Module 5: Discrete Distributions

Module 5: Discrete Distributions

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Discrete vs. Continuous • Discrete Random Variables:

• Result of “counting” something• They must have “space” between them

• Continuous Random Variables:• Results from a “measurement”• Assumes one of an infinite number of values, within a relevant range

• Random Variable:• The particular outcome of an experiment, determined by chance

• Discrete Probability Distribution: • Displays all possible outcomes and the probability of each outcome• Sum of the probabilities of all possible outcomes must equal 1.• The individual P(X) must be between 0 and 1• The outcomes must not overlap (discrete outcomes)

Mean or Expectation• Mean (Expectation) of a Discrete Distribution

• Measures the center of the distribution• Represents the theoretical long-run average of the random variable• Weighted average where each outcome is weighted by the

probability of the outcome or X * P(X)• Also known as the “Expected Value” or “Expectation.”• Calculated as follows:

]

X P(X) X * P(X) $ 200 0.30 $ 60.00 $ 25 0.50 $ 12.50 $ (85) 0.20 $(17.00)

μ = $ 55.50

Mean of Discrete Distribution

Variance of Discrete Distribution• The variance of the distribution describes the “spread”• The standard deviation of the distribution can be used to

compare two distributions with the same mean to determine which is the most variable

𝜎 2=∑ [ (𝑥−𝜇 ) 2¿𝑥 𝑃 (𝑋 )]¿X P(X) X * P(X) X - μ (X - μ)2 (X - μ)2 x P(X)

$ 200 0.30 $ 60.00 $ 144.50 $20,880.25 $ 6,264.08 $ 25 0.50 $ 12.50 $ (30.50) $ 930.25 $ 465.13 $ (85) 0.20 $(17.00) $(140.50) $19,740.25 $ 3,948.05

$ 10,677.25 Variance

μ = $ 55.50 $ 103.33 Standard Deviation

The Binomial DistributionThe binomial experiment is a probability experiment that satisfies these requirements:

1. Each trial can have only two possible outcomes—success or failure.

2. There must be a fixed number of trials.

3. The outcomes of each trial must be independent of each other.

4. The probability of success must remain the same for each trial.

!

- ! ! X n XnP X p q

n X X

In a binomial experiment, the probability of exactly X successes in n trials is

The Binomial Distribution

• n = the number of trials• X = number of successes desired• p = probability of success• q = probability of failure• Where p + q = 1

3 25!3 0.30 0.702!3!

P

5, 0.30,"at least 3" 3, 4,5 n p X

0.132

4 15!4 0.30 0.701!4!

P 0.028

5 05!5 0.30 0.700!5!

P 0.002

The Binomial DistributionA local textbook broker determines that 30% of all students purchase used textbooks for their classes. If 5 students come into the store, what is the probability that at least 3 will purchase used books?


3 0.1320.0280.002

0.162

P X

To determine the probability of “at least” three, determine the cumulative probabilities of: [P(X) = 3] + [P(X) = 4] + [P(X) = 5]

Answer: The probability that at least 3 out of 5 students will purchase used textbooks is 16.2%.


3 0.1320.0280.002

0.162

P X

To determine the probability of “at least” three, determine the cumulative probabilities of: [P(X) = 3] + [P(X) = 4] + [P(X) = 5]

Answer: The probability that at least 3 out of 5 students will purchase used textbooks is 16.2%.


1. Open MINITAB2. Select “Calc” from Menu3. Select “Probability

Distributions”4. Select “Binomial”


1. Select “Cumulative Distribution”

2. Number of Trials = “n”3. Event probability = “p”4. Input Constant – The value

and all below that value to be EXCLUDED from the calculation of the Binomial Probability.

MINTAB RESULTS: P(X = 0, 1 and 2) = 0.83692Solving for P(X ≥ 3) as 1.00 – 0.83692 = 0.16308 ≈ 16.31% probability that at least 3 students out of the 5 students rented their textbooks.


1. Select “Probability”2. Number of Trials = “n”3. Event probability = “p”4. Input Constant – The single

value for which you are finding the probability such as P(X = 3), or the probability of exactly 3.

