Chapter 6

The Normal Probability Distribution

CONTINUOUS RANDOM VARIABLES Continuous random variables can assume the

infinitely many values corresponding to points on a

line interval. This means that the range of the variable

contains an interval of real numbers.

Examples:

Heights, weights

length of life of a particular product

experimental laboratory error

Recall that the probability distribution p(x) of a

discrete random variable is just the histogram of

its values. We had the following examples in

chapter 5:

1. Toss a fair coin three times and let x be the

number of heads. Then x is binomial and

x p(x)

0 1/8

1 3/8

2 3/8

3 1/8

We can create what we call the probability

polygon h(x) by starting from the midpoint of the

left side of the first bar and connecting the

midpoints of the upper sides of the subsequent

bars by straight lines and finishing at the

midpoint of the right side of the last bar.

2. A marksman hits a target 80% of the time. He

fires five shots at the target. What is the probability

that exactly 3 shots hit the target? If x is the number

of shots that hit the target, then x is a binomial and

Again we can draw the probability polygon h(x) in

the same fashion.

k p =

.80

0 .000

1 .007

2 .058

3 .263

4 .672

5 1.000

Note that the total area under the probability

histogram is 1 and that the area under the

probability polygon is very close to 1

Now let x be a continuous random variable and

consider a set of n measurements of x. we can

construct the relative frequency histogram of

these values and the corresponding probability

polygon h(x) . As mentioned before, the area

under h(x) is very close to 1. We can use this

polygon approximate as the area

under h(x) over the interval .

0

2

4

6

8

10

0.835 0.865 0.895 0.925 0.955 0.985 1.015 1.045 1.075 More

rel freq

rel freq

As (the number of measurements

increases without bound) the relative frequency

polygon h(x) approaches the graph of a smooth

function whose area under the graph is one.

This function is called the probability density

function, pdf, of the random variable x.

0

2

4

6

8

10

0.835 0.865 0.895 0.925 0.955 0.985 1.015 1.045 1.075 More

rel freq

rel freq

h(x)

The pdf of x can be used, among other

things, to find the exact value of

which is the area under the graph of over

the interval which is, from calculus,

a b

The probability density function for the random variable x describes the probability distribution of the continuous random variable x. This function is sometime referred to as the depth of probability.

•In many situations the density function is given

by a mathematical formula where as in many

many situations finding a nice formula may not

be possible.

PROPERTIES OF CONTINUOUSPROBABILITY

DISTRIBUTIONS

1.

2. The total area under the curve =

3. =area under the curve between a and b

Exercise: Show that there is no probability attached to any single

value of x. That is, P(x = a) = 0.

There are many different types of continuous random

variables. Here are some that occur often:

1. Uniform Random variable

2. Exponential Random Variable

Some other random variable include The Gamma, Chi

Square, and Bata random variables. The most important

random variable is the Normal Random variable.

To choose a model for real life situation, we try to pick a model that

Fits the data well

Allows us to make the best possible inferences using the data.

As mentioned, one important continuous random variable is the normal random variable and we will discuss that in more detail.

THE NORMAL DISTRIBUTION

deviation. standard andmean population theare and

1416.3 7183.2

for 2

1)(

2

2

1

e

xexf

x

• The shape and location of the normal

curve changes as the mean and standard

deviation change.

• The formula that generates the

normal probability distribution is:

THE STANDARD NORMAL

DISTRIBUTION

To find P(a < x < b), we need to find the

area under the appropriate normal curve.

To simplify the tabulation of these areas,

we standardize each value of x by

expressing it as a z-score, the number of

standard deviations it lies from the mean

.

xz

THE STANDARD

NORMAL (Z)

DISTRIBUTION

Mean = 0; Standard deviation = 1

When x = , z = 0

Symmetric about z = 0

Values of z to the left of center are negative

Values of z to the right of center are positive

Total area under the curve is 1.

USING TABLE 3

The four digit probability in a particular row

and column of Table 3 gives the area under

the z curve to the left that particular value of

z.

Area for z = 1.36

P(z 1.36) = .9131

P(z >1.36)

= 1 - .9131 = .0869

P(-1.20 z 1.36) =

.9131 - .1151 = .7980

EXAMPLE

Use Table 3 to calculate these probabilities:

To find an area to the left of a z-value, find the area

directly from the table.

To find an area to the right of a z-value, find the area

in Table 3 and subtract from 1.

To find the area between two values of z, find the two

areas in Table 3, and subtract one from the other.

P(-1.96 z 1.96)

= .9750 - .0250 =

.9500

P(-3 z 3)

= .9987 -

.0013=.9974

Remember the Empirical Rule:

Approximately 99.7% of the

measurements lie within 3

standard deviations of the mean.

USING TABLE 3

Remember the Empirical Rule:

Approximately 95% of the

measurements lie within 2

standard deviations of the mean.

1. Look for the four digit area

closest to .2500 in Table 3.

2. What row and column does

this value correspond to?

WORKING BACKWARDS

Find the value of z that has area .25 to its left.

4. What percentile

does this value

represent? 25th percentile,

or 1st quartile (Q1)

3. z = -.67

1. The area to its left will be 1 - .05

= .95

2. Look for the four digit area

closest to .9500 in Table 3.

Find the value of z that has area .05 to its right.

3. Since the value .9500 is

halfway between .9495 and

.9505, we choose z halfway

between 1.64 and 1.65.

4. z = 1.645

FINDING PROBABILITIES FOR THE

GENERAL NORMAL RANDOM

VARIABLE

To find an area for a normal random variable x

with mean and standard deviation , standardize

or rescale the interval in terms of z.

Find the appropriate area using Table 3.

Example: x has a normal

distribution with = 5 and =

2. Find P(x > 7).

1587.8413.1)1(

)2

57()7(

zP

zPxP

1 z

EXAMPLE The weights of packages of ground beef are

normally distributed with mean 1 pound and

standard deviation .10. What is the probability

that a randomly selected package weighs

between 0.80 and 0.85 pounds?

)85.80(. xP

)5.12( zP

0440.0228.0668.

What is the weight of a package

such that only 1% of all packages

exceed this weight?

233.11)1(.33.2?

33.21.

1? 3, Table From

01.)1.

1?(

01.?)(

zP

xP