Continuous Probability Distribution (1)

BMS1024 MANAGERIAL STATISTICS

BMS1024BMS1024MANAGERIAL MANAGERIAL

STATISTICSSTATISTICS

Continuous Probability Distribution


Continuous Probability Distributions

A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values) thickness of an item time required to complete a task temperature of a solution height

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


The Normal DistributionProperties

‘Bell Shaped’ Symmetrical (Mean = Median = Mode) Location is characterized by the mean, μ Spread is characterized by the standard deviation, σ The right-side of the curve is the mirror image of the left-side of the curve (vice-versa) The random variable has an infinite theoretical range: - to +

Mean = Median = Mode

f(X)

μ

σ


The Normal DistributionDensity Function

2μ)(X

2

1

e2π

1f(X)

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


The Normal DistributionShape

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



Changing μ shifts the distribution left or right.

X

f(X)

μ μμ



Changing σ increases or decreases the spread.

X

f(X)

μ

σ


The Standardized Normal Distribution

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

Need to transform X units into Z units (Z scores). The standardized normal distribution has a mean of 0 and a

standard deviation of 1.

X ~ N (, σ²)


The Standardized Normal Distribution

σ

μXZ

Translate from X to the standardized normal (the “Z” distribution) by subtracting the mean of X and dividing by its standard deviation:

Transformation Formula:


The Standardized Normal Distribution: Shape

Z

f(Z)

0

1

Also known as the “Z” distribution Mean is 0 Standard Deviation is 1

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


The Standardized Normal Distribution: Probabilities Under The Curve

Z

f(Z)

-3 -2 -1 0 1 2 3

Total probabilities within 3 are approximately 1.0.

P(-3 < Z < 3) 1


The Standardized Normal Distribution: Example

2.050

100200

σ

μXZ

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 ~ N ( = 100, σ = 50)


The Standardized Normal Distribution: Example

Z100

2.00200 X (μ = 100, σ = 50)

(μ = 0, σ = 1)

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


Normal Probabilities

a b

f(X)

(Note that the probability of any individual value is zero)

Probability is measured by the area under the curve

P(a ≤ X ≤ b)


Normal Probabilities

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

f(X)

0.50.5

1.0)XP(

0.5)XP(μ 0.5μ)XP(


Normal Probability Tables

Example:

P(Z < 2.00) = .97725

The probabilities can be found in Table 4: The Normal Distribution Function from the Cambridge Statistical Tables. It gives the probability less than a desired value for Z (i.e., from negative infinity to Z)

Z0 2.00

.97725



Z values

Probabilities (shaded area

under the curve)



x(P) columns are the Z values

Areas/Probabilities (shaded tail area expressed in %)

one-tail area expressed in %


Finding Normal ProbabilityProcedure

Draw the normal curve for the problem in terms of X.

Translate X-values to Z-values (use the transformation formula).

Use the Normal Distribution Function Table to find the probability.

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


Finding Normal ProbabilityExample

Let X represent the time it takes (in seconds) to download an image file from the internet.

Suppose X is normal with mean 8.0 and standard deviation 5.0

Find P(X < 8.6)

X

8.6

8.0

X ~ N ( = 8, σ = 5)



0.125.0

8.08.6

σ

μXZ

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

Z0.12 0X8.6 8

μ = 8 σ = 10

μ = 0σ = 1

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

X ~ N ( = 8, σ = 5)



Table 4: Normal Distribution Function Table (Portion)

Z0.12 0

μ = 0σ = 1

P(X < 8.6) = P(Z < 0.12) = 0.5478



Find P(X > 8.6)…

P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z < 0.12)

= 1.0 - .5478 = .4522

Z

0.12

0

.5478

1.0 - .5478 = .4522


Finding Normal ProbabilityBetween 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)

05

88

σ

μXZ

1205

868.

.

σ

μXZ

Calculate Z-values:

Z0.12 0

X8.6 8

X ~ N ( = 8, σ = 5)


Finding Normal ProbabilityBetween Two Values

Z0.12

.0478

0.00

= P(0 < Z < 0.12)

P(8 < X < 8.6)

= P(Z < 0.12) – P(Z < 0)

= .5478 - .5000 = .0478

.5



Given Normal Probability, Find the X Value

Let X represent the time it takes (in seconds) to download an image file from the internet.

Suppose X is normal with mean 8.0 and standard deviation 5.0 Find X such that 20% of download times are less than X.

X? 8.0

.2

Z? 0

X ~ N ( = 8, σ = 5)



First, find the Z value corresponds to the known probability using the table.

X? 8.0

.2

Z-0.84 0


P(Z < 0.84) = .7995 .80

The closest to 80% area

P(Z > -0.84) = 0.7995 0.8

Z

.2

Mirror Image of Table 4 (portion)

0 0.84



Second, convert the Z value to X units using the following transformation formula.

seconds 80.3

0.5)84.0(0.8

ZσμX

So 20% of the download times from the distribution with mean 8.0 and standard deviation 5.0 are less than 3.80 seconds.

σ

μXZ


Working In-Class Example

A fast food restaurant sells hamburgers and chicken sandwiches. On a typical weekday the demand for hamburgers is normally distributed with a mean of 313 and standard deviation of 57. The demand for chicken sandwiches is also normally distributed with mean 93 and standard deviation 22.

a) How many hamburgers must the restaurant stock to be 98% sure of not running out on a given day?

b) How many chicken sandwiches must the restaurant stock to be 98% sure of not running out on a given day?


Working In-Class Example

c) If the restaurant stocks 400 hamburgers and 150 chicken sandwiches for a given day, what is the probability that the restaurant will run out of hamburgers or chicken sandwiches or both on that day. (Assume that the demand for hamburgers and the demand for chicken sandwiches are statistically independent).

d) Why is the independence assumption in part (c) probably not realistic? Using a more realistic assumption, do you think the probability requested for in part (c) would increase or decrease?


At the end of this lesson, you should be able to:

Understand on how to use the Normal Probability Distribution Function Table

Understand the properties of a Normal Probability Distribution

Transform X units to Z units (or Z-scores) using the transformation formula

Compute probabilities from the Normal Distribution Apply distribution to decision problems

Continuous Probability Distribution (1)

Documents

Transcript of Continuous Probability Distribution (1)