Normal Distribution
See how to find the probability of a normal distribution
(z-score).
6 examples and their solutions.
Normal Distribution: Definition
Definition
N(x, σ2)
x: Mean σ: Standard deviation
If the shape of a histogram looks like this,
then the values show normal distribution.
It can be found when n is big enough.
(test score, fruit size, sleeping time, ...)
Standard Normal Distribution
Formula
N(0, 12)
Standard normal distribution is usedto compare and analyze a normal distribution easily.
Each normal distribution has different mean and standard deviation.
But their shapes are all the same.
→ Standardized: Standard normal distribution
Z-Score
Formula
Z = X - xσ
The z-score is usedto change N(x, σ2) to N(0, 12).
How to Use
P(0 ≤ Z ≤ z): Area (known)
you can find the area (= probability) under the curve.
For each z-score,
the area under the curve is known.
(total area) = 1
(left half area) = (right half area) = 0.5
The curve is symmetric to the center axis (Z = 0).
Other values: Z-score table
Example
P(0 ≤ Z ≤ 1) = ?
Solution | z | P(0 ≤ Z ≤ z) |
|---|---|
| 1 | 0.3413 |
| 2 | 0.4771 |
| 3 | 0.4987 |
P(0 ≤ Z ≤ 1) = 0.3413
Close
Example
P(Z ≥ -2) = ?
Solution | z | P(0 ≤ Z ≤ z) |
|---|---|
| 1 | 0.3413 |
| 2 | 0.4771 |
| 3 | 0.4987 |
P(Z ≥ -2)
= P(-2 ≤ Z ≤ 0) + P(Z ≥ 0)
= P(0 ≤ Z ≤ 2) + P(Z ≥ 0) - [1]
= 0.4771 + 0.5
= 0.9771
[1]
The curve is symmetric to the center axis.
→ P(-2 ≤ Z ≤ 0) = P(0 ≤ Z ≤ 2)
→ P(-2 ≤ Z ≤ 0) = P(0 ≤ Z ≤ 2)
Close
Example
P(1 ≤ Z ≤ 3) = ?
Solution | z | P(0 ≤ Z ≤ z) |
|---|---|
| 1 | 0.3413 |
| 2 | 0.4771 |
| 3 | 0.4987 |
P(1 ≤ Z ≤ 3)
= P(0 ≤ Z ≤ 3) - P(0 ≤ Z ≤ 1)
= 0.4987 - 0.3413
= 0.1514
Close
Normal Distribution
How to Solve
N(x, σ2)
↓
N(0, 1)
1. Standardize the given normal distribution. ↓
N(0, 1)
N(x, σ2) → N(0, 12)
2. Use the z-score table to find the probability.
Example
The test scores of 1,000 students are normally distributed.
Mean: 70. Standard deviation: 7.
About how many students score between 63 and 84?
Solution Mean: 70. Standard deviation: 7.
About how many students score between 63 and 84?
| z | P(0 ≤ Z ≤ z) |
|---|---|
| 1 | 0.3413 |
| 2 | 0.4771 |
| 3 | 0.4987 |
N(70, 72)
P(63 ≤ X ≤ 84)
Z = 63 - 707
= -77
= -1
Z = 84 - 707
= 147
= 2
= P(-1 ≤ Z ≤ 2) - [1]
= P(-1 ≤ Z ≤ 0) + P(0 ≤ Z ≤ 2)
= P(0 ≤ Z ≤ 1) + P(0 ≤ Z ≤ 2) - [2]
= 0.3413 + 0.4771
= 0.8184
E(X) = 1000⋅0.8184 - [3]
= 818.4
→ 818 - [4]
P(63 ≤ X ≤ 84)
Z = 63 - 707
= -77
= -1
Z = 84 - 707
= 147
= 2
= P(-1 ≤ Z ≤ 2) - [1]
= P(-1 ≤ Z ≤ 0) + P(0 ≤ Z ≤ 2)
= P(0 ≤ Z ≤ 1) + P(0 ≤ Z ≤ 2) - [2]
= 0.3413 + 0.4771
= 0.8184
E(X) = 1000⋅0.8184 - [3]
= 818.4
→ 818 - [4]
[1]
N(70, 72)
P(63 ≤ X ≤ 84)
↓
N(0, 12)
P(-1 ≤ Z ≤ 2)
P(63 ≤ X ≤ 84)
↓
N(0, 12)
P(-1 ≤ Z ≤ 2)
[2]
The curve is symmetric to the center axis.
