Normal Distributions Q U A N T T E C H I N T E U Q I A S E V I T 1 0 S

Continuous Random Variable Discrete Variables Bernoulli Binomial Poisson Random variables can take a fixed number of values, like ... Integers Range of integers Any arbitrary FINITE set For each value there is a probability of occurrence Continuous Exponential Normal Random value can take any value, including fractional values, in a range. This is potentially INFINITE Between 1 and 2 1.1, 1.2, 1.3 .... 1.9, 2.0 1.1, 1.11, 1.12, 1.13 .. 1.111, 1.112, 1.113 ... 1.1111, 1.1112, 1.1113 ..

Discrete Variables

Bernoulli

Binomial

Poisson

Random variables can take a fixed number of values, like ...

Integers

Range of integers

Any arbitrary FINITE set

For each value there is a probability of occurrence

Continuous

Exponential

Normal

Random value can take any value, including fractional values, in a range.

This is potentially INFINITE

Between 1 and 2

1.1, 1.2, 1.3 .... 1.9, 2.0

1.1, 1.11, 1.12, 1.13 ..

1.111, 1.112, 1.113 ...

1.1111, 1.1112, 1.1113 ..

Probability of a Continuous Variable Since an infinite number of random variables are possible, in ANY given range, the probability of any single variable is ZERO! P( X = x) = 0 !! However the probability of the variable lying in a range is not zero P(x < X <= x+ d x) = f(x) Where d x is small range Example : Suppose Z is the mm of rain that falls in Calcutta Probability (Z = 144.5 mm ) = 0 Probability (Z between 144.49 and 144.51) = non zero

Since an infinite number of random variables are possible, in ANY given range, the probability of any single variable is ZERO!

P( X = x) = 0 !!

However the probability of the variable lying in a range is not zero

P(x < X <= x+ d x) = f(x)

Where d x is small range

Example :

Suppose Z is the mm of rain that falls in Calcutta

Probability (Z = 144.5 mm ) = 0

Probability (Z between 144.49 and 144.51) = non zero

Distribution Functions Discrete Random Variable Bernoulli, Binomial, Poisson p(N = n) is defined S P(N = n i ) = 1 Continuous Random Variable Exponential, Normal p(X = x) = 0 f(x) is defined as P(x < X <= x+dx) Were dx is a very small number = 1 The sum of the probabilities of ALL possible values of the random variable MUST be equal to 1

Discrete Random Variable

Bernoulli, Binomial, Poisson

p(N = n) is defined

S P(N = n i ) = 1

Continuous Random Variable

Exponential, Normal

p(X = x) = 0

f(x) is defined as

P(x < X <= x+dx)

Were dx is a very small number

= 1

Density Function Exponential Variable P(x) = l r . e -( l r x) Normal Distribution Normal Distribution The equation looks very complex .. but it is rarely used in this form. We use a tool to calculate the values ... Spreadsheets allow us to calculate values of P(x) in terms of x, m , s Printed tables are available that show the values

Exponential Variable

P(x) = l r . e -( l r x)

Normal Distribution

Normal Distribution

The equation looks very complex .. but it is rarely used in this form.

We use a tool to calculate the values ...

Spreadsheets allow us to calculate values of P(x) in terms of x, m , s

Printed tables are available that show the values

What do these graphs mean ? Probability of X lying in the range x -> x+ d x when the underlying distribution has the following parameters A : m = 0.0, s = 0.1 B : m = 00.0, s = 0.4 C : m = 0.75, s = 0.1 D : m = 0.75, s = 0.4

Probability of X lying in the range x -> x+ d x when the underlying distribution has the following parameters

A : m = 0.0, s = 0.1

B : m = 00.0, s = 0.4

C : m = 0.75, s = 0.1

D : m = 0.75, s = 0.4

Characteristics of a Normal Distribution The curve has a single peak. It is Unimodal Has a bell shape The mean of a normally distributed population lies at the centre of the of the normal curve Because of the symmetry of the normal distribution, the median and the mode is also at the centre. The mean, median and mode are equal in value The two tails of the normal distribution extend indefinitely and (theoretically) should never touch the horizontal axis

The curve has a single peak.

