Z-Scores (Normal Distribution)

Z-Scores (Normal Distribution)

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

Normal Distribution

If the shape of a histogram looks like this, then that data shows a normal distribution. The left half and the right half, which are cut by the mean, show the reflection of each other.

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

Frequency histogram

The left side and the right side of the curve,
which are cut by the mean (x),
have the same shape.

Z-Scores: Formula

Z = (x - x bar)/sigma, x: Value of the data, x bar: Mean of the data, sigma: Standard deviation

Each normal distribution has
different mean and standard deviation.
But their shapes all show the same curve.

So, to analyze the normally distributed data,
you can use the z-score.

The z-score is
the coefficient of the standard deviation.

So the z-score of x can be found
by using this formula:
Z = (x - x)/σ

By using the z-score,
you can find out where the value is located.

For the z-score curve,
the mean is 0,
and the standard deviation is 1.

Z-Scores: Meaning

The z-score shows where the value is located in the normal distribution curve. So, by finding the z-score, you can find the percentage area between 0 and Z.

The z-score shows
where the value is located
in the z-score curve.

So, by finding the z-score,
you can find the percentage areas.

The area of [Z: 0 ~ 1]
is about 34%. (= 0.3413...)

So the area of [Z: -1 ~ 1]
is about 34⋅2 = 68%. (= 0.6826...)

The area of [Z: 1 ~ 2]
is about 13.5%. (= 0.1359...)

So the area of [Z: -2 ~ 2]
is about (34 + 13.5)⋅2 = 95%. (= 0.9544...)

The area of [Z: 2 ~ 3]
is about 2%. (= 0.0215...)

So the area of [Z: -3 ~ 3]
is about (34 + 13.5 + 2)⋅2 = 99%. (= 0.9974...)

To see more percentage areas,
use the z-score table.

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.

The mean is 70.
And the standard deviation is 7.

Find the z-score of 56
by using the formula.

Z = (56 - 70)/7

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?

Find the percentage of 63 ~ 84 points:
P(63 ≦ X ≦ 84).

The test scores are normally distributed.

So find the z-scores of 63 and 84:
Z = -1, 2.

So P(63 ≦ X ≦ 84)
= P(-1 ≦ Z ≦ 2).

Draw the z-score curve.

Mark the percentage areas between -1 and 2:
34%, 34%, and 13.5%.

So P(-1 ≦ Z ≦ 2) = 0.68 + 0.135
= 0.815.

Find the expected value.

E(X) = xP
= 1000⋅0.815
= 815

So about 815 students score between 63 and 84.

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?

Find the percentage of 0 ~ 77 points:
P(X ≦ 77).

The test scores are normally distributed.

So find the z-score of 77:
Z = 1.

So P(X ≦ 77)
= P(Z ≦ 1).

Draw the z-score curve.

Mark the percentage areas:
50% (left half) and 34%.

So P(Z ≦ 1) = 0.5 + 0.34
= 0.84.

Find the expected value.

E(X) = xP
= 1000⋅0.84
= 840

So about 840 students score at or below 77.