Normal Distribution

α = tail area	central area = 1 – 2α	zα
0.10	0.80	z.10 = 1.28
0.05	0.90	z.05 = 1.645
0.025	0.95	z.025 = 1.96
0.01	0.98	z.01 = 2.33
0.005	0.99	z.005 = 2.58

Normal Distribution Worked Examples

Percentile Rank is the percentage of scores at or below a particular value.

Example 1:

SAT Scores are normally distributed with a  Mean of 500 and a Standard Deviation of 100. What is the Percentile Rank of a person with score 650?

The problem can be written as what is P(X<=650) when X has the N(500,100) distribution?

We need the find the percentage of scores at or below 650.

With R, this is simple:

> pnorm(650, 500, 100)

The output: [1] 0.9331928

So, the answer is: 93.32%

Example 2:

SAT Scores are normally distributed with a  Mean of 500 and a Standard Deviation of 100. What is the Percentile Rank of a person with score 433?

The problem can be written as what is P(X<=433) when X has the N(500,100) distribution?

Type at R prompt:

> pnorm(433,500,100)

[1] 0.2514289

So, the answer is: 25.14%

Inverse

qnorm R function calculates the inverse c. d. f. F-1 of the normal distribution. The c. d. f. and the inverse c. d. f. are related by

p = F(x)
x = F-1(p)


So given a number p between zero and one, qnorm looks up the p-th quantile of the normal distribution. As with pnorm, optional arguments specify the mean and standard deviation of the distribution.

Example 3:

SAT Scores are normally distributed with a  Mean of 500 and a Standard Deviation of 100. What is the 93.32 nd Percentile score? (relate this to Example 1)

>qnorm(0.9332, 500,100)

[1] 650.0056

So, the answer is 650

Example 4:

A soda manufacturer sets its bottling machine to put soda in each bottle with a mean of 16.5 oz and a standard deviation of 0.5 oz. Six bottles are chosen at random and packaged in a carton. What percentage of 6-pack bottle has mean of 17 oz or more soda?

Answer: This question is related to sampling distribution. Mean (μ) of the sampling distribution is 16.5 oz, and standard error = standard deviation of the population divided by square root of the sample size.

Hence, Standard Error (SE) = 0.5/ (square root of Sample Size)
                                          = 0.5/(square root of 6)
                                          = 0.2
P(X>=17) when X has the distribution N(16.5, 0.2)

In R:

> 1-pnorm(17,16.5,.204)
[1] 0.007123386

So, the percentage is  0.7%.