval is centered 0.4 oz above the mean, so about half of bottles fall in it.
5. The interval is outside 1 standard deviation of the mean, so it contains only a small minority of bottles.

Explanation: This question requires interpreting an interval on a normal distribution. With μ = 20.0 oz and σ = 0.4 oz, we need to analyze the interval from 19.6 oz to 20.4 oz. First, find how many standard deviations each endpoint is from the mean: (19.6 - 20.0)/0.4 = -0.4/0.4 = -1 and (20.4 - 20.0)/0.4 = 0.4/0.4 = 1. The interval spans from 1 standard deviation below to 1 standard deviation above the mean. According to the empirical rule for normal distributions, approximately 68% of all values fall within 1 standard deviation of the mean. This means a majority of bottles (about 68%) fall in this interval, but certainly not "almost all" of them. Understanding that μ ± 1σ captures about 68% of the distribution is crucial for interpreting normal models.

Question 12

  Normal curve (mL)
  |
  |                 .-''''-.
  |              .-'        '-.
  |            .'              '.
  |           /                  \
  |----------|----|----|----|----|----|----|----> x
           476  484  492  500  508  516  524
                          \mu=500      <u>516</u>

1. A fill volume of 516 mL is about 2 standard deviations above the mean, so it is unusually high compared with most bottles. (correct answer)
2. A fill volume of 516 mL is about 2 standard deviations below the mean, so it is unusually low compared with most bottles.
3. A fill volume of 516 mL is about 1 standard deviation above the mean, so it is very typical.
4. A fill volume of 516 mL is about 2 standard errors above the mean, so it is unusually high.
5. A fill volume of 516 mL is about 2 standard deviations above the mean, so most bottles have more than 516 mL.

Explanation: This question checks assessing fill volumes in normal distributions using AP Statistics. Volumes are normal with μ = 500 mL and σ = 8 mL. For 516 mL, z = (516 - 500) / 8 = 2, or 2 standard deviations above the mean. Per the empirical rule, this is beyond 95% of data, so unusually high. Choice E distracts by saying most have more, but only about 2.5% exceed it in the upper tail. Normal models interpret positions relative to μ and σ, using rules to estimate rarity without full calculations.

Question 13

  Normal curve (minutes)
  |
  |               .-''''-.
  |            .-'  <u>####</u>  '-.
  |          .'     <u>####</u>     '.
  |         /       <u>####</u>       \
  |--------|----|----|----|----|----|----> x
          1.5  2.5  3.5  4.5  5.5  6.5  7.5
                <u>[3.5, 5.5]</u>  \mu=4.5

1. 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. (correct answer)
2. The interval from 3.5 to 5.5 minutes is about $\pm 2\sigma$ from the mean, so it contains nearly all checkout times.
3. The interval from 3.5 to 5.5 minutes is about $\pm 1$ standard error from the mean, so it contains about 68% of checkout times.
4. The interval from 3.5 to 5.5 minutes is centered above the mean, so it mostly covers the right tail.
5. The interval from 3.5 to 5.5 minutes is about $\pm 3\sigma$ from the mean, so it contains about 99.7% of checkout times.

Explanation: 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.

Question 14

1. A wait time of 7.5 minutes is about 1 standard deviation above the mean, so it is somewhat high but not unusual. (correct answer)
2. A wait time of 7.5 minutes is about 1 standard deviation below the mean, so it is somewhat low but not unusual.
3. A wait time of 7.5 minutes is about 2 standard deviations above the mean, so it is relatively unusual.
4. A wait time of 7.5 minutes is about 1 standard error above the mean, so it indicates an unusually long wait for the population.
5. A wait time of 7.5 minutes is near the mean because it is within 2 minutes of 6.0.

Explanation: The skill here is interpreting a wait time's location in a normal distribution with μ = 6.0 minutes and σ = 1.5 minutes. For 7.5 minutes, z = (7.5 - 6.0) / 1.5 = 1, so it's 1 standard deviation above the mean, somewhat high but not unusual as it's within the central 68% range. This positions it where about 84% of wait times are shorter. Choice C distracts by saying it's 2 SDs above, possibly from doubling the difference incorrectly. Mini-lesson: Normal distributions are symmetric around μ, with σ measuring spread; intervals like μ ± σ cover ~68%, μ ± 2σ ~95%, helping judge if a value is typical or outlier-like without precise calculations.

