# 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, σ

x: Mean ^{2})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, 1

Standard normal distribution is used^{2})to 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, σ

^{2}) to N(0, 1

^{2}).

### How to Use

P(0 ≤ Z ≤ z) = (area, known)

^{2}) to N(0, 1

^{2}),

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, σ

↓

N(0, 1)

1. Standardize the given normal distribution. ^{2})↓

N(0, 1)

N(x, σ

^{2}) → N(0, 1

^{2})

2. Use the z-score table to find the probability.

### Example

The test scores of 1,000 stuents 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, 7

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]

^{2})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, 7

P(63 ≤ X ≤ 84)

↓

N(0, 1

P(-1 ≤ Z ≤ 2)

^{2})P(63 ≤ X ≤ 84)

↓

N(0, 1

^{2})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 stuents 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, 7

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

^{2})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, 7

P(X ≤ 77)

↓

N(0, 1

P(Z ≤ 1)

^{2})P(X ≤ 77)

↓

N(0, 1

^{2})P(Z ≤ 1)

Close

## Binomial Distribution to Normal Approximation

### How to Solve

B(n, p)

↓

N(np, npq)

↓

N(0, 1

1. Find x = np, σ = √npq. ↓

N(np, npq)

↓

N(0, 1

^{2})If n is big enough,

you can approximate B(n, p) = N(np, npq).

Binomial Distribution

2. Standardize.

N(np, npq) → N(0, 1

^{2})

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, 10

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, 10

^{2}) - [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, 10

x = 200

σ = 10

↓

N(200, 10

^{2})x = 200

σ = 10

[4]

N(200, 10

P(185 ≤ X ≤ 210)

↓

N(0, 1

P(-1.5 ≤ Z ≤ 1)

^{2})P(185 ≤ X ≤ 210)

↓

N(0, 1

^{2})P(-1.5 ≤ Z ≤ 1)

Close