C 4 + c 3 Random Variable and Its Probability Distribution: Part I: Discrete Random Variable




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Ch5.4 + Ch1.3 Random Variable and Its Probability Distribution:

Part I: Discrete Random Variable

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Topics: Random Variable (§5.4)

Probability Distribution of a discrete random variable (§5.4, §1.3)

Mean and Variance of a discrete random variable (§5.4, §2.1, §2.2)

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    • Random Variable (r.v.)

  • A random variable is a real-valued variable whose value depends on the outcome of an experiment.

Ex. Toss a coin twice, S={HH, HT, TH, TT }

We can define a r.v. x = # of heads. It takes 0, 1, 2. three different values



Ex. (continuous r.v.) Define a r.v. to for the height (in foot or meter) of a student as

x = the height

  • We can think that an r.v. is any rule that associates each outcome in an experiment with a real number. That is, a r.v. is real-valued function defined on the sample space of an experiment.

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    • Discrete r.v.

The possible values of the r.v. are isolated points along the number line.

Ex. x = # of Heads of tossing 2 coins x, =0,1,2

Ex. x = # of cars passing a bus stop from 8:00 to 8:30, x =0,1, 2, 3,……

(c.f. Continuous r.v.: The possible values forms an interval along the real line)



    • Probability Distribution of a Discrete r.v.

  1. The probability distribution of a discrete r.v. x, denoted as p(x), describes the probabilities that the r.v. x takes all possible values. The function p(x) is called the probability mass function.

Ex. x = # of head in tossing a fair coin. Then the probability distribution of x is

P(0) = P[x=0] = P[T] = 0.5

P(1) = P[x=1] = P[H] = 0.5

Ex. x = result of tossing a fair dice. The probability distribution of x is

P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6



2. In general, the probability that x gets a value c, P(x=c), is defined as the sum of all corresponding outcomes in S (i.e., the sample space) that are assigned to the value x.

Ex. x = # of heads in tossing 2 fair coins.



3. There are 3 ways to display a probability distribution for a discrete r.v.:

  1. through a density plot

  2. through a table

  3. through a formula

Ex. Toss a coin 3 times, and let x= # of heads. Then the probability distribution of x is:



(1) Density plot



If we convert the density plot in (1) we got:

(2) Table

x

0

1

2

3

P(x)

0.1

0.4

0.3

0.2

(3) Formula

Such as the one we gave for x = # of heads from tossing a coin 3 times.

 From the probability distribution given in (2), we can calculate

P( x = 3 ) = 0.2

P( x < 2 ) = P(x=0) + P(x=1) = 0.1 + 0.4 = 0.5

P( x 2 ) = P(x<2) + P(x=2) = 0.5 + 0.3 = 0.8.

P( x > 0 ) = 1 – P(x=0) = 1 – 0.1 = 0.9.


  1. For any probability distribution P(x), (recall the axiom of probabilities…)





Ex. (1) Find the value of c so that the following function is a probability distribution of a r.v. x:

P(0)+P(1)+P(2)+P(3)=1 (Note that x can only take 0, 1, 2, 3)

2c+3c+4c+5c=1, so c=1/14.

(2) For this probability distribution, find P(x2)

P(x2) =P(x=0) + P(x=1) + P(x=2) = 2c+3c + 4c = 9c = 9/14

Alternatively, P(x2) = 1- P(x>2) = 1- P(x=3) = 1- 5c = 1- 5/14=9/14.



Mean and Variance of a discrete r.v. with probability distribution

The mean

(The mean of a r.v. is also called as the expected value.)

The variances

The standard deviation =

Ex. Toss a coin twice, x = # of heads. Find and .

The probability distribution is



x

0

1

2

P(x)

0.25

0.5

0.25

(the number of heads we can expect to get if we toss a coin twice).

.

Ex. A contractor is required by a county planning department to submit from 1 to 5 different forms, depending on the nature of the project. Let x = # of forms required of the next contractor, and for x=1,2,3,4,5.

  1. What is the value of k?

From the form of P(x), we have

(Note x can only take 1, 2, 3, 4, 5)

That is, 1k + 2k + 3k + 4k + 5k = 1. So k = 1/15 (since P(x) = x/15 are between 0 and 1 for x = 1, 2, 3, 4, 5 and they sum to 1, k=1/15 is a valid solution.



  1. What is the probability that at most 3 forms are required?

P(x3) = p(x=1) + P(x=2)+P(x=3) = 1k + 2k + 3k = 6k = 6/15 = 2/5

  1. What is the expected number (i.e., mean) of forms required ?

1P(1) + 2P(2)+3P(3)+4P(4)+5P(5) = 1k + 2*2k + 3*3k + 4*4k+ 5*5k = 55k = 55/15=3.67

  1. What is the SD of the number of forms required?









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