
Standard Normal Distribution (Z-Scores)
A continuous random variable Z follows a standard normal distribution if it's a special case of the normal distribution that is centered at 0 (i.e. μ=0) and has a standard deviation of σ=1.

- This is a VERY useful distribution, so we give it a special variable Z
- We denote this by Z ∼ N(0, 1)
- We use z-scores (z) to denote the number of standard deviations a value is away from the mean
- If z=0, the value equals the mean
- If z>0, the value is to the right of the mean
- If z<0, the value is to the left of the mean
Z-Tables
Standard normal distributions are very useful because we have a z-table (a.k.a. standard normal table) that gives us the area (i.e. probability) from the far left of the curve up to the z-value.
Examples
P(Z<1.32)= ________
P(Z>1.32)=________

Examples
P(Z<−1.32)= ________
P(Z>−1.32)= ________


Standardizing a Normal Random Variable
Standard normal random variables are very useful since they allow us to use the z-table to solve probability problems.
What if you were asked to find the probability of a value associated with just a Normal random variable?
Standardization
Given a value associated with a normal random variable X, we can find the z-score that corresponds to the standard normal random variable Z by standardizing it using this formula:
- z is the z-score that follows the standard normal distribution Z
- We can then use the z-table to solve probability problems!
- x is the given value that follows the normal distribution X
- X has a mean of μ
- X has a standard deviation of σ
Example
Suppose X has a normal distribution with a mean of 500 and standard deviation of 250. What is the probability that X is 830 or higher?

Using the standardization formula:
This z-score of 1.32 tells us that value of 830 is 1.32 standard deviations above the mean, which is 500.
Using the z-table, the area (probability) to the left of 1.32 is 0.9066.
So,
P(Z>1.32)
=1−0.9066
=0.0934
Therefore, the probability that X has a value of 830 or higher is 0.0934.

Practice: Standardizing
Contestants at a talent show were given scores between 0 and 100. The mean score is 65 with a standard deviation of 8. Assume the scores are normally distributed.
(a) Only the top 10% will compete in the finals. What is the minimum score to qualify?

P(Z<1.282)=0.9
z=σX−μ
1.282=8X−65
X=75.256
(b) What percent of the contestants received a score of 57 or lower?

z=σX−μ=857−65=−1
z-table: P(Z<−1)=0.1587
Empirical Rule: P(Z<−1)=21−0.68=0.16
(c) What percent of the contestants received a score between 57 and 89 percent?
See graph above.
z=σX−μ=889−65=3
z-table: P(Z>3)=1−0.9987=0.0013
P(57<X<89)=P(−1<Z<3)=(1−0.0013)−0.1587=0.84
Empirical Rule: P(Z<−1)=21−0.997=0.0015
P(57<X<89)=P(−1<Z<3)=1−0.0015−0.16=0.8385

Skee Ball: In order to win a $1000 prize, Billy has to roll five balls and either score below 150 or over 250. Suppose the average player scores 210 with a standard deviation of 20. Assume Bill is an average player. What is the probability that he will not win a prize?
Enter answer with at least 2 decimal places (e.g. 0.88)