Standardizing and the Standard Normal Curve

Standardizing and the Standard Normal Curve AP Statistics

The Standardized Value (the Z score) • If x is an observation from a distribution that has mean and standard deviation , the standardized value of x is

z 

x



Example 2.4 p. 94 The heights of young women are approximately normal with mean



=64.5 inches

• A woman 68 inches

tall for example has a standardized height of

68  64.5 z  1.4 2.5

• this means that the

woman is 1.4 standard deviations from the mean.

 =2.5 inches.

The heights of young women are approximately normal with mean  =64.5 inches  =2.5 inches. • Now, another woman is 5 feet (60 inches tall). The standardized height of this girl is 60  64.5 z  1.8 2.5

• this means that the

woman is 1.8 standard deviations less than the mean height.

The Standard Normal Distribution • The Standard Normal Distribution is the normal distribution N(0,1) with mean 0 and standard deviation 1. • If a variable x has any normal distribution N( ,  ) with mean  and standard deviation then the standardized variable

,

z 

x  



• has the standard normal distribution.

• Table A is a table for

the areas under the standard normal curve. The table entry for each value of z is the area under the curve to the left of z

Finding Normal Proportions • Step 1: State the problem in terms of the observed

value x. Draw a picture of the distribution and shade the area of interest under the curve. Step 2: Standardize x to restate the problem in terms of a standard normal variable z. Draw a picture to show the area of interest under the standard normal 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: Writ your conclusion in the context of the problem

The Dummy Rules: Dummy Rule 1 • If we want the probability that x is less than a number, then it is written #  P ( x  #)  P ( z  )



• Look up your z score on table A at this

point • Example say that the distribution of the weights of dogs is N(25, 10) and we want the probability that a dog weighs less than 15 pounds. then… 15  25 P ( x  15)  P( z  )  P( z  1) 10

• Now look up -1 on table A and

you will see that the probability is • 0.1587

The Dummy Rules: Dummy Rule 2 • If we want the probability that x is greater than a number, then it is written

P( x  #)  1  P( x  #)  1  P( z 

#  ) 

• Look up your z score on table A at this

•

point Example say that the distribution of the weights of dogs is N(25, 10) and we want the probability that a dog weighs more than 30. then…

P( x  30)  1  P( x  30)  1  P( z 

30  25 )  1  P( z  0.5) 10

• Now look up 0.5 on table A and you •

will see that the probability is 0.6915

The Dummy Rules: Dummy rule 3 • If we want the probability that x is in between two numbers, then it is written (Let our numbers be represented by A and B)

P( A  x  B)  P( x  B)  P( x  A)  P( z 

B A  )  P( z  )  

• Look up both your z score on table A •

and subtract Example say that the distribution of the weights of dogs is N(25, 10) and we want the probability that a dog weighs less than 30 but more than 2. then…

P(2  x  30)  P( x  30)  P( x  2)  P( z 

• Now look up 0.5 and -2.3 on table A • •

30  25 2  25 )  P( z  )  P( z  0.5)  P( z  2.3) 10 10

then subtract 0.6915 – 0.0107 = 0.6808 So the probability that a dog weighs less than 30 but more than 2 is 0.6808

Example 2.9 Working with an interval The distribution of blood cholesterol levels is roughly normal for 14-year old boys with mean =170 and =30





• What percent of 14-year old boys have

blood cholesterol between 170 and 240?

• Step 1: State the problem: We want the proportion of boys with..

170  x  240

Continued

Example 2.9 Working with an interval The distribution of blood cholesterol levels is roughly normal for 14-year old boys with mean =170 and =30





• Step 2: Standardize and draw a picture.

P (170  x  240) P ( x  240)  P ( x  170) 240  170 170  170 P( z  )  P( z  ) 30 30 P ( z  2.33)  P ( z  0)

Continued

Example 2.9 Working with an interval The distribution of blood cholesterol levels is roughly normal for 14-year old boys with mean =170 and =30



• Step 3: Use the

table: Look up 2.33 and 0 on the table:

P( z  2.33)  P( z  0) 0.9901  0.5000 0.4901



• Step 4: State your •

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