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high school Normal Distribution

Data that is common in nature, such as height or weight, piles up in a bell shape that bulges in the middle, and the distribution that describes this shape is the normal distribution. The mean μ sets where the middle of the bell is and the standard deviation σ sets its width; for any normal distribution, about 68.3% falls within μ ± σ, about 95.4% within μ ± 2σ and about 99.7% within μ ± 3σ. Normal distributions with different means and standard deviations all gather into one standard normal distribution, with mean 0 and standard deviation 1, once you change X to Z = (X − μ)/σ, so a single table gives their probabilities. Move the mean slider μ and the bell slides sideways, move σ and it grows narrow and tall or wide and flat, and in the lower figure moving z lets you read the area from 0 to z straight off as a probability.

The Bell-Shaped Curve

🔔 What the normal distribution is
①It is the bell-shaped distribution followed by data built up from many small effects added together, such as height, weight or measurement error
②It is symmetric about the vertical line through the mean μ
③The total area under the curve is 1, and the area over an interval is the probability of landing in it
④Because the area is always 1, a small σ makes it narrow and tall and a large σ makes it wide and low

The N(μ, σ²) Curve

50
10
Probability density of the normal distribution
f(x) = 1σ√(2π) exp(-(x-μ)²/(2σ²))
The probability density of N(μ, σ²). It fixes the shape of the curve; probabilities are read from a table after standardizing
📏 The 68 · 95 · 99.7 rule
①Within μ ± σ lies about 68.3% of the whole
②Within μ ± 2σ lies about 95.4%
③Within μ ± 3σ lies about 99.7%
④At the point μ + σ, where the white arrow ends, the curve changes the way it bends; σ is the distance from the middle out to the shoulder of the bell

The Standard Normal Distribution Z

1.5
Standardization
Z = X - μσ
Changes X following N(μ, σ²) into Z following N(0, 1)
💡 What standardizing means
①Z tells how many standard deviations X lies from the mean
②Any normal distribution, once standardized, gathers into the single N(0, 1)
③So one standard normal table is enough to compute probabilities for every normal distribution
④P(a ≤ X ≤ b) = P((a − μ)/σ ≤ Z ≤ (b − μ)/σ)
⑤As the second line of the figure shows, symmetry makes P(−z ≤ Z ≤ z) twice P(0 ≤ Z ≤ z); for z = 1 it is about 0.6826, matching the 68.3% rule

Binomial and Normal Distributions

Normal approximation to the binomial
B(n, p) ≈ N(np, np(1−p)) (large n)
When n is large enough, the binomial distribution B(n, p) gets close to a normal distribution with mean np and variance np(1 − p)

Worked Examples

Example 1
X follows the normal distribution N(50, 10²). Standardize X = 70 and find its value of Z.
1
Write the standardization formula.
Z = (X − μ)/σ
2
Put in μ = 50 and σ = 10.
Z = (70 − 50)/10 = 2
▸ Z = 2
Z = 2 means that 70 is 2 standard deviations above the mean 50.
Example 2
X follows N(50, 10²). Find P(50 ≤ X ≤ 70), given that P(0 ≤ Z ≤ 2) = 0.4772.
1
X = 50 becomes Z = 0 and X = 70 becomes Z = 2.
P(50 ≤ X ≤ 70) = P(0 ≤ Z ≤ 2)
2
Read the value from the standard normal table.
= 0.4772
▸ 0.4772
Probabilities for a normal distribution are read as areas in the standard normal table after standardizing.

Summary

Key results
Z = X-μσ, X ∼ N(μ, σ2)
The normal distribution and standardization
KICE mock exam Math (Probability and Statistics) type
X follows the normal distribution N(60, 8²). What is P(X ≥ 60)?
①0.1587
②0.3413
③0.5
④0.6587
⑤0.8413
▸ ③ 0.5
1
A normal distribution is symmetric about its mean μ.
P(X ≥ μ) = 0.5
2
Since μ = 60, P(X ≥ 60) is the right half of the area, 0.5.
P(X ≥ 60) = 0.5
🎯 Exam Points
①Standardize with Z = (X − μ)/σ
②Read P(0 ≤ Z ≤ z) from the standard normal table
③Remember the 68 · 95 · 99.7 rule
④For a > 0, use symmetry to rewrite P(Z ≥ a) = 0.5 − P(0 ≤ Z ≤ a)
⑤When n is large, approximate the binomial distribution by a normal one
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