1. What is a distribution?
A distribution describes how values are spread across their possible range.
For example, a dataset of measurements may contain:
- Many values near the center.
- Fewer values moderately far from the center.
- Very few values far from the center.
A normal distribution is a mathematical model with this general pattern.
Its graph is a smooth, symmetric, bell-shaped curve.
A model, not a universal rule
Some measurements can be modeled approximately by a normal distribution under suitable conditions.
However:
- Not every dataset is normally distributed.
- A symmetric dataset is not automatically normal.
- Knowing the mean and standard deviation does not prove normality.
Use normal-distribution properties only when normality is stated or reasonably established.
2. Random variables and notation
A random variable assigns a numerical value to the result of a random experiment.
For example:
X = the measured length of a randomly selected component
A normal random variable is continuous. Its mathematical model allows values throughout the real number line.
We write:
X ~ N(μ, σ²)
Read this as:
“X follows a normal distribution with mean μ and variance σ².”
Here:
- μ, read as “mu,” is the mean.
- σ², read as “sigma squared,” is the variance.
- σ, read as “sigma,” is the standard deviation.
- σ > 0.
Important parameter convention
The second parameter in N(μ, σ²) is the variance, not the standard deviation.
Worked example 1: Reading the notation
Suppose:
X ~ N(50, 16)
Then:
- Mean: μ = 50.
- Variance: σ² = 16.
- Standard deviation: σ = √16 = 4.
The standard deviation is not 16.
3. Main properties of a normal curve
A normal curve:
- Is bell-shaped.
- Is symmetric about its mean.
- Has its highest point at the mean.
- Has total area 1 beneath it.
- Extends indefinitely in both directions.
- Approaches the horizontal axis without touching it.
- Is determined by its mean and standard deviation.
For a normal distribution:
Mean = median = mode = μ
Properties reference
| Property | Meaning |
|---|---|
| Symmetry | The two halves mirror each other about x = μ. |
| Center | Mean, median, and mode all equal μ. |
| Total area | The area beneath the entire curve is 1. |
| Spread | The standard deviation σ controls horizontal spread. |
| Tails | The curve extends indefinitely on both sides. |
Important: These statements describe the theoretical distribution. A finite sample drawn from it need not have exactly equal sample mean, median, and mode.
4. How the mean changes the graph
The mean determines the horizontal location of the curve.
If two normal distributions have the same standard deviation but different means:
- Their shapes and spreads are the same.
- Their centers are different.
- One curve is a horizontal shift of the other.
Worked example 2
Compare:
X ~ N(40, 9)
Y ~ N(55, 9)
Both have standard deviation 3.
The curve for Y has the same shape as the curve for X, shifted 15 units to the right.
Conclusion: A larger mean does not automatically mean greater dispersion.
5. How standard deviation changes the graph
For normal distributions with the same mean:
- Smaller σ: narrower curve with a higher peak.
- Larger σ: wider curve with a lower peak.
The total area remains 1 in both cases.
Worked example 3
Compare:
A ~ N(20, 4)
B ~ N(20, 25)
Both have mean 20.
Their standard deviations are:
- A: σ = 2.
- B: σ = 5.
Therefore:
- A is more tightly concentrated around 20.
- B is more spread out.
- A has the higher peak.
Why the peak changes
A narrower normal curve must be taller to retain total area 1.
Warning: A higher peak does not mean the probability of exactly the mean is greater than zero. Probability is measured by area, not by the height at a point.
6. Probability as area
For a continuous random variable, probabilities correspond to areas beneath its density curve.
For a normal random variable X:
P(a < X < b)
is the area under the curve between a and b.
Reading common probability expressions
| Expression | Area represented |
|---|---|
| P(X < a) | Area to the left of a |
| P(X > b) | Area to the right of b |
| P(a < X < b) | Area between a and b |
| P(X < a or X > b) | Combined area in the two outside regions, when a < b |
Probability at one exact value
For a normal random variable:
P(X = a) = 0
A single point has zero width, so it contributes zero area.
Therefore:
P(X < a) = P(X ≤ a)
and:
P(a < X < b) = P(a ≤ X ≤ b)
Probability zero does not always mean impossible
In a continuous model, an exact value can have probability zero even though it belongs to the possible range.
This differs from the finite positive-probability models used in earlier lessons.
Recorded measurements
A displayed measurement such as 10.0 may represent a rounded interval rather than one exact mathematical value.
This lesson treats X as the underlying continuous quantity unless stated otherwise.
7. Symmetry about the mean
A normal distribution is symmetric about μ.
Exactly half the area lies on each side:
P(X < μ) = 1/2
P(X > μ) = 1/2
Because P(X = μ) = 0, the same statements hold with ≤ or ≥.
