The Normal Distribution, Revisited
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AP Statistics › The Normal Distribution, Revisited
Scores on a standardized reading test for a large school district are approximately Normal with mean $\mu=500$ and standard deviation $\sigma=100$. Without calculating any exact probabilities, which comparison is supported by the symmetry of the Normal curve about the mean?
The area between 450 and 550 is less than the area between 400 and 600 because the first interval is closer to the mean.
The area between 300 and 500 equals the area between 500 and 700 because the curve is taller on the left side.
The area between 400 and 500 is greater than the area between 500 and 600 because 400 is farther from the mean.
The area below 400 equals the area above 600 because both are within one standard deviation of the mean.
The area between 400 and 500 equals the area between 500 and 600 because the intervals are symmetric about 500.
Explanation
This question tests qualitative reasoning about the symmetry of the normal distribution in AP Statistics, specifically how areas under the curve compare without calculations. The normal curve is symmetric about its mean, μ=500, meaning areas equidistant from the mean are equal. Choice B correctly states that the areas between 400-500 and 500-600 are equal because these intervals are symmetric about 500, each spanning 100 units away from the mean on either side. A common distractor like choice A might mislead by suggesting distance affects area unequally, but symmetry ensures equality regardless of the curve's height. In a mini-lesson on qualitative normal reasoning, remember that the bell shape peaks at the mean and mirrors perfectly on both sides, so probabilities for mirrored intervals are identical. This symmetry is key for comparing tails or intervals without z-scores or tables.
Heights of a certain plant species are approximately Normal with mean $\mu=30$ cm and standard deviation $\sigma=4$ cm. Consider the shaded region shown from 26 cm to 34 cm. Without calculating probabilities, which comparison is supported by the Normal curve’s shape and symmetry?
The area from 26 to 30 equals the area from 30 to 34 because the intervals are symmetric about 30.
The shaded area from 26 to 34 is greater than the area from 22 to 30 because 22 to 30 is farther from the mean.
The area from 29 to 31 is less than the area from 26 to 34 because the curve is lower near the mean.
The area from 26 to 34 equals the area outside 26 to 34 because the interval is two standard deviations wide.
The area below 26 is greater than the area above 34 because the left tail is longer in a Normal distribution.
Explanation
This AP Statistics question focuses on the normal distribution's symmetry and shape for comparing areas without computations, with heights ~N(30,4) and a shaded region from 26 to 34. The shaded interval is symmetric about μ=30, spanning 4 cm (one σ) on each side. Choice B correctly notes that areas from 26-30 and 30-34 are equal due to symmetry about 30. Distractor choice C wrongly claims the left tail is longer, but normal curves are perfectly symmetric with equal tails. In a mini-lesson on qualitative reasoning, the normal curve's bell shape means intervals of equal length symmetric about the mean have identical areas, aiding in quick comparisons. Referencing the shaded region, its total area is about 68% by the empirical rule, but symmetry splits it evenly.
The weights of packages shipped from a warehouse are approximately Normal with mean $\mu=12$ lb and standard deviation $\sigma=1.5$ lb. Without calculating probabilities, which comparison is supported by symmetry?
The proportion of packages between 10.5 lb and 12 lb is less than the proportion between 12 lb and 13.5 lb.
The proportion of packages above 13.5 lb is greater than the proportion below 10.5 lb because the right tail is longer.
The proportion of packages weighing less than 10.5 lb equals the proportion weighing more than 13.5 lb.
The proportion of packages below 12 lb is less than 0.50 because the Normal curve is bell-shaped.
The proportion of packages between 11 lb and 13 lb equals the proportion outside 11 lb to 13 lb.
Explanation
In AP Statistics, this revisits normal symmetry for package weights ~N(12,1.5), asking for supported comparisons without probabilities. Symmetry about μ=12 ensures equal areas for points equidistant from the mean. Choice A is correct: proportion <10.5 equals >13.5, both 1.5 lb (one σ) away. Distractor choice B wrongly compares intervals by assuming unequal areas due to position, but heights differ while symmetry holds for tails. Mini-lesson: Qualitative reasoning uses the bell curve's mirror symmetry across μ, equating opposite tail probabilities. No shaded region specified, but envision tails beyond 10.5 and 13.5 as equal shaded areas.
Daily low temperatures in a city during January are approximately Normal with mean $\mu=20^\circ!F$ and standard deviation $\sigma=6^\circ!F$. Without calculating probabilities, which comparison is supported by the fact that the Normal curve is highest at the mean and decreases as you move away from the mean?
The area between $18^\circ!F$ and $22^\circ!F$ is greater than the area between $8^\circ!F$ and $12^\circ!F$ because the first interval is closer to the mean.
The area between $8^\circ!F$ and $12^\circ!F$ is greater than the area between $18^\circ!F$ and $22^\circ!F$ because both intervals have width 4.
The area below $14^\circ!F$ is greater than the area above $26^\circ!F$ because $14$ is closer to the mean.
The area between $14^\circ!F$ and $20^\circ!F$ is greater than the area between $20^\circ!F$ and $26^\circ!F$ because the curve is higher on the left side.
The area between $18^\circ!F$ and $22^\circ!F$ equals the area between $14^\circ!F$ and $26^\circ!F$ because both are centered at the mean.
Explanation
In AP Statistics, this question tests qualitative understanding of the normal curve's peak at the mean and decreasing height away from it, for temperatures ~N(20,6). The curve is highest at μ=20, so intervals closer to the mean have more area for the same width. Choice B correctly states the area between 18-22 is greater than between 8-12, as 18-22 is nearer the mean. Distractor choice A wrongly suggests the left side is higher, but the curve is symmetric and decreases equally on both sides. Mini-lesson on qualitative reasoning: The normal density decreases as you move from the mean, so for fixed-width intervals, those centered at μ have the largest area, decreasing outward. No specific shaded region, but imagine shading these intervals to compare heights visually.
