Normal Curves: Compare Means, Widths and Peaks

1. Caption evidence
   Observation

   The problem text on screen states "The average score of students from School A is 60".

2. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Editorial A/B labels denote the schools, distinct from diagram-option A/B.

μ_A = 60

The central position of the normal distribution of scores for School A, corresponding to the horizontal axis position of the curve's peak.

Mean parameter on the score scale

1. Caption evidence
   Observation

   The problem text on screen states "standard deviation is 10" after the conditions for School A.

2. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Editorial A/B labels denote the schools, distinct from diagram-option A/B.

σ_A = 10

The dispersion of the score distribution for School A; a larger standard deviation results in a wider, more spread-out curve.

Dispersion parameter on the score scale

1. Caption evidence
   Observation

   The problem text on screen states "The average score of students from School B is 65".

2. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Editorial A/B labels denote the schools, distinct from diagram-option A/B.

μ_B = 65

The center of School B’s normal curve:65 exceeds60, placing its center to the right of School A’s center, without separating their supports.

Mean parameter on the score scale

1. Caption evidence
   Observation

   The problem text on screen states "standard deviation is 5" after the conditions for School B.

2. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Editorial A/B labels denote the schools, distinct from diagram-option A/B.

σ_B = 5

The dispersion of the score distribution for School B; a smaller standard deviation results in a narrower, taller, and more concentrated curve.

Dispersion parameter on the score scale

1. Caption evidence
   Observation

   The problem text states "If a thick line represents the score distribution curve of students from School A".

2. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Editorial A/B labels denote the schools, distinct from diagram-option A/B.

A

The line style representing the normal distribution curve of School A in the option diagrams.

Legend line style

1. Caption evidence
   Observation

   The problem text states "a thin line represents the score distribution curve of students from School B".

2. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Editorial A/B labels denote the schools, distinct from diagram-option A/B.

B

The line style representing the normal distribution curve of School B in the option diagrams.

Legend line style

1. Caption evidence
   Observation

   The problem text states 'The average score of students from School A is 60'.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

60

Average test score of students from School A

Score

1. Caption evidence
   Observation

   The problem text states 'The standard deviation is 10'.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

10

Standard deviation of test scores for students from School A

Score

1. Caption evidence
   Observation

   The problem text states 'The average score of students from School B is 65'.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

65

Average test score of students from School B

Score

1. Caption evidence
   Observation

   The problem text states 'The standard deviation is 5'.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

5

Standard deviation of test scores for students from School B

Score

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

2. Caption evidence
   Observation

   The problem text explains that the score distributions of both schools are "very close to a normal distribution."

The video describes the normal distribution as a bell-shaped curve and explicitly states that its "position" is determined by the mean and its "shape" is determined by the standard deviation. Therefore, when comparing two normal curves, one can first look at the mean to determine horizontal shift, and then look at the standard deviation to determine width and height.

1. Distribution is close to a normal distribution

2. Comparison objects are two normal curves

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

2. Caption evidence
   Observation

   The problem states School A's mean score is 60 and School B's mean score is 65.

School B’s mean65 exceeds School A’s60, so its normal center, symmetry axis and peak location are to the right. This compares centers, not disjoint supports; the schools also have different standard deviations, so their curves are not simple rigid translations.

1. Horizontal axis represents scores

2. Both schools' distributions are approximately normal

1. Position and shape of the normal distribution curve are determined by the mean and standard deviation

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

2. Caption evidence
   Observation

   The problem states School A's standard deviation is 10 and School B's standard deviation is 5.

The larger the standard deviation, the more dispersed the data, and the wider and flatter the curve; the smaller the standard deviation, the more concentrated the data, and the narrower and taller the curve. In this problem, School A's standard deviation of 10 is greater than School B's standard deviation of 5, so School A's curve should be wider and shorter than School B's.

1. Both curves are normal distributions

2. Comparing dispersion under the same score scale

1. Position and shape of the normal distribution curve are determined by the mean and standard deviation

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

The normalized normal densities in this question have total area1. Scaling the same normal family horizontally by its standard deviation inversely scales density height, with peak1/(σ√(2π)); arbitrary densities of area1 need not have this fixed width-height relation.

1. Normalized normal location-scale family, positive standard deviation, same coordinate scale

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

The larger the standard deviation, the more dispersed the data, resulting in a wider and lower normal distribution curve; the smaller the standard deviation, the more concentrated the data, resulting in a narrower and higher normal distribution curve.

1. Total area is always 1

2. Compare the same normal family, not arbitrary densities.

1. Relationship between Area and Shape of Normal Distribution Curve

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

The mean of a normal distribution sets its symmetry center. A pure horizontal translation holds only when the standard deviation is fixed and the mean alone changes; the schools have different standard deviations, so their complete curves are not simple translations.

