# Normal Distribution

How to find the probability of a normal distribution: formula, 2 examples, and their solutions.

## Normal Distribution

If the shape of a histogram looks like this,

then the data shows a normal distribution.

x: mean

σ: Standard deviation

The left side of the mean

and the right side of the mean

are the same.

For a normal distribution,

if the mean is x

and if the standard deviation is σ,

then the normal distribution can be written as

N(x, σ^{2}).

## Standard Normal Distribution

The standard normal distribution is

N(0, 1^{2}).

The mean is 0.

The standard deviation is 1.

Its shape is the same as

the other normal distribution curves.

There's a reason

why the standard normal distribution is important.

It's because

the area below the curve is known.

The total area below the curve is 1.

The left half is 0.5.

The right half is also 0.5.

The other areas can be found

by using the z-score table.

The area of 0 ≤ Z ≤ 1,

(blue),

is 0.3413.

The area of 0 ≤ Z ≤ 2,

(blue) + (green),

is 0.4771.

The area of 0 ≤ Z ≤ 3,

(blue) + (green) + (brown),

is 0.4987.

## Formula

Each normal distribution has

different mean and standard deviation.

But their shapes are all the same.

So, to solve a normal distribution problem,

change the normal distribution N(x, σ^{2})

to the standard normal distribution N(0, 1^{2})

by using this formula.

Z = [x - x]/σ

Z: Z-score of x

x: Given value

x: Mean

σ: Standard deviation

## Example

It says

the data is normally distributed

with a mean of 70

and a standard deviation of 7.

So this data is N(70, 7^{2}).

The probability (proportion) of 63 ~ 84 points

is P(63 ≤ X ≤ 84).

Find the z-score of 63.

N(70, 7^{2})

x = 70

σ = 7

Then Z = [63 - 70]/7.

63 - 70 = -7

-7/7 = -1

So the z-score of X = 63 is

Z = -1.

Find the z-score of 84.

N(70, 7^{2})

x = 70

σ = 7

Then Z = [84 - 70]/7.

84 - 70 = 14

14/7 = 2

So the z-score of X = 84 is

Z = 2.

X = 63 is Z = -1.

X = 84 is Z = 2.

So

P(63 ≤ X ≤ 84) = P(-1 ≤ Z ≤ 2).

By changing X to Z,

N(70, 7^{2}) changed to N(0, 1^{2}).

This process is called [standardizing].

Draw the normal distribution curve

like this.

Color the region

under the curve -1 ≤ Z ≤ 2.

The blue colored area is

P(-1 ≤ Z ≤ 0).

The left side and the right side are the same.

So P(-1 ≤ Z ≤ 0) = P(0 ≤ Z ≤ 1).

See the given z-score table.

P(0 ≤ Z ≤ 1) = 0.3413

So the blue colored area is

0.3413.

The green colored area is

P(0 ≤ Z ≤ 2).

See the given z-score table.

P(0 ≤ Z ≤ 2) = 0.4771

So the green colored area is

0.4771.

P(-1 ≤ Z ≤ 2) is the colored area

under the curve.

So

P(-1 ≤ Z ≤ 2)

= 0.3413 + 0.4771

= 0.8184.

See the given example.

There are 1,000 students.

So n = 1000.

P = 0.8184

So the expected value is

E(X) = 1000⋅0.8184.

1000⋅0.8184 = 818.4

E(X) is the numebr of students.

So round 818.4.

Then 818.4 → 818.

So 818 is the answer.

## Example

It says

the data is normally distributed

with a mean of 70

and a standard deviation of 7.

So this data is N(70, 7^{2}).

The probability (proportion) of

at or below 77 points

is P(X ≤ 77).

Find the z-score of 77.

N(70, 7^{2})

x = 70

σ = 7

Then Z = [77 - 70]/7.

77 - 70 = 7

7/7 = 1

X = 77 is Z = 1.

So

P(X ≤ 77) = P(Z ≤ 77).

By changing X to Z,

N(70, 7^{2}) changed to N(0, 1^{2}).

Draw the normal distribution curve

like this.

Color the region

under the curve Z ≤ 1.

Find the colored area.

The left half is 0.5.

The blue colored area is

P(0 ≤ Z ≤ 1).

See the given z-score table.

P(0 ≤ Z ≤ 1) = 0.3413

So the blue colored area is

0.3413.

P(Z ≤ 1) is the colored area

under the curve.

So

P(Z ≤ 1)

= 0.5 + 0.3413

= 0.8413.

See the given example.

There are 1,000 students.

So n = 1000.

P = 0.8413

So the expected value is

E(X) = 1000⋅0.8413.

1000⋅0.8413 = 841.3

E(X) is the numebr of students.

So round 841.3.

Then 841.3 → 841.

So 841 is the answer.