Distinguish between discrete and continuous random variables.

Module 5
This module addresses STA 2023 Competency #6: Have a basic understanding with respect to random variables.
• Distinguish between discrete and continuous random variables
• Construct a probability distribution for a discrete random variable and be able to compute its mean and standard deviation
• Compute probabilities for random variables having a binomial distribution

Section 5-2 Probability Distributions

At the end of this section, you will be able to:
• Construct a Probability Distribution Table
• Find the mean, variance and standard deviation of a Probability Distribution Table.
• Find the Expected Value
Sec. 5-2
There are two types of Random Variables, Discrete and Continuous.
Discrete Random Variables have a finite number of values or a countable number of values. For example, the number of outcomes when you roll a die is 6.
Continuous Random Variables have a infinite number of values. For example, temperature. If you think of the temperature at 7AM and then you measure the temperature at 3PM, the temperature probably gradually changed, and temperature is always changing.
Every problem in chapter 5 involves Discrete Variables, while every problem on Chapter 6, 7, and 8 involves continuous variables.
If the variable at hand is discrete, then we can construct a Probability Distribution Table for all the possible outcomes.
Construct a probability distribution table when you roll a die.
The table will consist of two columns, the first one includes a list of all the possible outcomes while the second column includes the probability for each outcome.
X P (X)
1 1/6
2 1/6
3 1/6
4 1/6
5 1/6
6 1/6
Note: The sum of all the probabilities equals one. It always equals one in a probability distribution table. That is:

Also, each probability is a number between or including 0 to 1. That is:

To find the mean, evaluate

So, go back to the table, and create a third column showing the product of x and P (X).
X P (X) x . P(x) x2 x2 . P(x)
1 1/6 1/6 1 1/6
2 1/6 2/6 4 4/6
3 1/6 3/6 9 9/6
4 1/6 4/6 16 16/6
5 1/6 5/6 25 25/6
6 1/6 6/6 36 36/6

Now, add the fractions on the third column

So, the mean is 2.33.
To figure out the standard deviation, evaluate

So, go back to the table and add a column for the x’s squared and a column where you multiply the x’s squared times the probabilities. See the columns that were added in blue.
Now add the fractions on the last column, the sum is 91/6.
Go to the formula, and plug in the numbers.

Suppose a couple decides to have three babies, show the probabilities of getting boys.
Well, the couple may have 0, 1, 2, or 3 boys. Those are the possible outcomes.
So, the table is:
X P (X)
0 1/8
1 3/8
2 3/8
3 1/8
You may be wandering where I got the 1/8, 3/8 and so on.
I used a tree diagram to come up with the sample space.
To find the mean and the standard deviation, revisit the table and add the necessary columns to help us evaluate the formulas.
X P (X) x . P(x) x2 x2 . P(x)
0 1/8 0 0 0
1 3/8 3/8 1 3/8
2 3/8 6/8 4 12/8
3 1/8 3/8 9 9/8

Section 5-3 Binomial Probability Distribution

At the end of this section, you will be able to:
• Identify whether or not we have a Binomial Distribution
• Apply the Binomial Formula
• Use the Binomial Probability Table
Sec. 5-3
Requirements for a Binomial Distribution are the following,
1. Fixed number of trials
2. Trials must be independent
3. Each trial only has two outcomes
4. Probability of success remains the same in all trials

For example, if you toss a coin 20 times, that follows a Binomial Distribution because
1. There are exactly 20 trials
2. Each time you toss the coin is independent from the next time you toss it
3. Every time you toss the coin there are only two possible outcomes, heads or tails
4. The probability of getting heads is 1/2, and every time you toss the coin the chances of getting heads remains at 1/2.
See how the example satisfies all four requirements for a Binomial Distribution.
Now that we know that tossing a coin 20 times follows a Binomial Distribution, let’s find the probability of getting heads on 8 of the 20 tosses.
To do this, we will be using the Binomial Formula
There are four variables in this formula, they are n, x, p and q.
n: stands for the total number of trials: 20
x: what you are looking for: 8
p: stands for the probability of success, success being the probability of getting heads when I toss the coin: 1/2
q: stands for the complement of p : 1-p : 1-1/2 :1/2
So, please plug in these values into the formula:

So, the chances that we get 8 heads when tossing the coin 20 times is 12.01%.
Another Example: If an instructor gives a pop quiz consisting of 5 multiple choice questions, where each question has four choices a through d, what is the probability of passing the quiz if you guess the answers? Passing means getting four of the five questions correct.
So, the problem follows a binomial distribution because there are a fixed number of trials, 5. And, you will answer each question correct or not, so there are only two possible outcomes.
n = 5 questions
x = 4 you need 4 correct answers to pass
p = 1/4 that’s the probability thatyou answer the question correctly
q = 1 – 1/4 = 3/4 the chances that you answer the question wrong.
So, use the binomial formulas.

So, a student has a 1.46% chance of passing the pop quiz.
Another way of getting the same answer:
You can use the Binomial Table that is found in Appendix A page 769 through page 771.
Search for the n value of 5 on the left column of the table.
Then search for x of 4.
Then, search for p=.25 on the top row.
You will not find p=.25 on the top row.
So you are stuck! You can’t use the table.
If you can find the n, x and p value on the table, you can use the table to find the answer instead of using the formula. In this example, you have to use the formula!

Section 5-4 Parameters for Binomial Distributions

At the end of this section, you will be able to:
Find the mean, variance, and Standard Deviation for the Binomial Distribution
Sec. 5-4
To find the mean use the following formula,

To find the standard deviation,

So, if you have a Binomial Distribution where n=5 and p=1/4, then and

So, the mean is 1.25.
The standard deviation is .97.
The variance is the standard deviation squared, that is .9375.

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