Worked example 4
Suppose:
X ~ N(70, 25)
Find:
- P(X < 70).
- P(X ≥ 70).
Since the mean is 70:
P(X < 70) = 0.5
P(X ≥ 70) = 0.5
The variance is not needed for these two calculations.
8. Equal distances from the mean
Points equally far from the mean have matching reflected areas.
For d > 0:
P(X < μ − d) = P(X > μ + d)
Also:
P(μ − d < X < μ) = P(μ < X < μ + d)
Worked example 5: Matching tails
A normal distribution has mean 50.
Given:
P(X < 42) = 0.12
find:
- P(X > 58).
- P(42 < X < 58).
Step 1: Check the distances
50 − 42 = 8
58 − 50 = 8
The boundaries are symmetric about the mean.
Step 2: Match the tails
P(X > 58) = 0.12
Step 3: Find the central area
The two tails have total probability:
0.12 + 0.12 = 0.24
Therefore:
P(42 < X < 58) = 1 − 0.24 = 0.76
Key lesson: Equal-tail conclusions require boundaries equally distant from the mean.
9. Central intervals and tail probabilities
Suppose a symmetric interval around the mean contains probability q.
The probability outside it is:
1 − q
By symmetry, each tail has probability:
(1 − q)/2
Worked example 6
A normal distribution has mean 100.
It is given that:
P(90 < X < 110) = 0.80
Find:
- P(X < 90).
- P(X > 110).
- P(100 < X < 110).
- P(X < 110).
Step 1: Find the outside area
Outside probability = 1 − 0.80 = 0.20
Step 2: Split the tails equally
P(X < 90) = 0.10
P(X > 110) = 0.10
Step 3: Split the central area equally
P(100 < X < 110) = 0.80/2 = 0.40
Step 4: Find the area left of 110
P(X < 110) = 1 − 0.10 = 0.90
Verification
From left to right, the four regions have probabilities:
0.10, 0.40, 0.40, 0.10
Their total is 1.
10. Measuring distance in standard deviations
A standard deviation provides a natural unit for describing distance from the mean.
For a value x:
Standardized distance = (x − μ)/σ
This is often called a z-score:
z = (x − μ)/σ
Interpretation
- z = 0: x equals the mean.
- z = 1: x is one standard deviation above the mean.
- z = −2: x is two standard deviations below the mean.
Worked example 7
Suppose:
X ~ N(60, 16)
Then:
μ = 60
σ = 4
For x = 68:
z = (68 − 60)/4 = 2
So 68 is two standard deviations above the mean.
For x = 56:
z = (56 − 60)/4 = −1
So 56 is one standard deviation below the mean.
Standard normal distribution
If X is normal, the transformed variable:
Z = (X − μ)/σ
has mean 0 and variance 1.
It is called the standard normal distribution:
Z ~ N(0, 1)
This lesson uses standardization only to locate values and identify familiar symmetric intervals. Normal tables are not needed.
11. Approximate central-area benchmarks
The following are useful properties of a normal distribution:
| Interval | Approximate probability | Approximate probability in each outside tail |
|---|---|---|
| μ − σ to μ + σ | 68.27% | 15.865% |
| μ − 2σ to μ + 2σ | 95.45% | 2.275% |
| μ − 3σ to μ + 3σ | 99.73% | 0.135% |
These are often rounded to the 68–95–99.7 rule.
They are approximations, not exact percentages. They apply to normal distributions, not arbitrary datasets.
For exercises in this lesson, any required central probability is supplied. Use the supplied precision consistently.
Worked example 8: Applying a benchmark
Suppose:
X ~ N(100, 25)
Use:
P(μ − 2σ < X < μ + 2σ) ≈ 0.9545
Find:
- The corresponding central interval.
- P(X > 110).
Step 1: Find σ
σ = √25 = 5
Step 2: Find the interval
μ − 2σ = 100 − 10 = 90
μ + 2σ = 100 + 10 = 110
The interval is:
90 < X < 110
Step 3: Find one tail
P(X > 110) ≈ (1 − 0.9545)/2
P(X > 110) ≈ 0.02275
Answer: Approximately 2.275%.
12. Comparing normal distributions
Worked example 9: Same mean, different spread
Two distributions are:
X ~ N(50, 4)
Y ~ N(50, 16)
Which has the greater probability of lying between 46 and 54?
Step 1: Find the standard deviations
For X:
σ = 2
For Y:
σ = 4
Step 2: Interpret the interval
For X, the interval 46 to 54 is:
μ − 2σ to μ + 2σ
For Y, it is:
μ − σ to μ + σ
The normal central-area benchmarks give:
- X: approximately 95.45%.
- Y: approximately 68.27%.
Therefore, X has the greater probability of lying between 46 and 54.
Important condition
This comparison uses the same interval centered on a shared mean.
