+

The Normal Distribution

Chapter 6

+Applications of the Normal Distribution Section 6-2

+The Standard Normal Distribution and Practical Applications

!  We can convert any variable that in normally distributed into a standard normal distribution variable (z-value ) by using the following formula.

!  Consider a standardized test that has a mean of 100 and a standard deviation of 15. When the scores are transformed to z-values, the two distributions coincide.

z =X − µ

σ

+Steps For Finding the Area Under Any Normal Curve

Step 1: Draw a normal curve and

shade the desired area.

Step 2: Convert the values of X

(data values) to z-values using the

formula.

Step 3: Find the corresponding

area, using Table E on page 788.

z =X − µ

σ

+Example of an Application

+Example of an Application

+Example of an Application

+Finding Data Values Given Specific Probabilities

!  Formula for finding a z-value is

!  Formula for find a data value (X) is

z =X − µ

σ

X = z ⋅ σ + µ

+Finding Data Values for Specific Probabilities

Step 3: Calculate the X value by using the formula X = zσ + μ

Step 2: Find the z-value from Table E that corresponds to the desired probability (area).

Step 1: Draw a normal curve and shade the desired area that represents the probability, proportion or percentile.

+Example of an Application

+Example of an Application

+Determining Normality

!  We may want to check if a distribution that we are working with is a normal distribution so that we can use the methods learned in this chapter.

!  One way to check for normality is to draw a histogram for the data and check its shape. !  If the distribution is not approximately bell-shaped, the data are not

normally distributed.

!  If the data is skewed, then it is NOT normally distributed. There is also a test to check if the data is skewed…

!  Look for outliers. Recall that outliers are outside the range of Q1 – 1.5(IQR) and Q3 + 1.5(IQR) where the IQR = Q3 – Q1. The existence of outliers can have a big effect on normality.

+The Pearson Coefficient

!  The Pearson coefficient (PC) can be calculated as follows:

A Test for Skewness

PC =3 X − median( )

s

PC ≥ 1 Data significantly skewed right

PC ≤ -1 Data significantly skewed left

PC = 0 Data not skewed

+Example of Determining Normality

+Example of Determining Normality

+The Central Limit Theorem Section 6-3

+A Distribution of Sample Means

!  Recall that a population is the totality of all subjects being studied. !  As we have learned about the normal distribution, we have been

talking about populations.

!  Recall that a sample is a group of subjects selected from a population. !  Now we are going to discuss taking a sample from that

population.

!  In fact, we want to consider what will happen if we take numerous samples of the same size (with replacement) from that population.

+A Distribution of Sample Means

!  If we take numerous samples of the same size (with replacement) from a single population, we can compare the different means of the different samples. They will not all be the same.

!  A sampling distribution of sample means is a distribution using the means computed from all possible random samples of a specific size taken from a population. !  The various sample means will be different from the population mean μ.

!  Sampling error is the difference between the sample measure and the corresponding population measure due to the fact that the sample is not a perfect representation of the population.

+Properties of the Distribution of Sample Means

!  When all possible samples of a specific size are selected with replacement from a population, the distribution of the sample means for a variable has the following properties:

1.  The mean of the sample means will be the same as the population mean.

2.  The standard deviation of the sample means will be smaller than the standard deviation of the population, and it will be equal to the population standard deviation divided by the square root of the sample size.

µX

= µ

σX

=σ

n

+An Example of a Distribution of Sample Means

!  Consider a teacher gives an 8-point quiz to a class of four students. The scores on the four quizzes were 2, 4, 6 and 8.

!  This is a population and the mean is 5: µ =2 + 4 + 6 + 8

4=20

4= 5

We call this a uniform distribution.

+An Example of a Distribution of Sample Means

+An Example of a Distribution of Sample Means

!  Below is a graph of the distribution of the sample means. Notice that this is a normal distribution.

!  Also notice that the mean of the sample means equals the population mean.

µX

= µ = 5

+An Example of a Distribution of Sample Means

!  The standard deviation of the sample means (also called the standard error of the mean) does NOT equal the standard deviation of the population. It is smaller.

σX

=σ

n

Because 2 is the sample size.

+The Central Limit Theorem

1.  When the original variable is normally distributed, the distribution of the sample means will be normally distributed, for any sample size n.

2.  When the distribution of the original variable is not normal, a sample size of 30 or more is needed to use a normal distribution to approximate the distribution of the sample means. The larger the sample, the better the approximation will be

!  In either case, the mean of the sample means equals the population mean.

!  The standard deviation of the sample means is not equal to the standard deviation of the population.

Relating to a Distribution of Sample Means

µX

= µ

σX

=σ

n

+The Central Limit Theorem Relating to a Distribution of Sample Means

A distribution of the data values.

A distribution of the sample means.

+The Central Limit Theorem

!  When we use the Central Limit Theorem to answer questions about sample means in the same manner that a normal distribution can be used to answer questions about individual values.

!  The only difference is that a new formula must be used for z-values.

Relating to a Distribution of Sample Means

z =X − µ

σ n

is the sample mean, and n is the sample size. X

+An Example

!  The training heart rates of all 20-year-old athletes are normally distributed, with a mean of 135 beats per minute and a standard deviation of 18 beats per minute. Random samples of size 4 are drawn from the population, and the mean of each sample is determined. Find the mean and standard deviation of the sampling distribution of sample means. Sketch a graph of the sampling distribution.

Interpreting the Central Limit Theorem

+An Example: The Central Limit Theorem

µX

= µ

σX

=σ

n

z =X − µ

σ n

+An Example: The Central Limit Theorem

z =X − µ

σ n

σX

=σ

n

µX

= µ

+An Example: The Central Limit Theorem