STANDARD NORMAL CALCULATIONSSection 2.2

NORMAL DISTRIBUTIONS

Can be compared if we measure in units of size σ about the mean µ as center.

Changing these units is called standardizing.

STANDARDIZING AND Z-SCORES

If x is an observation from a distribution that has mean µ and standard deviation σ, the standardized value of x is

A standardized value is often called a z-score.

HEIGHTS OF YOUNG WOMEN The heights of young women are

approximately normal with µ = 64.5 inches and σ = 2.5 inches. The standardized height is

𝑧=h h𝑒𝑖𝑔 𝑡−64.5

2.5

HEIGHTS OF YOUNG WOMEN A woman’s standardized height is the

number of standard deviations by which her height differs from the mean height of all women. For example, a woman who is 68 inches tall has a standardized height

𝑧=68−64.5

2.5=1.4

or 1.4 standard deviations above the mean.

HEIGHTS OF YOUNG WOMEN A woman who is 5 feet (60 inches) tall has a

standardized height

𝑧=60−64.5

2.5=−1.8

or 1.8 standard deviations less than the mean.

STANDARD NORMAL DISTRIBUTION

The normal distribution N(0,1) with mean 0 and standard deviation 1.

If a variable x has any normal distribution N(µ,σ) with mean µ and standard deviation σ, then the standardized variable

has the standard normal distribution.

THE STANDARD NORMAL TABLE

Table A (front of your book) is a table of areas under the standard normal curve. The table entry for each value z is the area under the curve to the left of z.

USING THE Z TABLE

Back to our example of women 68 inches or less. We had a z-score of 1.4. To find the proportion of observations

from the standard normal distribution that are less 1.4, locate 1.4 in Table A.

USING THE Z TABLE

Back to our example of women 68 inches or less. We had a z-score of 1.4. To find the proportion of observations from the

standard normal distribution that are less 1.4, locate 1.4 in Table A.

What does this mean?

About 91.92% of young women are 68 inches or shorter.

Find the proportion of observations from the standard normal distribution that are greater than -2.15.

Find the proportion of observations from the standard normal distribution that are greater than -2.15.

Proportion = 0.0158 Remember, Table A gives us what is less than

a z-score.

1 – 0.0158 = .9842

STEPS FOR FINDING NORMAL DISTRIBUTION

Step 1: State the problem in terms of the observed variable x.

Step 2: Standardize x to restate the problem in terms of a standard normal curve. Draw a picture of the distribution and shade the area of interest under the curve.

Step 3: Find the required area under the standard normal curve, using Table A and the fact that the total area under the curve is 1.

Step 4: Write your conclusion in the context of the problem.

CHOLESTEROL PROBLEM The level of cholesterol in the blood is important because high cholesterol

levels may increase the risk of heart disease. The distribution of blood cholesterol levels in a large population of people of the same age and sex is roughly normal. For 14-year old boys, the mean is µ = 170 milligrams of cholesterol per deciliter of blood (mg/dl) and the standard deviation is σ = 30 mg/dl. Levels above 240 mg/dl may require medical attention. What percent of 14-year-old boys have more than 240 mg/dl of cholesterol?

Step 1: State the Problem.

Level of cholesterol = x x has the N(170,30) distributionWant the proportion of boys with cholesterol level x > 240

CHOLESTEROL PROBLEM The level of cholesterol in the blood is important because high cholesterol

levels may increase the risk of heart disease. The distribution of blood cholesterol levels in a large population of people of the same age and sex is roughly normal. For 14-year old boys, the mean is µ = 170 milligrams of cholesterol per deciliter of blood (mg/dl) and the standard deviation is σ = 30 mg/dl. Levels above 240 mg/dl may require medical attention. What percent of 14-year-old boys have more than 240 mg/dl of cholesterol?

Step 2: Standardize x and draw a picture𝑥>240

𝑥−17030

>240 −170

30

𝑧>2.33

CHOLESTEROL PROBLEM Step 3: Use the Table (z > 2.33)

0.9901 is the proportion of observations less than 2.33.

1 – 0.9901 = 0.0099

About 0.01 or 1%

CHOLESTEROL PROBLEM Step 4: Write your conclusion in the context of the

problem.

Only about 1% of boys have high cholesterol.

WORKING WITH AN INTERVAL

What percent of 14-year-old boys have blood cholesterol between 170 and 240 mg/dl?Step 1: State the problem

We want the proportion of boys with Step 2: Standardize and draw a picture

170 240x

170 ≤ 𝑥≤ 240170 −170

30≤𝑥−170

30≤

240 − 17030

0 ≤ 𝑧≤ 2.33

WORKING WITH AN INTERVAL

What percent of 14-year-old boys have blood cholesterol between 170 and 240 mg/dl?Step 3: Use the table

z < 0 0.5000 z < 2.33 0.9901

0 < z < 2.33 0.9901 – 0.5000 = 0.4901

0 ≤ 𝑧≤ 2.33

WORKING WITH AN INTERVAL

What percent of 14-year-old boys have blood cholesterol between 170 and 240 mg/dl?Step 4: State your conclusion in context

About 49% of boys have cholesterol levels between 170 and 240 mg/dl.

FINDING A VALUE GIVEN A PROPORTION

Scores on the SAT Verbal test in recent years follow approximately the N(505,110) distribution. How high must a student score in order to place in the top 10% of all students taking the SAT?

Find the SAT score x with 0.1 to its right under the normal curve. (Same as finding an SAT score x with 0.9 to its left)

µ = 505, σ = 110

FINDING A VALUE GIVEN A PROPORTION

Scores on the SAT Verbal test in recent years follow approximately the N(505,110) distribution. How high must a student score in order to place in the top 10% of all students taking the SAT?

USE THE TABLE!!! Go backwards.

FINDING A VALUE GIVEN A PROPORTION

Scores on the SAT Verbal test in recent years follow approximately the N(505,110) distribution. How high must a student score in order to place in the top 10% of all students taking the SAT?

FINDING A VALUE GIVEN A PROPORTION

Scores on the SAT Verbal test in recent years follow approximately the N(505,110) distribution. How high must a student score in order to place in the top 10% of all students taking the SAT?z-score = 1.28Unstandardize:

𝑥=505+ (1.28 ) (110 )=645.8

Homework

• 2.19 p. 95• 2.21, 2.22, 2.23 p. 103