Question 15

1. A high of 74°F is about 1 standard deviation above the mean, so it is warmer than average but not unusual. (correct answer)
2. A high of 74°F is about 2 standard deviations above the mean, so it is in the extreme right tail.
3. A high of 74°F is about 1 standard deviation below the mean, so it is cooler than average but not unusual.
4. A high of 74°F is about 1 standard error above the mean, so it suggests the population mean is greater than 68°F.
5. A high of 74°F is near the center because it is less than 10°F from the mean.

Explanation: This question evaluates placing a temperature in a normal distribution with μ = 68°F and σ = 6°F. For 74°F, z = (74 - 68) / 6 = 1, 1 standard deviation above the mean, warmer than average but not unusual. About 84% of days are cooler. Choice B is a distractor, stating 2 SDs above, maybe from doubling the difference. Mini-lesson: Normal models help forecast with z-scores; values within ±1σ are common, covering ~68%, while beyond ±2σ are rare, assisting weather pattern analysis.

Question 16

1. The interval $490$ to $510$ is within about 1 standard deviation of the mean, so it should include about two-thirds of bottles. (correct answer)
2. The interval $490$ to $510$ is within about 2 standard deviations of the mean, so it should include about two-thirds of bottles.
3. The interval $490$ to $510$ is within about 1 standard error of the mean, so it should include about two-thirds of bottles.
4. The interval $490$ to $510$ is 10 mL below the mean to 10 mL above the mean, so it should include nearly all bottles.
5. The interval $490$ to $510$ is centered above the mean, so it includes mostly overfilled bottles.

Explanation: This question tests understanding of intervals and the empirical rule in normal distributions. With μ = 500 mL and σ = 10 mL, the interval 490 to 510 mL extends from (490-500)/10 = -1 to (510-500)/10 = +1 standard deviation from the mean. According to the empirical rule, approximately 68% (about two-thirds) of values in a normal distribution fall within one standard deviation of the mean. The distractors incorrectly state this is 2 standard deviations, confuse standard deviation with standard error, or misinterpret the interval's position. The interval is perfectly centered on the mean, not shifted above it.

Question 17

1. The marked score is 1 standard deviation above the mean, so it is above average but not especially rare.
2. The marked score is 2 standard deviations above the mean, so it is in the upper tail and would be relatively uncommon. (correct answer)
3. The marked score is 2 standard deviations below the mean, so it is in the lower tail and would be relatively uncommon.
4. The marked score is 2 standard errors above the mean, so it is uncommon only for large samples.
5. The marked score is at the mean, so about half of students score above it.

Explanation: This problem requires locating a SAT score on a normal distribution. Given μ = 520 and σ = 80, we need to find where a score of 680 falls. Calculate the standardized value: (680 - 520)/80 = 160/80 = 2. This means 680 is exactly 2 standard deviations above the mean. In a normal distribution, values that are 2 standard deviations above the mean fall in the upper tail and are relatively uncommon, occurring in only about 2.5% of cases. This represents a high SAT Math score that would be achieved by relatively few students. The distinction between standard deviations (which measure individual values' distances from μ) and standard errors (which relate to sampling distributions) is crucial. For interpreting individual scores, we always use standard deviations, not standard errors.

Question 18

1. The marked value is 2 standard deviations above the mean, so it should be among the longest-lasting bulbs.
2. The marked value is 1 standard deviation above the mean, so it is somewhat above average but still fairly typical. (correct answer)
3. The marked value is 1 standard deviation below the mean, so it is somewhat below average but still fairly typical.
4. The marked value is 1 standard error above the mean, so it indicates the sample mean is unusually high.
5. The marked value is at the mean, so about 50% of bulbs last longer than this.

Explanation: 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.

Question 19

1. The marked use is 2 standard deviations above the mean, so it would be quite high compared with most households.
2. The marked use is 1 standard deviation above the mean, so it is higher than average but still fairly typical. (correct answer)
3. The marked use is 1 standard deviation below the mean, so it is lower than average but still fairly typical.
4. The marked use is 1 standard error above the mean, so it would be unusual only for small samples.
5. The marked use is at the mean, so about half of households use more than this.

Explanation: 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.

Question 20

1. The interval is within 1 standard deviation of the mean, so it contains a majority of ball bearings.
2. The interval is within 2 standard deviations of the mean, so it contains almost all ball bearings. (correct answer)
3. The interval is within 3 standard deviations of the mean, so it contains nearly all ball bearings.
4. The interval is within 2 standard errors of the mean, so it contains about 95% of sample means for any sample size.
5. The interval is far from the mean, so it contains only a small minority of ball bearings.

Explanation: 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.