(12)continuous distributions

7/30/2019 (12)Continuous Distributions

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Applied Statistics and Computing Lab

CONTINUOUS DISTRIBUTIONS


Indian School of Business


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Probability Distribution of a Continuous

Random Variable

Weights of new born babies is an example of continuous randomvariable.

What is the probability of the weight of a new born baby beingexactly 3.54689 kgs, given that the weights of all new born babies

lie in the range 2.5-4 kgs? Answer is 0- probability of one event occurring out of an

uncountably infinite number of possibilities

Concept of pmf not valid- pmf is the probability of X assuming aparticular value x

But what if you are asked the probability of weight lying in theinterval 3.5-4, which includes the number 3.54689?

Concept of probability density function- based on finding theprobability of a continuous random variable lying in a particularinterval

2


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Probability Density Function

3

For a continuous random variable, if there exists a function f(x), such that, for

ab ,

( This gives the probability that X lies in the interval

[a,b])

A function of this type satisfying the conditionsa) f(x)0, for all x

b)

Is called the probability density function of x

( ) 1f x dx

( ) ( )b

a

f x dx P a x b


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Revisiting the new born baby example

Now,

We are interested in finding P(axa)=

P(X=a)=

Hence the probability of a continuous variable

taking a single value is 0

Thus, the probability of the weight of a newborn baby being exactly 3.54689 kgs is 0

4

( ) 0

a

a

f x dx

( ) ( )b

a

f x dx P a x b


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Cumulative Distribution function, Expectation,

Variance of a Random Variable: The Continuous

counterpart All the concepts and definitions are the same as those for discrete random variable In all the definitions, the summation sign is simply replaced by its continuous

counterpart- the integral sign

Cumulative Distribution Function:

F(x)= P(X x)=

Expectation: E(X)=

Variance: v(X)=

5

x

(- )

( )f x dx

x

(- )

( )xf x dx

x

2

(- )

( ( )) ( )x E X f x dx


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Applications A clock can stop at any time during the day. Let X be the time (hours plus

fractions of hours) at which the clock stops. The pdf for X isf (x) =(1/24), 0x24

=0, otherwise

What is the probability that the clock will stop between 2:00pm and 2:45pm?

P(14X14.75)=

Let the pdf of a continuous random variable x be

f(x) = (1/2)-(x/8), 0x4

=0, otherwise

a) What is P(2x+3>5) b) What is E(X)?(Hint: a) P(2x+3>5) is equivalent to P(x>1).

b) E(X)= )

6

14.75

14

1 1

24 32dx

4

0

1( )

2 8

xx dx


7/7Applied Statistics and Computing Lab

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(12)continuous distributions

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