1 JTE9 Chapter 6 ~ Normal Probability Distributions

1

JTE9

xa b

P a x b( )

Chapter 6 ~ Normal Probability Distributions

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JTE9 Chapter Goals

• Learn about the normal, bell-shaped, or Gaussian distribution

• How probabilities are found

• How probabilities are represented

• How normal distributions are used in the real world

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JTE9 6.1 ~ Normal Probability Distributions

• The normal probability distribution is the most important distribution in all of statistics

• Many continuous random variables have normal or approximately normal distributions

• Need to learn how to describe a normal probability distribution

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JTE9 Normal Probability Distribution

1. A continuous random variable

2. Description involves two functions:a. A function to determine the ordinates of the graph picturing the distributionb. A function to determine probabilities

4. The probability that x lies in some interval is the area under the curve

3. Normal probability distribution function:

This is the function for the normal (bell-shaped) curvef x e

x

( )( )

12

2

2

1

5

JTE9

2 3 2 3

The Normal Probability Distribution

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JTE9

• Illustration

P a x b f x dxa

b( ) ( )

a b x

Probabilities for a Normal Distribution

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JTE9 Notes The definite integral is a calculus topic We will use the TI83/84 to find probabilities for normal distributions

We will learn how to compute probabilities for one special normal distribution: the standard normal distribution

We will learn to transform all other normal probability questions to this special distribution

Recall the empirical rule: the percentages that lie within certain intervals about the mean come from the normal probability distribution

We need to refine the empirical rule to be able to find the percentage that lies between any two numbers

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JTE9 Percentage, Proportion & Probability

• Basically the same concepts

• Percentage (30%) is usually used when talking about a proportion (3/10) of a population

• Probability is usually used when talking about the chance that the next individual item will possess a certain property

• Area is the graphic representation of all three when we draw a picture to illustrate the situation

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JTE96.2 ~ The Standard Normal Distribution

• There are infinitely many normal probability distributions

• They are all related to the standard normal distribution

• The standard normal distribution is thenormal distribution of the standard variable z(the z-score)

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JTE9 Standard Normal Distribution

Properties:• The total area under the normal curve is equal to 1

• The distribution is mounded and symmetric; it extends indefinitely in both directions, approaching but never touching the horizontal axis

• The distribution has a mean of 0 and a standard deviation of 1

• The mean divides the area in half, 0.50 on each side

• Nearly all the area is between z = -3.00 and z = 3.00

Notes: Table 3, Appendix B lists the probabilities associated with the intervals

from the mean (0) to a specific value of z Probabilities of other intervals are found using the table entries,

addition, subtraction, and the properties above

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JTE9

0 z

Table 3, Appendix B Entries

• The table contains the area under the standard normal curve between 0 and a specific value of z

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JTE9

0 145. z

Example

Example: Find the area under the standard normal curve between z = 0 and z = 1.45

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06

1.4 0.4265

• A portion of Table 3:

P z( . ) .0 145 0 4265

...

...

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JTE9 Using the TI 83/84

• To find the area between 0 and 1.45, do the following:

• 2nd DISTR 2 which is normalcdf(

• Enter the lower bound of 0

• Enter a comma

• Then enter 1.45

• Close the parentheses if you like or hit “Enter”

• The value of .426 is shown as the answer!

• Interpretation of the result: The probability that Z lies between 0 and 1.45 is 0.426

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JTE9

P z( . ) . . . 145 0 5000 0 4265 0 0735

0 4265.

0 145.

Area asked for

z

Example

Example: Find the area under the normal curve to the right of z = 1.45; P(z > 1.45)

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JTE9 Using the TI 83/84

• To find the area between 1.45 and ∞, do the following:

• 2nd DISTR 2 which is normalcdf(

• Enter the lower bound of 1.45

• Enter a comma

• Then enter 1 2nd EE 99

• Close the parentheses if you like or hit “Enter”

• The value of .074 is shown as the answer!

• Interpretation of result: The probability that Z is greater than 1.45 is 0.074

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JTE9

P z( . ) . . . 145 0 5000 0 4265 0 9265

0 145.

0 5000. 0 4265.

z

Example

Example: Find the area to the left of z = 1.45; P(z < 1.45)

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JTE9 Using The TI 83/84

• To find the area between - ∞ and 1.45, do the following:

• 2nd DISTR 2 which is normalcdf(

• Enter the lower bound of -1 2nd EE 99

• Enter a comma

• Then enter 1.45

• Close the parentheses if you like or hit “Enter”

• The value of 0.926 is shown as the answer!

• Interpretation of result: The probability that Z is less than 1.45 is 0.926

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JTE9 Notes

The addition and subtraction used in the previous examples are correct because the “areas” represent mutually exclusive events

The symmetry of the normal distribution is a key factor in determining probabilities associated with values below (to the left of) the mean. For example: the area between the mean and z = -1.37 is exactly the same as the area between the mean and z = +1.37.