For intervals elsewhere on the number line, “smaller standard deviation means greater probability” is not a universal rule.
13. Normal models and real data
A model may be approximate
Physical measurements may have practical limits, while a normal model extends over all real numbers.
For example, a normal model for length assigns some mathematical probability to negative values. The model is useful only when such unrealistic probabilities are negligible for the intended application.
Do not assume normality from appearance alone
A rough bell shape may suggest a normal model, but it does not prove one.
Likewise, symmetry alone is insufficient. Other distributions can also be symmetric.
Sample percentages need not match exactly
Even if measurements follow a normal model:
- A particular sample need not contain exactly 68.27% within one standard deviation.
- Its sample mean need not equal the model mean exactly.
- Its sample histogram need not be perfectly symmetric.
The theoretical curve describes the probability model, not a fixed pattern every sample must reproduce.
14. Common mistakes and verification
Mistake 1: Reading variance as standard deviation
For N(30, 9):
Variance = 9
Standard deviation = 3
Mistake 2: Treating curve height as probability
Probability is area over an interval, not height at a single point.
Mistake 3: Assigning positive probability to one exact value
For a normal random variable:
P(X = a) = 0
Mistake 4: Splitting an asymmetric interval into equal halves
Symmetry applies to reflected intervals about μ, not arbitrary intervals.
Mistake 5: Forgetting to divide the outside area by 2
If a symmetric central interval has probability 0.90, the two tails total 0.10.
Each tail is 0.05.
Mistake 6: Treating approximate benchmarks as exact
Use ≈ when applying rounded normal-area values.
Mistake 7: Applying normal benchmarks to any dataset
The 68–95–99.7 rule requires a normal or approximately normal model.
Mistake 8: Assuming three standard deviations contain every value
Approximately 99.73% is not 100%. The tails continue beyond that interval.
Verification checklist
- Is a normal model stated?
- Have I identified μ and σ correctly?
- Is the second notation parameter a variance?
- Are the boundaries symmetric about μ?
- Am I calculating a central area, one tail, or both tails?
- Do the disjoint region probabilities add to 1?
- Is the result between 0 and 1?
- Have I distinguished exact symmetry results from approximations?
15. Independent practice
Complete these before reading the solutions.
Practice 1
Given:
X ~ N(40, 9)
State the mean, variance, standard deviation, median, and mode.
Practice 2
Given:
X ~ N(75, 16)
Find:
- P(X < 75).
- P(X ≥ 75).
- P(X = 75).
Practice 3
A normal distribution has mean 20.
Given:
P(X < 14) = 0.08
Find:
- P(X > 26).
- P(14 < X < 26).
Practice 4
A normal distribution has mean 50.
Given:
P(45 < X < 55) = 0.70
Find:
- P(X > 55).
- P(50 < X < 55).
- P(X < 55).
Practice 5
Given:
X ~ N(100, 36)
Find the values one standard deviation below and above the mean.
Practice 6
Given:
X ~ N(30, 16)
Find the standardized distances of:
- x = 38.
- x = 26.
Practice 7
For a normal random variable, use:
P(μ − σ < X < μ + σ) ≈ 0.6827
Find the approximate probability:
P(X > μ + σ)
Practice 8
Given:
X ~ N(60, 25)
Use the supplied approximation:
P(μ − 2σ < X < μ + 2σ) ≈ 0.9545
Find:
- P(50 < X < 70).
- P(X < 50 or X > 70).
- P(X > 70).
Practice 9
Compare:
A ~ N(10, 1)
B ~ N(10, 9)
Which curve has:
- The greater spread?
- The higher peak?
- The same center as the other?
Practice 10
A student says:
A normal variable must lie between μ − 3σ and μ + 3σ.
Explain the error.
16. Practice solutions
Solution 1
From X ~ N(40, 9):
- Mean = 40.
- Variance = 9.
- Standard deviation = 3.
- Median = 40.
- Mode = 40.
Solution 2
By symmetry:
P(X < 75) = 0.5
P(X ≥ 75) = 0.5
Because the distribution is continuous:
P(X = 75) = 0
Solution 3
The boundaries 14 and 26 are both 6 units from the mean 20.
Therefore:
P(X > 26) = 0.08
Central probability:
P(14 < X < 26) = 1 − 0.08 − 0.08 = 0.84
Solution 4
The central interval is symmetric.