It is Unimodal

Has a bell shape

The mean of a normally distributed population lies at the centre of the of the normal curve

Because of the symmetry of the normal distribution, the median and the mode is also at the centre.

The mean, median and mode are equal in value

The two tails of the normal distribution extend indefinitely and (theoretically) should never touch the horizontal axis

Three very similar terms ! Distribution A name which describes the nature of the underlying population from where a random variable is selected Bernoulli, Binomial, Poisson Exponential, Normal Density Function f(x) = probability P(x < X <= x+ d x) Probability that x lies in the small range x to x + d x Distribution Function F(x) = probability P (X <= x ) Probability that X is less than or equal to x

Distribution

A name which describes the nature of the underlying population from where a random variable is selected

Bernoulli, Binomial, Poisson

Exponential, Normal

Density Function

f(x) = probability P(x < X <= x+ d x)

Probability that x lies in the small range x to x + d x

Distribution Function

F(x) = probability P (X <= x )

Probability that X is less than or equal to x

From Density to Distribution Area under the curve of the density function to the left of the red line ( at some value of x) IS EQUAL TO Value of the distribution function at the same value of x

Area under the curve of the density function to the left of the red line ( at some value of x)

IS EQUAL TO

Value of the distribution function at the same value of x

Area under the Density curve Area under the curve of the density function represents the fraction of the observations that lie below ( less than) the corresponding value in the distribution function The final value of the distribution function is 1 Probability of all values being below this value is 1 Certainty

Area under the curve of the density function

represents the

fraction of the observations that lie below ( less than) the corresponding value in the distribution function

The final value of the distribution function is 1

Probability of all values being below this value is 1

Certainty

Area under the Density curve Area to left of A Values less than m 50% total area 50% of all observations are below this value Area to left of B Values less than m – s 16% of total area 16% of all observations are below this value Area to left of C Values less than m – 2s 2.25% of total area 2.25% of all observations are below this value m =1.5 s 2 s A B C 2.25% 16%

Area to left of A

Values less than m

50% total area

50% of all observations are below this value

Area to left of B

Values less than m – s

16% of total area

16% of all observations are below this value

Area to left of C

Values less than m – 2s

2.25% of total area

2.25% of all observations are below this value

The 2 s limit 95.5 % of all values are expected to Lie in region around the mean value m in range that lies between Lower bound : m – 2s Upper bound : m + 2s Similarly 68% in the region between m – s to m + s 99.7% in the region between m – 3s to m + 3s m =1.5 2 s A C1 2.25% C2 2 s 2.25%

95.5 % of all values are expected to

Lie in region around the mean value m in range that lies between

Lower bound : m – 2s

Upper bound : m + 2s

Similarly

68% in the region between m – s to m + s

99.7% in the region between m – 3s to m + 3s

Problem Type A Number of blue shirts sold at a store Average 30 Std Deviation 8 Number of defects in a batch of 1000 Average 50 Std Deviation 10 What is the probability that On a given date, the number of shirts sold will be Less than or equal to 20 More than 35 Between 32 - 27 In any given batch the number of defects will be Less than 40 More than 60 Between 55 and 60

Number of blue shirts sold at a store

Average 30

Std Deviation 8

Number of defects in a batch of 1000

Average 50

Std Deviation 10

What is the probability that

On a given date, the number of shirts sold will be

Less than or equal to 20

More than 35

Between 32 - 27

In any given batch the number of defects will be

Less than 40

More than 60

Between 55 and 60

Solution to Shirt Problem P( x <= 20) = 0.11 P (x > 35) = 1 – 0.73 = 0.27 P (27 < x <= 32) = 0.6 – 0.35 = 0.25

