Normal Distribution-week3 (2)(2)

Chapter 6

The Normal Distribution and Other Continuous Distributions

PowerPoint to accompany:

Continuous Probability Distributions

A continuous random variable is a variable that can assume any value on a continuum (can assume an infinite number of values)

These can potentially take on any value, depending only on the ability to measure accurately

• Thickness of an item• Time required to complete a task• Weight, in grams• Height, in centimetres

2 Copyright © 2013 Pearson Australia (a division of Pearson Australia Group Pty Ltd) – 9781442549272/Berenson/Business Statistics /2e

The Normal Distribution


Probability Distributions

Normal

Uniform

Exponential


The Normal Distribution


‘Bell-shaped

Symmetrical

Mean, median and mode are equal

Central location is determined by the mean, μ

Spread is determined by the standard deviation, σ

The random variable X has an infinite theoretical range: + to

Mean = Median = Mode

X

f(X)

μ

σ

Many Normal Distributions


By varying the parameters μ and σ, we obtain different normal distributions

The Normal Distribution Shape


X

f(X)

μ

σ

Changing μ shifts the distribution left or right

Changing σ increases or decreases the spread (variability)

The Normal Probability Density Function


The formula for the normal probability density function is

Where e = the mathematical constant approximated by 2.71828

Π = the mathematical constant approximated by 3.14159

μ = the population mean

σ = the population standard deviation

X = any value of the continuous variable

2μ)/σ](1/2)[(Xe2π

1f(X)

Translation to the Standardised Normal Distribution


Translate any X to the

Standardised Normal (the Z

distribution) by subtracting

from any particular X value

the population mean and

dividing by the population

standard deviation:

Any normal distribution (with any mean and standard deviation combination) can be transformed into the standardised normal distribution (Z)

X μZ

σ

The Standardised Normal Probability Density Function


The formula for the standardised normal probability density function is

Where e = the mathematical constant approximated by 2.71828

π = the mathematical constant approximated by 3.14159

Z = any value of the standardised normal distribution

2(1/2)Ze2π

1f(Z)

The Standardised Normal Distribution


Also known as the Z distribution

Mean is 0

Standard deviation is 1

Z

f(Z)

0

1

Values above the mean have positive Z-values Values below the mean have negative Z-values

Standardised Normal Distribution Example


If X is distributed normally with mean of 100 and standard deviation of 50, the Z value for X = 200 is

This says that X = 200 is two standard deviations (2 increments of 50 units) above the mean of 100

X μ 200 100Z 2.0

σ 50

Comparing X and Z Units


Z100

2.00

200 X

Note that the distribution is the same, only the scale has changed. We can express the problem in original units (X) or in standardised units (Z)

(μ = 100, σ = 50)

(μ = 0, σ = 1)

Finding Normal Probabilities


a b X

f(X) P a X b( )≤

Probability is measured by the area under the curve

≤

P a X b( )<<=

(Note that the probability of any individual value is zero since the X axis has an infinite theoretical range: + to

Probability as Area Under the Curve


f(X)

Xμ

0.50.5

The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below

1 .0)XP (

P( X μ) 0.5 P(μ X ) 0.5

Empirical Rules


μ ± 1σ encloses approx 68.26 % of observations

f(X)

Xμ μ+1σμ-1σ

What can we say about the distribution of values around the mean?

There are some general rules:

σσ

68.26%

Empirical Rules


μ ± 2σ covers approx 95.44% of observations

μ ± 3σ covers approx 99.73% of observations

xμ

2σ 2σ

xμ

3σ 3σ

95.44% 99.73%

The Standardised Normal Table


The Cumulative Standardised Normal table

in the textbook (Appendix Table E.2) gives

the probability less than a desired value for Z

Once Z<-6 the values listed become so small as

to be effectively 0 in area

Z0 2.00

0.9772

Example: P(Z < 2.00) = 0.9772

The Standardised Normal Table


The value within the

table gives the

probability from

Z = +6 down to the

desired Z value.9772

2.0P(Z < 2.00) = 0.9772

The row shows the value of Z to the first decimal point

The column gives the value of Z to the second decimal point

2.0

.

.

.