Outside probability:
1 − 0.70 = 0.30
Each tail:
0.30/2 = 0.15
Therefore:
P(X > 55) = 0.15
P(50 < X < 55) = 0.70/2 = 0.35
P(X < 55) = 1 − 0.15 = 0.85
Solution 5
σ = √36 = 6
One standard deviation below the mean:
100 − 6 = 94
One standard deviation above:
100 + 6 = 106
Solution 6
μ = 30, σ = 4
For x = 38:
z = (38 − 30)/4 = 2
For x = 26:
z = (26 − 30)/4 = −1
Solution 7
Both outside tails total:
1 − 0.6827 = 0.3173
Each tail:
0.3173/2 = 0.15865
P(X > μ + σ) ≈ 0.15865
Solution 8
μ = 60, σ = 5
The interval 50 to 70 is μ ± 2σ.
Therefore:
P(50 < X < 70) ≈ 0.9545
Both tails:
P(X < 50 or X > 70) ≈ 1 − 0.9545 = 0.0455
Right tail:
P(X > 70) ≈ 0.0455/2 = 0.02275
Solution 9
Standard deviations:
- A: 1.
- B: 3.
Therefore:
- B has greater spread.
- A has the higher peak.
- Both have center 10.
Solution 10
The interval μ ± 3σ contains approximately 99.73% of the probability, not all of it.
Approximately 0.27% lies outside, split equally between the two tails.
The normal curve continues indefinitely in both directions.
17. Lesson self-check
This is a formative check within 04.6, not a replacement for a separate scheduled normal-distribution assessment.
Check 1
Given:
X ~ N(80, 25)
What is its standard deviation?
A. 25
B. 5
C. 625
D. 80
Check 2
For a normal random variable with mean μ, which statement is correct?
A. P(X = μ) = 0.5
B. P(X < μ) = 1
C. P(X > μ) = 0
D. P(X < μ) = 0.5
Check 3
A normal distribution has mean 30.
Given:
P(X < 24) = 0.10
What is P(X > 36)?
A. 0.10
B. 0.20
C. 0.80
D. 0.90
Check 4
A symmetric interval around a normal distribution’s mean contains probability 0.90.
What is the probability in each outside tail?
A. 0.90
B. 0.10
C. 0.05
D. 0.45
Check 5
Two normal distributions have the same mean. Distribution A has standard deviation 2, and B has standard deviation 6.
Which statement is correct?
A. A is wider and lower.
B. B is wider and lower.
C. Their curves are identical.
D. B has total area 3.
Check 6
For:
X ~ N(50, 16)
how many standard deviations above the mean is 58?
A. 1/2
B. 1
C. 8
D. 2
18. Self-check answers and explanations
Check 1: B
The second parameter is the variance.
σ = √25 = 5
Check 2: D
Symmetry places half the probability below the mean.
P(X < μ) = 0.5
The probability at the exact mean is zero.
Check 3: A
The values 24 and 36 are both 6 units from the mean 30.
The reflected tails have equal probabilities.
P(X > 36) = 0.10
Check 4: C
Outside probability:
1 − 0.90 = 0.10
Each tail:
0.10/2 = 0.05
Check 5: B
A larger standard deviation produces a wider normal curve with a lower peak.
Both curves retain total area 1.
Check 6: D
σ = √16 = 4
z = (58 − 50)/4 = 2
The value 58 is two standard deviations above the mean.
19. Revision guidance
- Difficulty reading notation: Separate the mean, variance, and square root of the variance.
- Difficulty with symmetry: Mark the mean and compare boundary distances.
- Difficulty with tails: Identify whether the question asks for one tail or both.
- Difficulty with intervals: Divide a symmetric central area into two equal parts.
- Difficulty with standardized distances: Calculate σ before using z = (x − μ)/σ.
- Difficulty interpreting curves: Separate location, spread, height, and area.
- Difficulty with benchmarks: Use them only for a stated normal model and preserve approximate notation.
For each incorrect or guessed answer:
- State the mean and standard deviation.
- Identify the relevant region.
- State whether symmetry applies.
- Calculate the probability or standardized distance.
- Check that the result is reasonable.
20. Lesson review
You should now be able to:
- Recognize the basic shape of a normal distribution.
- Interpret X ~ N(μ, σ²).
- Distinguish variance from standard deviation.
- Explain the roles of μ and σ.
- Interpret probabilities as areas.
- Explain why exact points have probability zero.
- Use symmetry about the mean.
- Calculate equal tails from a supplied central probability.
- Describe distance from the mean in standard-deviation units.
- Apply supplied normal-area benchmarks.
- Distinguish theoretical properties from finite-sample results.
- Recognize the limits of a normal model.
Essential statements
X ~ N(μ, σ²), with σ > 0
Mean = median = mode = μ
Total area under the curve = 1
P(X < μ) = P(X > μ) = 0.5
P(X = a) = 0
For d > 0: P(X < μ − d) = P(X > μ + d)
Each outside tail = (1 − symmetric central probability)/2
z = (x − μ)/σ
Completion target: Interpret the parameters and curve correctly, use symmetry without a normal table, and independently correct every missed practice or self-check question.