Problem Type B Demand for blue shirts at a store Average 30 Std Deviation 8 Number of defects in a batch of 1000 Average 50 Std Deviation 10 Batch is rejected if number of defects is 52 What is the probability Of a 'stock-out' on any day if the store stocks 40 shirts How many shirts should be stocked So that the probability of stock out is 5% or less What is the probability of a batch being rejected ? To what average level of defects should the production be improved to ensure that probability of rejection is 5% or less

Demand for blue shirts at a store

Average 30

Std Deviation 8

Number of defects in a batch of 1000

Average 50

Std Deviation 10

Batch is rejected if number of defects is 52

What is the probability

Of a 'stock-out' on any day if the store stocks 40 shirts

How many shirts should be stocked

So that the probability of stock out is 5% or less

What is the probability of a batch being rejected ?

To what average level of defects should the production be improved to ensure that probability of rejection is 5% or less

Solution to Shirt Problem

Mostly 2 Categories of Problems Normal Distribution Mean is known Std Deviation is known Given : A Value Find Probability of Variable being less than or equal to this value Given : A Probability Find a value such that the probability of the variable being less than or equal to this value is equal to given probability Use Formula from Spreadsheet .. Since mathematical formula is very complex P = NORMDIST( V , m,s ) V = NORMINV( P , m,s )

Normal Distribution

Mean is known

Std Deviation is known

Given : A Value

Find Probability of Variable being less than or equal to this value

Given : A Probability

Find a value such that the probability of the variable being less than or equal to this value is equal to given probability

Use Formula from Spreadsheet ..

Since mathematical formula is very complex

P = NORMDIST( V , m,s )

V = NORMINV( P , m,s )

Plus the issue of “range” What is the probability of the random variable falling between 1 and 4 ? Area under the density curve between the two red lines Difference between the corresponding values in the distribution function as shown as the gap between the two green lines

What is the probability of the random variable falling between 1 and 4 ?

Area under the density curve between the two red lines

Difference between the corresponding values in the distribution function as shown as the gap between the two green lines

Before the Era of Spreadsheets Given a normal distribution N( m,s ) that has mean = m and standard deviation is s Calculation of either P = NORMDIST( X , m,s ) or X = NORMINV( P , m,s ) Can only be done through a computer OR ... By Looking up printed tables that list values of P and V together But for each combination of m and s there would be a different table ! How many tables would you need to keep at hand ? Strangely enough ... ONLY ONE !

Given a normal distribution N( m,s ) that has mean = m and standard deviation is s

Calculation of either

P = NORMDIST( X , m,s ) or

X = NORMINV( P , m,s )

Can only be done through a computer

OR ...

By Looking up printed tables that list values of P and V together

But for each combination of m and s there would be a different table !

How many tables would you need to keep at hand ?

Strangely enough ... ONLY ONE !

Standard Normal Distribution Suppose you have a variable X that follows a Normal distribution N( m,s ) Define a second variable Z such that Z = (X – m )/ s Then Z will follow a normal distribution N(0,1) where m z = 0 and s z = 1 So we can calculate .. P = NORMDIST( X , m,s ) = NORMDIST( Z ,0 ,1 ) AND Z = NORMINV( P ,0 ,1 ) & since Z = (X – m ) / s X = s * NORMINV( P ,0 ,1 ) + m If we have just ONE table that lists P and X values of the normal distribution N(0,1)

Suppose you have a variable X that follows a Normal distribution N( m,s )

Define a second variable Z such that

Z = (X – m )/ s

Then Z will follow a normal distribution N(0,1) where m z = 0 and s z = 1

So we can calculate ..

P = NORMDIST( X , m,s )

= NORMDIST( Z ,0 ,1 )

AND

Z = NORMINV( P ,0 ,1 )

& since Z = (X – m ) / s

X = s * NORMINV( P ,0 ,1 ) + m

If we have just ONE table that lists P and X values of the normal distribution N(0,1)

Identical Distribution Functions