When finding normal distribution probabilities, a sketch is always helpful

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JTE9

0 126. 126. z

Area asked for

P z( . ) . 126 0 0 3962

Example

Example: Find the area between the mean (z = 0) andz = -1.26

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JTE9 Using the TI 83/84

• Find the area to the left of z = -0.98

• Use -1E99 for - ∞ and enter 2nd DISTR • Normalcdf (-1e99, -0.98) which gives .164

00.98

Area asked for

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JTE9

0 180. 230.

0 4641.0 4893.

P z P z P z( . . ) ( . ) ( . )

. . .

230 180 2 30 0 0 180

0 4893 0 4641 0 9534

Example

Example: Find the area between z = -2.30 and z = 1.80

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JTE9 Using the TI 83/84

Find the area between z = -2.30 and z = 1.80• Enter 2nd DISTR, normalcdf (-2.3, 1.80) and press enter

• .953 is given as the answer.

• Remember, the function normalcdf is of the form:• Normalcdf(lower limit, upper limit, mean, standard deviation) and if

you’re working with distributions other than the standard normal (recall mean = 0, stddev = 1), you must enter the values for mean and standard deviation

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JTE9 Normal Distribution Note

The normal distribution table may also be used to determine a z-score if we are given the area (working backwards)

Example: What is the z-score associated with the 85th percentile?

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JTE9 Using the TI 83/84

• There is another function in the DISTR list that is used to find the value of z (or x) when the probability is given. For the previous problem, we are actually asking what is the value of z such that 85% of the distribution lies below it.

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JTE9 Using the TI 83/84

• Use 2nd DISTR invNorm( to calculate this value

• 2nd DISTR invNorm(.85) “ENTER” gives us a value of 1.036 which is shown

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JTE9 Example

Example: What z-scores bound the middle 90% of a standard normal distribution?

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JTE9 Using the TI 83/84

• The TI 83/84 calculates areas from -∞ to the value of z we are interested in. Therefore, we must get a little creative to solve some problems.

• Using the idea that the total area equals one comes in very handy here!

• For the example given, where we are interested in the value of z that bounds the middle 90%, the tails therefore represent a total of 10%. Divide this in two since it is symmetric and this gives 5% in each tail.

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JTE9 Using the TI 83/84

• Now use the 2nd DISTR invNorm with .05 in the argument like this:

• Which gives an answer of -1.645– Since the distribution is symmetric, the upper limit is

1.645, so 90% of the distribution lies between (-1.645, 1.645)

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JTE9 Using the TI 83/84

Now let’s work the problems on page 279

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JTE9 6.3 ~ Applications of Normal Distributions

• Apply the techniques learned for the z distribution to all normal distributions

• Start with a probability question in terms ofx-values

• Convert, or transform, the question into an equivalent probability statement involvingz-values

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JTE9

c x

0c

z

Standardization

• Suppose x is a normal random variable with mean and standard deviation

• The random variable has a standard normal distribution

zx

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JTE9 Example

Example: A bottling machine is adjusted to fill bottles with a mean of 32.0 oz of soda and standard deviation of 0.02. Assume the amount of fill is normally

distributed and a bottle is selected at random:

1) Find the probability the bottle contains between 32.00 oz and 32.025 oz

2) Find the probability the bottle contains more than 31.97 oz

When x z 32.0032.00 32.00 32.0

0.00;0.02

Solutions:

1)

When x z

32 02532.025 32 025 32.0

125. ;.

0.02.

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JTE9

P x Px

P z

( . )0. 0.

.

0.

( . ) .

32.0 32 02532.0 32.0

02

32.0

02

32 025 32.0

02

0 125 0 3944

32.0 32 025. x0 125. z

Area asked for

Solution Continued

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JTE9

P x Px

P z( . ).

( .

. . .

319732.0

0.023197 32.0

0.02150)

0 5000 0 4332 0 9332

32.03197. x

0 150. z

Example, Part 2

2)

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JTE9 Notes

• The normal table may be used to answer many kinds of questions involving a normal distribution

Example: The waiting time x at a certain bank is approximately normally distributed with a mean of 3.7 minutes and a standard deviation of 1.4 minutes. The bank would like to claim that 95% of all customers are waited on by a teller within c minutes. Find the value of c that makes this statement true.

• Often we need to find a cutoff point: a value of x such that there is a certain probability in a specified interval defined by x

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JTE9

P x c

Px c

P zc

( ) 0.

.

.

.

.0.

..

0.

95

3 7

14

3 7

1495

3 714

95

c

c

c

3 714

1645

1645 14 3 7 6 003

6

..

.

( . )( . ) . .

minutes

3 7. c x0 1645. z

0 5000. 0 4500.

0 0500.

