As a manager of an organization, what statistics from this module would you use and why if you wanted to estimate your annual employee turnover? In your post, be sure to specifically identify the statistical formulas and what additional data you would need to determine your estimate(s).Module 4: Discrete and Continuous Probability

1. Random Variables

A random variable provides a way to describe experimental outcomes using numerical values. These

values can be used for a variety of decision-making purposes. A random variable can be called a

continuous variable or a discrete variable. A discrete random variable can assume either a finite

number of values or an infinite sequence of values that can be listed one-by-one. A continuous random

variable may assume any numerical value in an interval or collection of intervals.

Similar to variables, probability distributions can be classified as discrete or continuous. If a random

variable is a discrete variable, its probability distribution is called a discrete probability distribution. For

example, if you flip a coin two times you can have four possible outcomes: HH, HT, TH, and TT. If the

random variable X represents the number of Heads that result from this experiment, then the random

variable X can only take on the values 0, 1, or 2, so it is a discrete random variable. Each possible value

of the discrete random variable can be associated with a non-zero probability. Therefore, a discrete

probability distribution can be presented in tabular form.

For any outcome value x in a discrete probability distribution we have 0 ≤ P(x) ≤1

The values of x in the table are exhaustive, meaning that the probability distribution that is shown

contains all possible values. The values of x are mutually exclusive, meaning only one value can occur

for a given experiment. Finally, the sum of all the probabilities in the table must equal 1.

General formulas for a discrete probability distribution are as follows:

Mean: µ = E(x)

or

µ =∑xiP(xi) for all possible values of x.

Where E(x) is the expected value of x and P(𝑥і ) is the probability of value xi.

Variance: σ2 = E[(x – μ)²]

or

σ2 = ∑(𝑥і – μ)²P(𝑥і )

Where E[(x – µ) ²] is the expected value of the squared difference between the x and its mean µ.

Standard Deviation: σ=√σ²

Standard deviation (σ) is the square root of the variance (σ²).

Example: Determine the mean, variance and standard deviation for the following discrete probability

distribution.

X

0

1

2

P(x)

0.6

0.3

0.1

Mean: µ =E(x) = ∑𝑥і P(𝑥і )= 0(0.60) + 1(0.30) + 2(0.10) = 0.50

Variance: σ2= E[(x – μ)²] =∑(𝑥і −μ)²P(𝑥і )=(0−0.5)²(0.60)+(1−0.5)²(0.30)+(2−0.5)²(0.10)

σ²=0.450

Standard deviation: σ=√σ²=√0.450=0.67

𝒙і

𝒙і 𝐏(𝒙і )

P(𝒙і )

(𝒙і − 𝒎)²P(𝒙і )

0

0.60

0.00

0.150

1

0.30

0.30

0.075

2

0.10

0.20

0.225

Sum

1.0

0.50

0.450

If a random variable is a continuous variable, its probability distribution is called a continuous

probability distribution. A continuous probability distribution differs from a discrete probability

distribution because the probability that a continuous random variable will assume a particular value is

zero. Therefore, a continuous probability distribution cannot be expressed in tabular form but rather an

equation or formula to describe a continuous probability distribution.

2. Binomial Probability Distribution

A binomial experiment is a statistical experiment that has the following properties:

Property 1: The experiment consists of n repeated trials.

Property 2: Each trial can result in just two possible outcomes. We call one of these outcomes a success

and the other, a failure.

Property 3: The probability of success, denoted by P, is the same on every trial.

Property 4: The trials are independent; that is, the outcome on one trial does not affect the outcome on

other trials.

For example, for a statistical experiment, you flip a coin two times and count the number of times the

coin lands on heads. This is a binomial experiment because:

The experiment consists of repeated trials. We flip a coin two times.

Each trial can result in just two possible outcomes — heads or tails.

The probability of success is constant (0.5) on every trial.

The trials are independent; that is, getting heads on one trial does not affect whether we get

heads on other trials.

A binomial random variable is the number of successes x in n repeated trials of a binomial experiment.

The probability distribution of a binomial random variable is called a binomial distribution (also known

as a Bernoulli distribution).

Suppose we flip a coin two times and count the number of heads. The binomial random variable is the

number of heads, which can take on values of 0, 1, or 2. The binomial distribution is:

Number of Heads

Probability

0

.25

1

.50

2

.25

The binomial probability refers to the probability that a binomial experiment results in an exact number

of successes. For example, in the above table, we see that the binomial probability of getting exactly one

head in two coin flips is 0.50.