1. Single-parameter horizontal translation requires fixed standard deviation.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

2. Caption evidence
   Observation

   School A's mean score is 60, School B's mean score is 65.

School B’s mean65 exceeds School A’s60, so its normal center and peak location lie to the right of School A’s center. Both normal densities still have support on the entire real line; the curves are not disjoint.

1. Score distributions of both schools are close to normal distributions

2. Horizontal axis is score and increases to the right

Holds for the normal distribution curves of Schools A and B in this problem.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

2. Caption evidence
   Observation

   School A's standard deviation is 10, School B's standard deviation is 5.

Because School A's standard deviation of 10 is greater than School B's standard deviation of 5, School A's normal curve should be wider, flatter, and more dispersed than School B's normal curve.

1. Score distributions of both schools are close to normal distributions

2. Comparing curve shapes under the same coordinate scale

Holds for the normal distribution curves of Schools A and B in this problem.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

2. Diagram
   Observation

   The screen displays five normal curve options (A) to (E), and the speaker makes circle and cross marks on the options.

1. The video does not read out the final judgment sentence for each option verbatim; it relies on screen markings and semantic inference.

2. The first-segment model misread D. Native55 shows C crossed out;91 shows A circled. A is the later complete-source answer, not a completed selection at53–61seconds.

1. Expression

   $$$$
   Explanation

   First read the problem setup: School A mean 60, standard deviation 10; School B mean 65, standard deviation 5; thick line represents School A, thin line represents School B.

   Justification

   From the problem text and the speaker's oral description.

   Shown in the video
2. Expression

   $$$$
   Explanation

   Compare positions by mean: 65 > 60, so School B's curve should be to the right of School A's.

   Justification

   The position of the normal curve is determined by the mean.

   Shown in the video
3. Expression

   $$$$
   Explanation

   Observe the five options; the speaker points out that they all roughly satisfy "School A is more to the left, School B is more to the right," so relying solely on the mean cannot distinguish the options.

   Justification

   From the speaker's description of the common position of the five graphs.

   Shown in the video
4. Expression

   $$$$
   Explanation

   Switch to comparing shapes by standard deviation: School A's standard deviation of 10 is larger, so School A's curve should be wider and more dispersed; School B's standard deviation of 5 is smaller, so School B's curve should be narrower and more concentrated.

   Justification

   The shape of the normal curve is determined by the standard deviation.

   Shown in the video
5. Expression

   $$$$
   Explanation

   Check the relative width and height of the thick and thin lines in the graphs item by item, eliminating options that do not fit "thick line is wider, thin line is narrower and shifted right."

   Justification

   Translate problem parameters into graphical features and compare.

   Shown in the video
6. Expression

   $$$$
   Explanation

   This segment begins eliminating mismatched shapes; the full source later confirms A, rather than showing a completed D selection here.

   Justification

   Actual55 and91 native frames distinguish the current stage from the complete-source conclusion.

   Supplementary explanation

This stage establishes location and spread checks; the later complete-source answer is A.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

1. Expression

   $$$$
   Explanation

   Observe options A and B; in both, the thick line (School A) is wider, and the thin line (School B) is narrower.

   Justification

   Visual observation

   Shown in the video
2. Expression

   $$$$
   Explanation

   According to the problem, School A's standard deviation (10) is greater than School B's standard deviation (5), so School A's curve should be wider and lower than School B's.

   Justification

   Standard deviation affects curve width and height

   Derived from the video
3. Expression

   $$$$
   Explanation

   In option B, the peak heights of the two curves are identical, which does not match the height difference caused by different standard deviations.

   Justification

   Peak height in the same normalized normal family is inversely proportional to positive standard deviation; the distinct10 and5 scales cannot give equal peaks.

   Supplementary explanation
4. Expression

   $$$$
   Explanation

   In option A, the thick line (School A) is wider and shorter, while the thin line (School B) is narrower and taller, consistent with the properties of standard deviation.

   Justification

   Intuitive comparison

   Shown in the video

The correct answer is (A).

1. Caption evidence
   Observation

   The screen fully presents the problem: Students from Schools A and B took a math ability test, and their scores are both close to a normal distribution; School A mean 60, standard deviation 10; School B mean 65, standard deviation 5; thick line represents School A, thin line represents School B, asking which distribution graph is more correct.

2. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. The first-segment model misread D. Native55 shows C crossed out;91 shows A circled. A is the later complete-source answer, not a completed selection at53–61seconds.

Students from Schools A and B, who have the same number of students, took a math ability test. The distribution of test scores for students from both schools is very close to a normal distribution. Among them, the average score of students from School A is 60, with a standard deviation of 10; the average score of students from School B is 65, with a standard deviation of 5. If a thick line represents the score distribution curve of students from School A, and a thin line represents the score distribution curve of students from School B, then which of the following distribution graphs is more correct?