MINTAB RESULTS: P(X = 3) = 0.1323 ≈ There is a 13.23% probability that exactly 3 students out of the 5 students rented their textbooks.


Find the probability for the following:

1. n = 10 p = .87 P(X = 6)

2. n = 15 p = .20 P(4 < X < 7)

3. n = 15 p = .20 P(X = 0)

4. n = 18 p = .30 P(X > 8)

5. n = 20 p = .90 P(X < 16)

6. n = 20 p = .20 P(X > 8)

7. n = 20 p= .70 P(X < 12)


Mean: np2Variance: npq

The mean, variance, and standard deviation of a variable that follows a binomial distribution :

Standard Deviation: npq

The Automobile Dealers Association reports that in 2013 of the first 8,000 hybrid vehicles produced 2% of them had a fault in the recharger circuit. What is the mean, variance and standard deviation for this data?

8000 0.02 160 np

2 8000 0.02 0.98 156.8 157 npq

8000 0.02 0.98 12.5 13 npq


The Poisson distribution describes events that will occur over an interval: time, distance, area or volume. In a Poisson Distribution:

1. The events are independent

2. The probability is proportional to the interval or “the longer the interval the higher the probability.”

It is also considered the “probability of improbable events” where n is large and p is small

The Poisson Distribution

In a Poisson Distribution, the probability of exactly X events over the interval of interest is:


• X = number of occurrences of events or “successes” desired

• λ = mean number of occurrences over the interval

• e = Base of Napierian Logarithm, a constant equal to 2.71828

;!

XeP X

X

The mean (λ) of a Poisson Distribution is found as:


n = number of trials and p = probability of success

• The variance of a Poisson Distribution is equal to the mean

• The random variable X in a Poisson Distribution can assume an infinite number of values, thereby having no upper limit

• The Poisson Distribution is positively skewed

The mean (λ) of a Poisson Distribution is proportional to the interval of occurrence, which requires an adjustment of the value of λ for intervals OTHER than the “original” interval.

Example: λ = 10 defects per 5,000 units produced

to find P(Defects) per 2,500 units produced: λ = 5to find P(Defects) per 1,000 units produced: λ = 2to find P(Defects) per 10,000 units produced: λ = 20to find P(Defects) per 15,000 units produced: λ = 30


If there are 200 typographical errors randomly distributed in a 500-page manuscript, find the probability that a given page contains exactly 3 errors.

First, find the mean number of errors. With 200 errors distributed over 500 pages, each page has an average of how many errors per page.

;!

XeP X

X

30.4 0.43!

e

Thus, there is less than 1% chance that any given page will contain exactly 3 errors.

0.0072



1. Open MINITAB2. Select “Calc” from Menu3. Select “Probability

Distributions”4. Select “Poisson”


1. Select “Cumulative Distribution”

2. Mean = “λ”3. Input Constant – The value

and all below that value to be EXCLUDED from the calculation of the Poisson Probability.

MINTAB RESULTS: P(X = 0, 1 and 2) = 0.99274 Solving for P(X ≥ 3) as 1.00 – 0.99274 = 0.00726 ≈ 0.726% probability that there will be at least 3 errors on a given page of a 500 page manuscript.


1. Select “Probability”2. Mean = λ3. Input Constant – The single

value for which you are finding the probability such as P(X = 3), or the probability of exactly 3.

MINTAB RESULTS: P(X = 3) = 0. 0071501 ≈ There is a .715 % probability that there will be exactly 3 errors on any given page of a 500 page manuscript.

The Poisson Distribution• The Securities and Exchange Commission has determined that the number

of companies listed in NYSE declaring bankruptcy is approximately a Poisson distribution with a mean of 2.6 per month. Find the probability that exactly 4 bankruptcies occur next month.

• Assume the number of trucks passing an intersection has a Poisson distribution with mean of 5 trucks per minute. What is the probability of 0 or 1 trucks in one minute?

• During off hours, cars arrive at a tollbooth on the East-West toll road at an average rate of 0.5 cars per minute. The arrivals are distributed according to the Poisson distribution. What is the probability that during the next five minutes three cars will arrive?

Module 5: Discrete Distributions

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