Z 0.00 0.01 0.02

0.0

0.1

General Procedure for Finding Probabilities

To find P(a < X < b) when X is distributed normally:

1. Draw the normal curve for the

problem in terms of X

2. Translate X-values to Z-values and

put Z values on your diagram

3. Use the Standardised Normal Table


Finding Normal Probabilities


Z scale0.12X scale8.6 8

μ = 8 σ = 10

Therefore, P(X < 8.6) if the same as P(Z < 0.12)

0

Suppose X is normally distributed with mean 8.0 and standard deviation 5.0. Find P(X < 8.6)

X μ 8.6 8.0Z 0.12

σ 5.0

Solution: Finding P(Z < 0.12)


Z scale0.12

Z .00 .01

0.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

.5478.02

0.1 .5478

Standardised Normal Probability Table (Portion)

0.00

= P(Z < 0.12)P(X < 8.6)

Upper Tail Probabilities


Z scale0.12 0

0.5478 0.4522

We already know P(Z < 0.12 = 0.5478) We also know that total area under the curve can only ever be 1.0

(100%)

Thus, to find P(Z > 0.12) = 1.0 – P(Z < 0.12) = 1.0 – 0.5478 = 0.4522

Total area under curve is 1.0

Suppose X is normally distributed with mean 8.0

and standard deviation 5.0

Now find P(X > 8.6) = P (Z > 0.12)

Probability Between Two Values


Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(8 < X < 8.6)

P(8 < X < 8.6)

= P(0 < Z < 0.12)

Z0.12 0

X8.6 8

Calculate two sets of Z-values (at X=8 and at X=8.6)

1

X μ 8 8Z 0

σ 5

2

X μ 8.6 8Z 0.12

σ 5

Solution: Finding P(0 < Z < 0.12)


Z scale0.120.00

= P(0 < Z < 0.12)

P(8 < X < 8.6)

= P(Z < 0.12) – P(Z ≤ 0)

= 0.5478 - .5000 = 0.0478

0.5000

Z .00 .01

0.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

.02

0.1 .5478

Table E.2 (Portion)

Probabilities in the Lower Tail


X7.4 8.0

P(7.4 < X < 8)

= P(-0.12 < Z < 0)

= P(Z < 0) – P(Z ≤ -0.12)

= 0.5000 - 0.4522 = 0.0478

0.4522

Z-0.12 0

The normal distribution is symmetric, so this probability is the same as P(0 < Z < 0.12)

Suppose X is normally distributed with mean 8.0 and standard deviation 5.0

Now find P(7.4 < X < 8)

Finding the X Value for a Known Probability


1. Draw a normal curve placing all known

values on it such as mean of X and Z

2. Shade in area of interest and find

cumulative probability

3. Find the Z value for the known probability

4. Convert to X units using the formula

Note* This formula is simply our Z formula rearranged in terms of X

ZσμX

Finding the X Value for a Known Probability Example


Suppose X is normal with mean 8.0 and standard deviation 5.0

Now find the X value so that only 20% of all values are below this X

Steps 1 and 2

X? 8.0

0.2000

Z? 0



20% area in the lower tail is consistent with a Z value of -0.84 (closest)

Z .03

-0.9 .1762 .1736

.2033

-0.7 .2327 .2296

.04

-0.8 .2005

Table E.2 (Portion)

.05

.1711

.1977

.2266

…

…

…

…X? 8.0

Z-0.84 0

Step 3. Find the Z value for the known probability



Step 4. Convert to X units using the formula

80.3

0.5)84.0(0.8

ZσμX

So 20% of the values from a distribution with mean 8.0 and standard deviation 5.0 are less than 3.80

Note Z = -0.84 (not +0.84) since we are dealing with the left hand side of the curve

The Uniform Distribution



Probability Distributions

Normal

Uniform

Exponential

The uniform distribution is a probability distribution that has equal probabilities for all possible outcomes of the random variable

Also called a rectangular distribution

The Uniform Distribution


The Continuous Uniform Distribution

otherwise 0

bXaifab

1

wheref(X) = value of the density function at any X valuea = minimum value of Xb = maximum value of X

f(X) =

Characteristics of the Uniform Distribution


The mean of a uniform distribution is

The standard deviation is

a bμ

2

2(b-a)σ

12

Uniform Distribution Example for the Range 2≤X≤6


2 6

0.25

f(X) = = 0.25 for 2 ≤ X ≤ 66 - 21

X

f(X)

a b 2 6μ 4

2 2

2 2(b-a) (6-2)

σ 1.154712 12

Uniform Distribution Example to find P(3 ≤ X ≤ 5)


2 6

0.25

P(3 ≤ X ≤ 5) = (Base)(Height) = (2)(0.25) = 0.5

X

f(X)

3 54

Normal Approximation to the Binomial Distribution

The binomial distribution is a discrete distribution whereas the normal distribution is continuous



General rule: The normal distribution can be used to approximate the binomial distribution if

• np ≥ 5 and n(1 – p) ≥ 5

The closer p is to 0.5, the better the normal approximation to the binomial

The larger the sample size n, the better the normal approximation to the binomial




The mean and standard deviation of the binomial distribution are

μ = np

Transform binomial to normal using the formula:

p)np(1

npX

σ

μXZ

p)np(1σ

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