MA 320-001: Introductory Probabilitydmu228/ma320/lectures/sec5.1.pdfThe Uniform Distribution Because X is a continuous-type random variable, F0(x) is equal to the p.d.f. of X whenever

MA 320-001: Introductory Probability

David Murrugarra

Department of Mathematics,University of Kentucky

http://www.math.uky.edu/~dmu228/ma320/

Spring 2017

David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring 2017 1 / 18

mailto:[email protected]

http://www.math.uky.edu/

http://www.uky.edu/

http://www.math.uky.edu/~dmu228/ma320/


The Uniform Distribution

Let the random variable X denote the outcome when point is selectedat random from an interval [a,b],−∞ < a < b <∞. If the experimentis performed in a fair manner, it is reasonable to assume that theprobability that the point is selected from the interval [a, x ],a ≤ x < b,is (x − a)/(b − a). That is, the probability is proportional to the lengthof the interval, so the distribution function of X is

F (x) =

0, x < a,x−ab−a , a ≤ x < b,1, b ≤ x .



The Uniform Distribution

Because X is a continuous-type random variable, F ′(x) is equal to thep.d.f. of X whenever F ′(x) exists; thus, when a < x < b, we havef (x) = F ′(x) = 1/(b − a).

The random variable X has a uniform distribution if its p.d.f. is equalto a constant on its support. In particular, if the support is the interval[a,b], then

f (x) =1

b − a, a ≤ x ≤ b.



Uniform Distribution PDF

Figure: Uniform Distribution PDF



Uniform Distribution CDF

Figure: Uniform Distribution c.d.f.



Binomial Distribution

f (x) =(

nx

)px(1− p)n−x , x = 0,1,2, . . . ,n.

These probabilities are called binomial probabilities, and the randomvariable X is said to have a binomial distribution.

A binomial distribution will be denoted by the symbol b(n,p), and wethat the distribution of X is b(n,p).The constants n and p are called theparameters of the binomial distribution.

∑x∈S

f (x) = 1




f (x) =(

nx

)px(1− p)n−x , x = 0,1,2, . . . ,n.



∑x∈S

f (x) = 1




f (x) =(

nx

)px(1− p)n−x , x = 0,1,2, . . . ,n.



∑x∈S

f (x) = 1




Figure: Binomial density function.




f (x) =(

nx

)px(1− p)n−x , x = 0,1,2, . . . ,n.

µ = E(X ) = np.

σ2 = npq




f (x) =(

nx

)px(1− p)n−x , x = 0,1,2, . . . ,n.

µ = E(X ) = np.

σ2 = npq



Cumulative Distribution Function

The cumulative distribution function or, more simply, thedistribution function of the random variable X is

F (x) = P(X ≤ x), −∞ < x <∞,

For the binomial distribution the distribution function is defined by

F (x) = P(X ≤ x) =bxc∑y=0

(ny

)py (1− p)n−y

where bxc is the floor or greatest integer less than or equal to x .




Figure: Binomial distribution cdf



The Poisson Distribution

DefinitionLet the number of changes that occur in a given continuous interval becounted. Then we have an approximate poisson process withparameter λ > 0 if the following conditions are satisfied:

1 The number of changes occurring in nonoverlapping intervals areindependent.

2 The probability of exactly one change occurring in a sufficientlyshort interval of length h is approximately λh.

3 The probability of two or more changes occurring in a sufficientlyshort interval is essentially zero.




The random variable X has a Poisson distribution if its densityfunction is of the form

f (x) =λxe−λ

x!, x = 0,1,2, ...,

where λ > 0.

In this case, µ = σ2 = λ.




The Poisson distribution has been used to model:

1 The number of chocolate chips in a cookie.2 The number of calls coming into a call centre.3 The number of deaths from horse kicks in the Prussian army




Figure: Poisson distribution pmf




Figure: Poisson distribution cdf




If events in a poisson process occur at a mean rate of λ per unit, theexpected number of occurrences in an interval of length t is λt .

For example, if phone calls arrive at a switchboard following a Poissonprocess at a mean rate of three per minute, then the expected numberof phone calls in a 5-minute period is (3)(5) = 15.

Moreover, the number of occurrences say, x , in the interval of length thas the Poisson density function,

f (x) =(λt)xe−λt

x!, x = 0,1,2, ...




ExampleTelephone calls enter a college switchboard on the average of twoevery 3 minutes. If one assumes an approximate Poisson process,what is the probability of five or more calls arriving in a 9 minutesperiod?

Let X denote the number of calls in a 9 minute period.




ExampleTelephone calls enter a college switchboard on the average of twoevery 3 minutes. If one assumes an approximate Poisson process,what is the probability of five or more calls arriving in a 9 minutesperiod?

Let X denote the number of calls in a 9 minute period.

We see that E(X ) = 6; that is, on the average, six calls will arriveduring a 9 minute period. Thus,

P(X ≥ 5) = 1− P(X ≤ 4) = 1−4∑

x=0

6xe−6

x!

= 1− 0.285 = 0.715David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring 2017 18 / 18


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MA 320-001: Introductory Probabilitydmu228/ma320/lectures/sec5.1.pdfThe Uniform Distribution Because X is a continuous-type random variable, F0(x) is equal to the p.d.f. of X whenever