1. School A mean score = 60

2. School A standard deviation = 10

3. School B mean score = 65

4. School B standard deviation = 5

5. Thick line represents School A

6. Thin line represents School B

7. Score distributions of both schools are close to normal distributions

Select the distribution graph that best fits the problem setup from the five normal curve graphs (A) to (E).

1. Expression

   $$$$
   Explanation

   First judge the horizontal position: Because School B's mean of 65 is greater than School A's mean of 60, School B's curve should be to the right of School A's.

   Justification

   The position of the normal distribution curve is determined by the mean.

   Shown in the video
2. Expression

   $$$$
   Explanation

   Then judge the shape: Because School A's standard deviation of 10 is greater than School B's standard deviation of 5, School A's curve should be wider, flatter, and more dispersed than School B's.

   Justification

   The shape of the normal distribution curve is determined by the standard deviation.

   Shown in the video
3. Expression

   $$$$
   Explanation

   Combine with line style conditions: The wider curve on the left should be the thick line (School A), and the narrower and taller curve on the right should be the thin line (School B).

   Justification

   The problem specifies that the thick line represents School A and the thin line represents School B.

   Shown in the video
4. Expression

   $$$$
   Explanation

   Combine the mean and standard-deviation conditions to compare the options; the full source later confirms A’s broad low thick curve left and narrow high thin curve right.

   Justification

   The complete answer follows the actual later A marking; the first segment does not mark a final answer.

   Supplementary explanation

Full-source answer A; current0–61seconds establishes the comparison conditions.

Can be checked with two conditions: ① Is the peak of the thin line to the right of the thick line? ② Is the thick line wider and shorter than the thin line? Option (A) satisfies both.

1. Caption evidence
   Observation

   The full problem statement is displayed at the top of the screen.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

Schools A and B have an equal number of students taking a mathematics ability test. The distribution of test scores for students from both schools is close to a normal distribution. The average score for students from School A is 60, with a standard deviation of 10; the average score for students from School B is 65, with a standard deviation of 5. If the thick line represents the score distribution curve for School A and the thin line represents the score distribution curve for School B, which of the following distribution graphs is most correct?

1. School A average score = 60

2. School A standard deviation = 10

3. School B average score = 65

4. School B standard deviation = 5

5. Thick line represents School A, thin line represents School B

Select the correct distribution graph option (A)-(E)

1. Expression

   $$$$
   Explanation

   Compare standard deviations: School A (10) > School B (5), so School A's curve should be wider and shorter than School B's.

   Justification

   Within the same normal family, larger standard deviation increases horizontal scale and lowers the peak.

   Derived from the video
2. Expression

   $$$$
   Explanation

   Compare means: School A (60) < School B (65), so the center of School A's curve should be to the left of School B's.

   Justification

   Mean determines the position of the curve's axis of symmetry

   Derived from the video
3. Expression

   $$$$
   Explanation

   Examine options: (C), (D), and (E) have been crossed out and eliminated. Between (A) and (B), (B) has two curves of equal height, which does not fit the standard deviation difference; (A) has a shorter and wider thick line, and a taller and narrower thin line, with the thick line on the left and the thin line on the right.

   Justification

   Graph feature matching

   Shown in the video

(A)

Option A satisfies both the width/height difference caused by standard deviation and the left/right position difference caused by the mean.

1. Diagram
   Observation

   The screen is a single-page problem with white text on a black background. At the top is a yellow title "CSAT Math 101 Multiple Choice 4 86;77;45;15", the middle section is the problem text, and the bottom has five normal curve graphs (A) to (E) arranged horizontally. In the lower left corner, there is a green option list (1)(A) to (5)(E).

1. Title bar

2. Problem text

3. Five normal curve graphs (A) to (E)

4. Option list in lower left corner

1. No scene switching; the entire segment maintains the same problem page.

1. Problem text and five option graphs remain visible throughout.

This is a static problem page where the solution is completed through oral explanation and subsequent marking, rather than explaining concepts through animation changes.

1. Animation
   Observation

   Green underlines appear sequentially below keywords such as "normal distribution", "average score is 60", and "average score is 65".

2. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Problem stem text

2. Green underlines

1. First marks "normal distribution", then marks School A's mean of 60 and School B's mean of 65.

1. Problem layout remains unchanged, only emphasis marks are added.

Visually separates the core parameters needed for solving: first confirm the distribution type, then confirm the means of the two schools.

1. Animation
   Observation

   Green ellipses circle "standard deviation is 10" and "standard deviation is 5".

2. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Standard deviation text in the problem stem

2. Green ellipse circling mark

1. After marking the means, further shifts attention to the standard deviation conditions.

1. The five option graphs remain unmodified.

This step shifts the focus of solving from "position" to "shape," corresponding to the real distinguishing point of this problem.

1. Animation
   Observation

   Actual native55: C is crossed out and A not yet circled. Later native91: A is circled, with B/C/D/E crossed out.

2. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Some intermediate marks appear for a short moment; the exact seconds of pen strokes can only be approximated.

2. The first-segment model misread D. Native55 shows C crossed out;91 shows A circled. A is the later complete-source answer, not a completed selection at53–61seconds.

1. Five option graphs (A) to (E)

2. Green crosses

3. Green circle marks

4. Option list in lower left corner

1. C has a green cross at this stage; A is circled later.

1. Problem stem text remains unchanged.

Native55 shows C crossed out,91 shows A circled; the first-segment mark is elimination, not selection of D.

1. Animation
   Observation

   Green pen marks draw an 'X' over option (B).

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

1. Graph for option (B)

1. Green 'X' appears

1. Other options remain unchanged

The speaker determines that option B does not conform to the relationship between area and height in a normal distribution and eliminates it.

1. Animation
   Observation

   Green pen marks circle the letter for option (A).

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

1. Letter for option (A)

1. Green circle appears

1. Other options remain unchanged

The speaker confirms A as the correct answer.

1. Animation
   Observation

   Green pen marks shade the area under the thick line in option (A) and draw vertical lines above the thin line.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

1. Curves in option (A)

1. Green shaded area appears under the thick line

2. Green vertical auxiliary lines appear next to the thin line

1. Curve shapes themselves remain unchanged

The source gives a area-1 intuition; the precise inverse width-height relation concerns scale changes within the same normalized normal family, not arbitrary densities.

1. Animation
   Observation

   Green pen marks draw a box around the answer area (1)(A) in the bottom left corner.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

1. Answer area (1)(A)

1. Green box appears

1. None

Emphasizes that the final selected answer is (1)(A).

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

Thinking that just judging "B is to the right of A" is enough to select the correct graph.

In the video, all five options roughly satisfy A on the left and B on the right, so one must further use the standard deviation to compare curve widths to filter further.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

It is easy to intuitively associate "larger value" with "taller curve," mistakenly thinking that a curve with a larger standard deviation is sharper and taller.

According to the video, a larger standard deviation means the data is more dispersed, and the curve should be wider and flatter; a smaller standard deviation means it is more concentrated, narrower, and taller. This item is an error-prone reminder organized by the analyst based on reasoning from the video.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

Believing that as long as the means are different or only the widths are different, the highest point (peak) of the curve can remain the same height.

For normalized normal densities on the same axes, the peak is1/(σ√(2π)); larger positive standard deviation gives a lower peak. Area1 alone does not imply this for arbitrary densities.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

Comparing means locates the relative normal centers: School B’s center is to the right of School A’s center, without separating the curves’ supports.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

The method "standard deviation determines width and height" is directly applied to this problem, yielding the conclusion that School A's curve is wider and more dispersed than School B's.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

The general property "normal curve is determined by mean and standard deviation" contains two sub-criteria, one of which is that the mean controls position.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

The general property also contains another sub-criterion, namely that the standard deviation controls shape and dispersion.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

Applies the properties of standard deviation to understand changes in the shape of the normal distribution curve.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

Uses the property that the mean determines curve position to solve the problem.

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. The mean determines the horizontal position of the normal curve
2. School B's curve is located to the right of School A's curve

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. The standard deviation determines the width and height of the normal curve
2. Looking only at the mean is not enough to distinguish the options in this problem

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. The standard deviation determines the width and height of the normal curve
2. School A's curve is wider and more dispersed than School B's curve

1. Diagram
   Observation

   The screen finally circles (A), and (1)(A) in the lower left corner is also circled.

1. The first-segment model misread D. Native55 shows C crossed out;91 shows A circled. A is the later complete-source answer, not a completed selection at53–61seconds.

1. Using mean and standard deviation to eliminate options and determine the answer
2. CSAT Math 101 Multiple Choice 4: Interpreting Normal Distribution Graphs of Two Schools' Scores

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

1. Relationship between Area and Shape of Normal Distribution Curve
2. Misconception that Standard Deviation Does Not Affect Curve Height

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

1. Effect of Standard Deviation on Normal Distribution Curve Shape

1. Audio
   Observation

   The narration compares centers by mean, then normal-curve width and peaks by standard deviation; the later area-1 explanation completes the selection of A.

1. Compare the same normalized normal family on the same coordinate scale; area1 alone does not determine width and peak height for arbitrary densities.

1. Effect of Mean on Normal Distribution Curve Position