Solution

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JTE9 Example

Example: A radar unit is used to measure the speed of automobiles on an expressway during rush-hour traffic. The speeds of individual automobiles are normally distributed with a mean of 62 mph. Find the standard deviation of all speeds if 3% of the automobiles travel faster than 72 mph.

62 72 x

0 188. z

0 4700.

0 0300.

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JTE9 Solution

P x( ) . 72 0 03

. 188 10

/ . . 10 188 532

P z( . ) . 188 0 03

zx

;72 621.88 =

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JTE9 Notation• If x is a normal random variable with mean and standard deviation , this is often

denoted:x ~ N(, )

Example: Suppose x is a normal random variable with = 35 and = 6. A convenient notation to identify this random variable is: x ~ N(35, 6).

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JTE9 6.4 ~ Notation

• z-score used throughout statistics in a variety of ways

• Need convenient notation to indicate the area under the standard normal distribution

• z() is the algebraic name, for the z-score (point on the z axis) such that there is of the area (probability) to the right of z()

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JTE9 Illustrations

z(0.10) represents the value of z such that the area to the right under the standard normal curve is 0.10

0 z

010.

z(0.10)

z(0.80) represents the value of z such that the area to the right under the standard normal curve is 0.80

0 z

0 80.

z(0.80)

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JTE9 Example

Example: Find the numerical value of z(0.10):

z(0.10) = 1.28

0 z

0.10 (area information from notation)

Table shows this area (0.4000)

z(0.10)

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JTE9 Example

Example: Find the numerical value of z(0.80):

• Use Table 3: look for an area as close as possible to 0.3000

• z(0.80) = -0.84

0 z

Look for 0.3000; remember that z must be negative

z(0.80)

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JTE9 Notes

• The values of z that will be used regularly come from one of the following situations:

1. The z-score such that there is a specified area in one tail of the normal distribution

2. The z-scores that bound a specified middle proportion of the normal distribution

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JTE9 Example

Example: Find the numerical value of z(0.99):

• Because of the symmetrical nature of the normal distribution, z(0.99) = -z(0.01)

0 z

0.01

z(0.99)

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JTE9 Example

Example: Find the z-scores that bound the middle 0.99 of the normal distribution:

z(0.005) = 2.575 and z(0.995) = -z(0.005) = -2.575

0or

0 495.0 495.0 005.0 005.

z(0.995)

-z(0.005)

z(0.005)

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JTE9 6.5 ~ Normal Approximation of the Binomial

• Recall: the binomial distribution is a probability distribution of the discrete random variable x, the number of successes observed in n repeated independent trials

• Binomial probabilities can be reasonably estimated by using the normal probability distribution

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JTE9 Background & Histogram

• Background: Consider the distribution of the binomial variable x when n = 20 and p = 0.5

The histogram may be approximated by a normal curve

• Histogram:

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 200.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0.18

x

P x( )

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JTE9

np

npq

( )( . )

( )( . )( . ) .

20 0 5 10

20 0 5 0 5 5 2 236

Notes

The normal curve has mean and standard deviation from the binomial distribution:

Can approximate the area of the rectangles with the area under the normal curve

The approximation becomes more accurate as n becomes larger

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JTE9 Two Problems

1. As p moves away from 0.5, the binomial distribution is less symmetric, less normal-looking

Solution: The normal distribution provides a reasonable approximation to a binomial probability distribution whenever the values of np and n(1 - p) both equal or exceed 5

2. The binomial distribution is discrete, and the normal distribution is continuous

Solution: Use the continuity correction factor. Add or subtract 0.5 to account for the width of each rectangle.

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JTE9 Example

Example: Research indicates 40% of all students entering a certain university withdraw from a course during their first year. What is the probability that

fewer than 650 of this year’s entering class of 1800 will withdraw from a class?

P xx

xx x( ) ( . ) ( . )

18000 4 0 6 1800 for 0, 1, 2, ... ,1800

• Let x be the number of students that withdraw from a course during their first year

• x has a binomial distribution: n = 1800, p = 0.4

• The probability function is given by:

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JTE9

np

npq

( )( . )

( )( . )( . ) .

1800 0 4 720

1800 0 4 0 6 432 20 78

P x P x x

P x x

Px

P z

( ) ( )

( . )

..

.

( . )

. . .

is fewer than 650 (for discrete variable )

(for a continuous variable )

650

649 5

72020 78

649 5 72020 78

3 39

0 5000 0 4997 0 0003

Solution

• Use the normal approximation method:

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JTE9 Random Number Generation

• With each rand execution, the TI-84 Plus generates the same random-number sequence

• for a given seed value. The TI-84 Plus factory-set seed value for rand is 0. To generate a

• different random-number sequence, store any nonzero seed value to rand. To restore

• the factory-set seed value, store 0 to rand or reset the defaults (Chapter 18).

• Note: The seed value also affects randInt(, randNorm(, and randBin( instructions.