Given x, n, and P, we can compute the binomial probability based on the following formula:

n=number of trials

k= number of successes

p = probability of success

Recall that nCk is n choose k: nCk = n!/( k! (n-k)!)

It turns out that the mean and standard deviation of the binomial distribution are:

Mean: μ= E(K) = np

Standard Deviation: σ = sqrt[ np(1-p) ]

(sqrt means the square root)

For example, if a die is tossed five times, what is the probability of getting exactly two sixes? This is a

binomial experiment in which the number of trials is equal to five, the number of successes is equal to

two, and the probability of success on a single trial is 1/6 or about 0.167. Therefore, the binomial

probability is:

b(2; 5, 0.167) = 5C2 * (0.167)2 * (0.833)3

b(2; 5, 0.167) = 0.161

You can also use tables to determine binomial probabilities. The tables for these probabilities include

data regarding the number of values, the number of successes, and the p-value.

For example, if you are trying to determine the probability that the number 5 would appear in ten rolls

of a die (n=10), and you know that the probability of success on any one trial is .2 (p=2) and that the

probability of x successes in a given number of trials is 3, then you could use the chart to determine that

the binomial probability is 0.2013.

Example: In the town of Hickoryville, an adult citizen has a 1\% chance of being selected for jury duty

next year. If jury members are selected at random, and the Anderson family includes three adults,

determine the probability that

1. None of the Andersons will be chosen.

2. Exactly one of the Andersons will be chosen.

3. Exactly two of the Andersons will be chosen.

4. All three of the Andersons will be chosen.

This is a binomial experiment with n = 3, π = 0.01 . Let x = number of Andersons chosen. We will need

the probability equation for all of these so

𝑃(𝑥) =

𝑛!

𝜋 𝑥 (1 − 𝜋)𝑛−𝑥

𝑥! (𝑛 − 𝑥)!

a. P( x = 0 ) is the probability that none of the Andersons will be chosen.

3!

𝑝(𝑥 = 0) = 0!(3−0)! 0.010 (1 − 0.01)3−0=0.9703

b. P( x = 1 ) is the probability that exactly one of the Andersons will be chosen.

3!

𝑝(𝑥 = 1) = 1!(3−1)! 0.011 (1 − 0.01)3−1=0.0294

We can continue the last two in the same manner by plugging in the appropriate values into P(x).

c. P(x = 2) = 0.0003

d. P(x = 3) = 0.0000

3. Poisson Distribution

If however, we have an experiment in which the occurrence or nonoccurrence in any interval is

independent of the occurrence or nonoccurrence in any other interval, and the probability of an

occurrence of the event is the same for any two intervals of equal length, we can use the Poisson

Probability Distribution. In other words, a Poisson experiment is a statistical experiment that has the

following properties:

Property 1: The experiment results in outcomes that can be classified as successes or failures.

Property 2: The average number of successes (μ) that occurs in a specified region is known.

Property 3: The probability that a success will occur is proportional to the size of the region.

Property 4. The probability that a success will occur in an extremely small region is virtually zero.

When we use the Poisson distribution we assume:

e: A constant equal to approximately 2.71828. (Actually, e is the base of the natural logarithm

system.)

μ: The mean number of successes that occur in a specified region.

x: The actual number of successes that occur in a specified region.

P(x; μ): The Poisson probability that exactly x successes occur in a Poisson experiment, when

the mean number of successes is μ.

where

f(x) = the probability of x occurrences in an interval

μ = expected value or mean number of occurrences in an interval

e = 2.71828

For example, the average number of clients your sales rep can see each day is two. What is the

probability that your sales rep can see three clients tomorrow?

Solution: This is a Poisson experiment in which we know the following:

μ = 2; since your sales rep can see two per day on average.

x = 3; since we want to find the likelihood that three clients will be seen tomorrow.

e = 2.71828; since e is a constant equal to approximately 2.71828.

We plug these values into the Poisson formula as follows:

P(3; 2) = (2.71828-2) (23) / 3!

P(3; 2) = (0.13534) (8) / 6

P(3; 2) = 0.180

Thus, the probability of your sale rep seeing 3 clients tomorrow is 0.180.

Discrete and continuous probability distributions, such as the Poisson Probability Distribution and

Binomial Probability Distribution, can provide information as to how often an event occurs. Your ability

to calculate these distributions can provide important insight needed for decisions in planning and those

that concern intervals of time and space.

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