A company tracks the number of seconds it takes a computer to boot. Boot times are approximately Normal with mean $\mu=42$ s and standard deviation $\sigma=6$ s. Which comparison is supported without computing any exact probabilities?
The area between 36 s and 48 s equals the area outside 36 s to 48 s because the interval spans one standard deviation on each side.
The area between 36 s and 42 s equals the area between 42 s and 48 s because the intervals are symmetric about the mean.
The area between 39 s and 45 s is less than the area between 33 s and 51 s because narrower intervals always have less area, even if centered at the mean.
The area between 30 s and 42 s is less than the area between 42 s and 54 s because the right tail is longer.
The area below 36 s is greater than the area above 48 s because values below the mean are more common.
Explanation
This AP Statistics problem tests qualitative comparisons using normal symmetry for boot times ~N(42,6). Intervals symmetric about μ=42 should have equal areas if equidistant. Choice A is correct: areas 36-42 and 42-48 are equal, both 6 s (one σ) spans symmetric about mean. Distractor choice E misapplies width, ignoring that narrower central intervals can have more area due to height, but here it's about symmetry. Mini-lesson: Qualitative reasoning leverages the curve's symmetry and decreasing density from μ, equating mirrored intervals. No shaded region mentioned, but consider 36-48 as shaded, split evenly by symmetry.
Two sections of the same course have exam scores that are approximately Normal with the same mean $\mu=75$. Section 1 has $\sigma=8$ and Section 2 has $\sigma=12$. A score is considered “high” if it is at least 90. Without computing any exact probabilities, which comparison is supported?
Both sections have the same proportion of scores at least 90 because 90 is 15 points above the mean in both sections.
Section 1 has a larger proportion of scores at least 90 because a smaller $\sigma$ produces heavier tails.
Both sections have the same proportion of scores at least 90 because they have the same mean.
Section 1 has a larger proportion of scores at least 90 because its Normal curve is wider.
Section 2 has a larger proportion of scores at least 90 because a larger $\sigma$ spreads scores out, putting more area in the tails.
Explanation
This AP Statistics question compares tail proportions qualitatively based on σ's effect, with scores ~N(75,σ) and 'high' as ≥90. Larger σ=12 in Section 2 spreads scores more, increasing tail areas beyond 90 (15 above μ). Choice B is correct: Section 2 has more high scores due to greater spread into tails. Distractor choice A reverses this, wrongly saying smaller σ produces heavier tails, but it actually concentrates near μ. Mini-lesson: Qualitatively, larger σ flattens the curve, pushing more probability into tails, while smaller σ peaks higher centrally. No shaded region, but imagine shading ≥90 to see Section 2's wider tail area.
The time (in minutes) it takes customers to complete an online checkout is approximately Normal with mean $\mu=8$ and standard deviation $\sigma=2$. Without doing any computations, which comparison is supported by the symmetry of the Normal distribution?
The area between 6 and 10 is less than the area between 7 and 9 because 6 and 10 are farther from the mean.
The area between 7 and 9 equals the area outside 7 to 9 because the interval is one standard deviation wide.
The area between 6 and 8 is greater than the area between 8 and 10 because the curve decreases as $x$ increases.
The area to the left of 6 is less than the area to the right of 10 because 6 is closer to the mean than 10.
The area to the left of 6 equals the area to the right of 10 because 6 and 10 are equally far from the mean.
Explanation
In AP Statistics, this problem revisits the normal distribution by emphasizing symmetry for qualitative comparisons of areas. With mean μ=8 and σ=2, the curve is symmetric about 8, so points equally distant from the mean have equal tail areas. Choice C is correct: the area left of 6 equals the area right of 10, as both are 2 units (one σ) away from 8 on opposite sides. A distractor like choice A incorrectly assumes closeness to the mean affects tail areas differently, ignoring symmetry. For a mini-lesson, qualitative reasoning involves visualizing the bell curve's mirror image across the mean, ensuring P(X < μ - k) = P(X > μ + k) for any k. This holds regardless of the distribution's spread, focusing purely on symmetry.
Two machines produce bolts whose lengths are approximately Normal. Machine A: $\mu=50$ mm, $\sigma=2$ mm. Machine B: $\mu=50$ mm, $\sigma=5$ mm. A bolt is considered “acceptable” if its length is between 48 mm and 52 mm. Without computing any exact probabilities, which comparison is supported?
Machine A has a greater proportion of acceptable bolts because a smaller $\sigma$ concentrates more area near the mean.
Both machines have the same proportion acceptable because the acceptable interval has the same width for both machines.
Both machines have the same proportion acceptable because they have the same mean.
Machine B has a greater proportion of acceptable bolts because a larger $\sigma$ makes the curve taller near the mean.
Machine B has a greater proportion acceptable because 48 to 52 is symmetric about the mean.
Explanation
This AP Statistics problem revisits the normal distribution by comparing proportions qualitatively based on standard deviation's effect, with acceptable bolts between 48-52 mm for both machines. Machine A has smaller σ=2, concentrating more probability near μ=50, while Machine B's larger σ=5 spreads it out. Choice B is correct: Machine A has more acceptable bolts as its narrower curve puts more area in the central interval. Distractor choice A reverses this, incorrectly stating larger σ makes the curve taller near the mean, but actually smaller σ does that. Mini-lesson: Qualitatively, smaller σ results in a taller, skinnier bell curve, increasing central probabilities, while larger σ flattens and widens it. No shaded region is mentioned, but visualize the acceptable interval as shaded to see A's higher density.