seegongsik
Saved words
Grade 11-12 (age 16-18)

Normal Distribution

Normal Distribution

Data that is very common in nature, like height or weight, piles up into a bell shape that bulges in the middle, and that is the normal distribution. The mean μ sets where the bell is centered and the standard deviation σ sets how wide it is, and about 68, 95, and 99.7% always fall within ±1σ, 2σ, and 3σ of the center. Even different normal distributions can be merged into one standard normal Z, with mean 0 and standard deviation 1, once you standardize them, so they become comparable. Use the mean μ and standard deviation σ sliders to watch the bell shift sideways and grow narrow or wide, and move the z value to see how the area (probability) changes.

Bell-shaped Curve
🔔 What is the Normal Distribution?
①Bell-shaped distribution that fits many natural and social phenomena
②Symmetric about the mean μ
③Smaller σ → tall and narrow; larger σ → short and wide
④Total area = 1 (sum of probabilities)
N(μ, σ²) Curve
50
10
Probability Density Function
f(x) = 1σ√(2π) e-\frac{(x-μ)^2}{2σ^2}
PDF of N(μ, σ²)
📏 68-95-99.7 Rule
①μ ± 1σ: ≈ 68.3%
②μ ± 2σ: ≈ 95.4%
③μ ± 3σ: ≈ 99.7%
④Larger σ → wider spread
Standard Normal Z
15
Standardization
Z = X - μσ
Convert N(μ,σ²) to N(0,1)
💡 Why Standardize
①Any normal distribution becomes N(0,1) via Z
②One standard table covers all normal probabilities
③P(a ≤ X ≤ b) = P(z₁ ≤ Z ≤ z₂)
Binomial vs Normal
Normal Approximation
B(n, p) ≈ N(np, np(1−p)) (large n)
For large n the binomial approaches the normal
Work It Out
Example 1
If X follows the normal distribution N(50, 10²), standardize X = 70 to find its Z value.
1
Write the standardization formula.
Z = (X − m)/σ
2
Substitute m = 50, σ = 10.
Z = (70 − 50)/10 = 2
Z = 2
Standardizing subtracts the mean and divides by the SD to get N(0, 1).
Example 2
If X follows N(50, 10²), find P(50 ≤ X ≤ 70). (Given P(0 ≤ Z ≤ 2) = 0.4772)
1
Standardize: X = 50 gives Z = 0, X = 70 gives Z = 2.
P(50 ≤ X ≤ 70) = P(0 ≤ Z ≤ 2)
2
Read the value from the standard normal table.
= 0.4772
0.4772
A normal probability is found by standardizing, then reading the standard-normal table (area).
Wrap-up
Core
Z = X-μσ, X ∼ N(μ, σ2)
Normal distribution and standardization
KICE mock-exam Math (Probability & Statistics) type
If X follows the normal distribution N(60, 8²), what is P(X ≥ 60)?
0.1587
0.3413
0.5
0.6587
0.8413
③ 0.5
1
A normal distribution is symmetric about its mean m.
P(X ≥ m) = 0.5
2
Since m = 60, it is exactly 0.5.
P(X ≥ 60) = 0.5
🎯 Exam Points
①Standardize via Z = (X−μ)/σ
②Read P(0≤Z≤z) from a Z-table
③68-95-99.7 rule
④P(Z≥a) = 0.5 − P(0≤Z≤a)
⑤Binomial → normal approximation when n is large
← Previous
Probability Distribution
Next →
Statistical Estimation
Was this helpful? Support seegongsik