Discrete

BA501 : The Class : Stats : Discrete : Discrete
Discrete Probability Distributions
Examples

§ Discrete Probability §

Lesson 2- Examples

§ Probability §

§ m, s, and s² for Any Discrete Random Variable §

§ Poisson Distribution §

Example 1

Probability

Relative Frequency approach to probability:

(a) frequency = f = m

(b) relative frequency = rf = f / n = m / n

(c) Example: n = 20

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Class	CW	f = m	m / n =	rf
1	0 - 10	7	7/10	0.35
2	10 - 20	3	3/20	0.15
3	20 - 30	6	6/20	0.30
4	30 - 40	4	4/20	0.20

{1} n = Sf = 20

Srf = 20/20 = 1.00

{2} P(10 - 20) = 3 / 20 = 0.15
(click me)

What is the probability that the data value is equal to or less than 20 i.e P(0 - 20)? 0.15 0.35 0.50 (click one)

Example 2

Mean, Variance and Standard Deviation of Any Discrete Random Variable

Some students in a statistics class at NAU read the editorial page in the Lumber Jack newspaper. If two students are chosen at random, the probability of choosing no students who read the editorial page is 0.75. The probability of choosing one student who reads the editorial page and one who does not read the editorial page is 0.20, and the probability of choosing two students who read the editorial page is 0.05. X is a random variable equal to the number of students who read the editorial page from the two chosen at random. Find the mean value of X and the variance of X.

X (random variable)= x₁ with P(x₁) or

= x₂ with P(x₂) or

= x₃ with P(x₃)

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X = (x₁ = 0 ) with P(x₁) = 0.75 or

= (x₂ = 1) with P(x₂) = 0.20 or

= (x₃ = 2 ) with P(x₃) = 0.05

SP[x_i] = 0.75 + 0.20 + 0.05 = 1.00

x	P[x_i]	x_iP[x_i]	x_i²	x_i²P[x_i]
0	0.75	0	0	0.00
1	0.20	0.20	1	0.20
2	0.05	0.10	4	0.20

m = Sx_iP[x_i] = 0.30

0.40 = Sx_i²P[x_i]

s² = Sx_i²P[x_i] - m² (click me)

= 0.40 - 0.30² = 0.40 - 0.09 = 0.31

s = Ö{Sx_i²P[x_i] - ²} = Ö[ 0.31 ] = 0.56

The variance should be used to interpret this problem. True False (click one)

Example 3

Poisson Distribution

The mail office at NAU sends out an average of nine special delivery packages daily. The number of special delivery packages is assumed to follow a Poisson distribution.

(a) What is the probability that for any day, the number of special delivery packages sent out will be more than five?

m = 9

Use Poisson Table

P[x > 5 ] = P[x ³ 6 ] = 1 - P[x £ 5 ] (click me)

= 1 - {P[x = 0 ] + P[x = 1 ]+ P[x = 2 ] (click me)

+ P[x = 3]+ P[x = 4] P[x = 5 ]}

= 1 - { 0.0001 + 0.0011 + 0.0050

+ 0.0150 + 0.0337 + 0.0607}

= 1 - 0.1156

= 0.8844

(b) What is the standard deviation of the number of special delivery packages sent out daily?

since m = s² for the Poisson distribution: (click me)

s = Ös² = Öm = Ö9 = 3.0

The Poisson distributions is used to measure the number of successes in the trials of an experiment. True False (click one)

You should now:

Go on to Excel and Equations
or

Go back to Lesson 1- Introduction
or
Go back to Discrete Distributions: Assignments and Activities

Please reference "BA501 (your last name) Assignment name and number" in the subject line of either below.

E-mail Dr. James V. Pinto at BA501@mail.cba.nau.edu
or call (928) 523-7356. Use WebMail for attachments.