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

Yüklə 32.33 Kb.
 tarix 18.04.2016 ölçüsü 32.33 Kb.
Ch5.4 + Ch1.3 Random Variable and Its Probability Distribution:

Part I: Discrete Random Variable

-----------------------------------------------------------------------------------------------------------

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)

-----------------------------------------------------------------------------------------------------------

• 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.

------------------------------------------------------------------------------------------------------

• 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…)

1. 2. 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(x 2)

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

Alternatively, P(x 2) = 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 .

 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(x 3) = 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?  Verilənlər bazası müəlliflik hüququ ilə müdafiə olunur ©azrefs.org 2016
rəhbərliyinə müraciət