Normal Distribution

Normal Distribution

How to solve the normal distribution problems: definition, z-scores, examples, and their solutions.

Normal Distribution

If the shape of a histogram shows this curve, then that data shows a normal distribution.

If the shape of a histogram shows this curve,
then that data shows a [normal distribution].

Frequency histogram

A normal distribution is found in real life
when n is very large.

Examples are
height, weight, IQ scores of people,
number of getting a head from tossing a coin,
number of getting a '1' from tossing a fair die.

Z-Score

For x = (mean) + Z*(standard deviation), the coefficient of the standard deviation is the z-score.

Recall that
the standard deviation σ means
how far the value is from the mean x.

So you can write a value x like this:
x + Z⋅σ = x.

The sign of the Z, the coefficient of σ,
determines where the value x is located.

The sign of the Z shows the direction:
left or right of the mean x.

The quantity of the Z shows
the amount of distance from the mean x.

x + Z⋅σ = x

Then Z = (x - x)/σ.

This Z, the coefficient of σ,
is the [z-score].

Standard Normal Distribution

Each normal distribution has different mean and standard deviation. But the shapes of their curves are all the same. So, by using the z-scores, you can change normal distribution curves to a standard normal distribution curve.

Each normal distribution has
different mean and standard deviation.

But the shapes of their curves are all the same
because they are all normally distributed.

So, by using the z-scores,
you can change normal distribution curves
to a [standard normal distribution] curve.

The mean x is 0.
And the standard deviation σ is 1.

Probabilities of the Normal Distribution

The percentage areas under the normal distribution curve are drawn.

The percentage areas
under the standard normal distribution curve
are found like this.

See the right half areas.

The percentage area of [0 ≤ Z ≤ 1]
is roughly 34%.

The percentage area of [1 ≤ Z ≤ 2]
is roughly 13.5%.

The percentage area of [2 ≤ Z ≤ 3]
is roughly 2%.

The curve is symmetric about the mean.
So the percentage areas of the left half
are also 34%, 13.5%, and 2%.

So the percentage area of [-1 ≤ Z ≤ 1]
is roughly 68%.
(34% + 34% = 68%)

The percentage area of [-2 ≤ Z ≤ 2]
is roughly 95%.
(13.5% + 34% + 34% + 13.5% = 95%)

And the percentage area of [-3 ≤ Z ≤ 3]
is roughly 99%.
(2% + 13.5% + 34% + 34% + 13.5% + 2% = 99%)

By using these percentage numbers,
you can find the probability
from a normal distribution data.

To find the precise percentage areas,
use the calculator or use the z-score table.

z-table.com

Example 1

The test scores of 1,000 students are normally distributed with a mean of 70 and a standard deviation of 7. 1. Find the z-score of a score of 56.

It says
the test scores are normally distributed
with a mean of 70 and a standard deviation of 7.

So write
x = 70 and σ = 7.

And it says
to find the z-score of 56.

So Z = (56 - 70)/7.

56 - 70 = -14

-14/7 = -2

So Z = -2.

Example 2

The test scores of 1,000 students are normally distributed with a mean of 70 and a standard deviation of 7. 2. About how many students score between 63 and 84?

First write the mean and the standard deviation:
x = 70
σ = 7.

Find the percentage of the students
that score between 63 and 84.

Think the percentage of the students
as the probability of choosing the wanted student.

Then it means
to find P(63 ≤ X ≤ 84)
from the given normally distributed students.

Find the z-score of 63.

x = 70
σ = 7

So Z = (63 - 70)/7.

63 - 70 = -7

-7/7 = -1

So, for 63 points, Z = -1.

Next, find the z-score of 84.

x = 70
σ = 7

So Z = (84 - 70)/7.

84 - 70 = 14

14/7 = 2

So, for 84 points, Z = 2.

For 63 points, Z = -1.
For 84 points, Z = 2.

So P(63 ≤ X ≤ 84)
= P(-1 ≤ Z ≤ 2).

Draw the standard normal distribution curve.

The percentage area of [-1 ≤ Z ≤ 0]
is 34%.

The percentage area of [0 ≤ Z ≤ 1]
is also 34%.

And the percentage area between [1 ≤ Z ≤ 2]
is 13.5%.

P(-1 ≤ Z ≤ 2) is the colored area.

0.34 + 0.34 = 0.68
0.68 + 0.135 = 0.815

So P(-1 ≤ Z ≤ 2) = 0.815.

n = 1000 (1000 students)
P = 0.815

So E(X) = 1000⋅0.815.

Expected value

1000⋅0.815 = 815

So E(X) = 815.

Example 3

The test scores of 1,000 students are normally distributed with a mean of 70 and a standard deviation of 7. 3. About how many students score at or below 77?

First write the mean and the standard deviation:
x = 70
σ = 7.

Find the percentage of the students
that score at or below 77.

Think the percentage of the students
as the probability of choosing the wanted student.

Then it means
to find P(X ≤ 77)
from the given normally distributed students.

Find the z-score of 77.

x = 70
σ = 7

So Z = (77 - 70)/7.

77 - 70 = 7

7/7 = 1

So, for 77 points, Z = 1.

Z = 1 for 77 points.

So P(X ≤ 77)
= P(Z ≤ 1).

Draw the standard normal distribution curve.

The left half is 50%.

And the percentage area of [0 ≤ Z ≤ 1]
is 34%.

P(Z ≤ 1) is the colored area.

0.50 + 0.34 = 0.84

So P(Z ≤ 1) = 0.84.

n = 1000 (1000 students)
P = 0.84

So E(X) = 1000⋅0.84.

1000⋅0.84 = 840

So E(X) = 840.