→ P(-1 ≤ Z ≤ 0) = P(0 ≤ Z ≤ 1)
→ P(-1 ≤ Z ≤ 0) = P(0 ≤ Z ≤ 1)
[4]
Round 818.4 to the nearest ones.
Close
Example
The test scores of 1,000 students are normally distributed.
Mean: 70. Standard deviation: 7.
About how many students score at or below 77?
Solution Mean: 70. Standard deviation: 7.
About how many students score at or below 77?
| z | P(0 ≤ Z ≤ z) |
|---|---|
| 1 | 0.3413 |
| 2 | 0.4771 |
| 3 | 0.4987 |
N(70, 72)
P(X ≤ 77)
Z = 77 - 707
= 77
= 1
= P(Z ≤ 1) - [1]
= P(Z ≤ 0) + P(0 ≤ Z ≤ 1)
= 0.5 + 0.3413
= 0.8413
E(X) = 1000⋅0.8413
= 841.3
→ 841
P(X ≤ 77)
Z = 77 - 707
= 77
= 1
= P(Z ≤ 1) - [1]
= P(Z ≤ 0) + P(0 ≤ Z ≤ 1)
= 0.5 + 0.3413
= 0.8413
E(X) = 1000⋅0.8413
= 841.3
→ 841
[1]
N(70, 72)
P(X ≤ 77)
↓
N(0, 12)
P(Z ≤ 1)
P(X ≤ 77)
↓
N(0, 12)
P(Z ≤ 1)
Close
Binomial Distribution to Normal Approximation
How to Solve
B(n, p)
↓
N(np, npq)
↓
N(0, 12)
1. Find x = np, σ = √npq. ↓
N(np, npq)
↓
N(0, 12)
If n is big enough,
you can approximate B(n, p) = N(np, npq).
Binomial Distribution
2. Standardize.
N(np, npq) → N(0, 12)
3. Use the z-score table to find the probability.
Example
A coin is tossed 400 times.
Find the probability of getting a head 185 ~ 210 times.
Solution Find the probability of getting a head 185 ~ 210 times.
| z | P(0 ≤ Z ≤ z) |
|---|---|
| 0.5 | 0.1915 |
| 1 | 0.3413 |
| 1.5 | 0.4332 |
| 2 | 0.4771 |
B(400, 12) - [1]
q = 1 - 12
= 22 - 12
= 12
N(400⋅12, 400⋅12⋅12) - [2]
= N(200, 100)
= N(200, 102) - [3]
P(185 ≤ X ≤ 210)
Z = 185 - 20010
= -1510
= -1.5
Z = 210 - 20010
= 1010
= 1
= P(-1.5 ≤ Z ≤ 1) - [4]
= P(-1.5 ≤ Z ≤ 0) + P(0 ≤ Z ≤ 1)
= P(0 ≤ Z ≤ 1.5) + P(0 ≤ Z ≤ 1)
= 0.4332 + 0.3413
= 0.7745
q = 1 - 12
= 22 - 12
= 12
N(400⋅12, 400⋅12⋅12) - [2]
= N(200, 100)
= N(200, 102) - [3]
P(185 ≤ X ≤ 210)
Z = 185 - 20010
= -1510
= -1.5
Z = 210 - 20010
= 1010
= 1
= P(-1.5 ≤ Z ≤ 1) - [4]
= P(-1.5 ≤ Z ≤ 0) + P(0 ≤ Z ≤ 1)
= P(0 ≤ Z ≤ 1.5) + P(0 ≤ Z ≤ 1)
= 0.4332 + 0.3413
= 0.7745
[1]
[3]
B(400, 1/2)
↓
N(200, 102)
x = 200
σ = 10
↓
N(200, 102)
x = 200
σ = 10
[4]
N(200, 102)
P(185 ≤ X ≤ 210)
↓
N(0, 12)
P(-1.5 ≤ Z ≤ 1)
P(185 ≤ X ≤ 210)
↓
N(0, 12)
P(-1.5 ≤ Z ≤ 1)
Close