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Wize AP Statistics Textbook > The Normal Distribution

Standardizing

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Standard Normal Distribution (Z-Scores)

A continuous random variable ZZ follows a standard normal distribution if it's a special case of the normal distribution that is centered at 0 (i.e. μ=0\mu=0) and has a standard deviation of σ=1\sigma=1.
  • This is a VERY useful distribution, so we give it a special variable ZZ
  • We denote this by Z ∼ N(0, 1)Z\ \sim\ N\left(0,\ 1\right)
  • We use z-scores (zz) to denote the number of standard deviations a value is away from the mean
  • If z=0z=0, the value equals the mean
  • If z>0z>0, the value is to the right of the mean
  • If z<0z<0, the value is to the left of the mean
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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)=
0.9066

P(Z>1.32)=P\left(Z>1.32\right)=
1 - 0.9066 = 0.0934



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Examples
P(Z<−1.32)=P\left(Z<-1.32\right)=
0.0934

P(Z>−1.32)=P\left(Z>-1.32\right)=
1 - 0.0934 = 0.9066


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Wize Tip
Since the curve is symmetrical, some profs will only give you the z-table with positive z-scores.

In this case, draw a picture to help you visualize the area that corresponds to the desired probability.

Watch Out!
Since ZZ is a continuous random variable, its probability distribution function cannot be entirely represented by a table of discrete values. That's why the z-table only shows some of the values for the distribution. Sometimes you may have to approximate the probability.

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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 XX, we can find the z-score that corresponds to the standard normal random variable ZZ by standardizing it using this formula:
z=x−μσ\boxed{\displaystyle z=\frac{x-\mu}{\sigma}}
  • z\colorFour z is the z-score that follows the standard normal distribution ZZ
  • We can then use the z-table to solve probability problems!
  • x\colorFour x is the given value that follows the normal distribution XX
  • XX has a mean of μ\colorFour \mu
  • XX has a standard deviation of σ\colorFour \sigma
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Example
Suppose XX has a normal distribution with a mean of 500 and standard deviation of 250. What is the probability that XX is 830 or higher?

Using the standardization formula:
z=x−μσ=830−500250=1.32z=\dfrac{x-\mu}{\sigma}=\dfrac{830-500}{250}=1.32

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)P(Z>1.32)
=1−0.9066=1-0.9066
=0.0934=0.0934

Therefore, the probability that XX has a value of 830 or higher is 0.0934.
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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)=0.9P\left(Z<1.282\right)=0.9

z=X−μσz=\frac{X-\mu}{\sigma}

1.282=X−6581.282=\frac{X-65}{8}

X=75.256X=75.256
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(b) What percent of the contestants received a score of 57 or lower?



z=X−μσ=57−658=−1\displaystyle{z=\frac{X-\mu}{\sigma}=\frac{57-65}{8}=-1}

z-table: P(Z<−1)=0.1587P\left(Z<-1\right)=0.1587

Empirical Rule: P(Z<−1)=1−0.682=0.16\displaystyle{P\left(Z<-1\right)=\frac{1-0.68}{2}=0.16}



(c) What percent of the contestants received a score between 57 and 89 percent?

See graph above.

z=X−μσ=89−658=3\displaystyle{z=\frac{X-\mu}{\sigma}=\frac{89-65}{8}=3}

z-table: P(Z>3)=1−0.9987=0.0013P\left(Z>3\right)=1-0.9987=0.0013

P(57<X<89)=P(−1<Z<3)=(1−0.0013)−0.1587=0.84P\left(57<X<89\right)=P\left(-1<Z<3\right)=\left(1-0.0013\right)-0.1587=0.84

Empirical Rule: P(Z<−1)=1−0.9972=0.0015\displaystyle{P\left(Z<-1\right)=\frac{1-0.997}{2}=0.0015}

P(57<X<89)=P(−1<Z<3)=1−0.0015−0.16=0.8385P\left(57<X<89\right)=P\left(-1<Z<3\right)=1-0.0015-0.16=0.8385

The class grade on the statistics final exam is normally distributed with a mean of 65 percent and a standard deviation of 8 percent.



Approximately 95% of all grades would lie between what two grades?


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)